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Determinants

Rajasthan Board · Class 12 · Mathematics

NCERT Solutions for Determinants — Rajasthan Board Class 12 Mathematics.

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72 Questions Solved · 6 Sections

Exercise 4.1

1Evaluate the determinant ∣2amp;4−5amp;−1∣\left|\begin{array}{cc}2 & 4\\-5 & -1\end{array}\right|Show solution
Given: Δ=∣2amp;4−5amp;−1∣\Delta = \left|\begin{array}{cc}2 & 4\\-5 & -1\end{array}\right|

Formula: For a 2×22\times2 determinant, ∣aamp;bcamp;d∣=ad−bc\left|\begin{array}{cc}a & b\\c & d\end{array}\right| = ad - bc

Working:
Δ=(2)(−1)−(4)(−5)=−2+20=18\Delta = (2)(-1) - (4)(-5) = -2 + 20 = 18

Answer: Δ=18\Delta = 18
2(i)Evaluate ∣cos⁡θamp;−sin⁡θsin⁡θamp;cos⁡θ∣\left|\begin{array}{cc}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{array}\right|Show solution
Given: Δ=∣cos⁡θamp;−sin⁡θsin⁡θamp;cos⁡θ∣\Delta = \left|\begin{array}{cc}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{array}\right|

Working:
Δ=(cos⁡θ)(cos⁡θ)−(−sin⁡θ)(sin⁡θ)\Delta = (\cos\theta)(\cos\theta) - (-\sin\theta)(\sin\theta)
=cos⁡2θ+sin⁡2θ=1= \cos^2\theta + \sin^2\theta = 1

Answer: Δ=1\Delta = 1
2(ii)Evaluate ∣x2−x+1amp;x−1x+1amp;x+1∣\left|\begin{array}{cc}x^2-x+1 & x-1\\x+1 & x+1\end{array}\right|Show solution
Given: Δ=∣x2−x+1amp;x−1x+1amp;x+1∣\Delta = \left|\begin{array}{cc}x^2-x+1 & x-1\\x+1 & x+1\end{array}\right|

Working:
Δ=(x2−x+1)(x+1)−(x−1)(x+1)\Delta = (x^2-x+1)(x+1) - (x-1)(x+1)
=(x3+x2−x2−x+x+1)−(x2−1)= (x^3 + x^2 - x^2 - x + x + 1) - (x^2 - 1)
=(x3+1)−(x2−1)= (x^3 + 1) - (x^2 - 1)
=x3+1−x2+1= x^3 + 1 - x^2 + 1
=x3−x2+2= x^3 - x^2 + 2

Answer: Δ=x3−x2+2\Delta = x^3 - x^2 + 2
3If A=[1amp;24amp;2]A = \begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}, then show that ∣2A∣=4∣A∣|2A| = 4|A|.Show solution
Given: A=[1amp;24amp;2]A = \begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}

Step 1: Find ∣A∣|A|.
∣A∣=(1)(2)−(2)(4)=2−8=−6|A| = (1)(2) - (2)(4) = 2 - 8 = -6

Step 2: Find 2A2A.
2A=[2amp;48amp;4]2A = \begin{bmatrix}2 & 4\\8 & 4\end{bmatrix}

Step 3: Find ∣2A∣|2A|.
∣2A∣=(2)(4)−(4)(8)=8−32=−24|2A| = (2)(4) - (4)(8) = 8 - 32 = -24

Step 4: Verify.
4∣A∣=4×(−6)=−24=∣2A∣✓4|A| = 4 \times (-6) = -24 = |2A| \quad \checkmark

Hence ∣2A∣=4∣A∣|2A| = 4|A| is proved.
4If A=[1amp;0amp;10amp;1amp;20amp;0amp;4]A = \begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}, then show that ∣3A∣=27∣A∣|3A| = 27|A|.Show solution
Given: A=[1amp;0amp;10amp;1amp;20amp;0amp;4]A = \begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}

Step 1: Find ∣A∣|A| (expanding along R1R_1).
∣A∣=1∣1amp;20amp;4∣−0+1∣0amp;10amp;0∣|A| = 1\left|\begin{array}{cc}1&2\\0&4\end{array}\right| - 0 + 1\left|\begin{array}{cc}0&1\\0&0\end{array}\right|
=1(4−0)−0+1(0−0)=4= 1(4-0) - 0 + 1(0-0) = 4

Step 2: Find 3A3A.
3A=[3amp;0amp;30amp;3amp;60amp;0amp;12]3A = \begin{bmatrix}3 & 0 & 3\\0 & 3 & 6\\0 & 0 & 12\end{bmatrix}

Step 3: Find ∣3A∣|3A| (expanding along R1R_1).
∣3A∣=3∣3amp;60amp;12∣−0+3∣0amp;30amp;0∣|3A| = 3\left|\begin{array}{cc}3&6\\0&12\end{array}\right| - 0 + 3\left|\begin{array}{cc}0&3\\0&0\end{array}\right|
=3(36−0)−0+3(0−0)=108= 3(36-0) - 0 + 3(0-0) = 108

Step 4: Verify.
27∣A∣=27×4=108=∣3A∣✓27|A| = 27 \times 4 = 108 = |3A| \quad \checkmark

Hence ∣3A∣=27∣A∣|3A| = 27|A| is proved. (Note: For an n×nn\times n matrix, ∣kA∣=kn∣A∣|kA| = k^n|A|; here n=3n=3, so ∣3A∣=33∣A∣=27∣A∣|3A|=3^3|A|=27|A|.)
5(i)Evaluate ∣3amp;−1amp;−20amp;0amp;−13amp;−5amp;0∣\left|\begin{array}{ccc}3 & -1 & -2\\0 & 0 & -1\\3 & -5 & 0\end{array}\right|Show solution
Expanding along R2R_2 (since it has two zeros):
Δ=−0⋅M21+0⋅M22−(−1)⋅M23\Delta = -0\cdot M_{21} + 0\cdot M_{22} - (-1)\cdot M_{23}
=0−0+1⋅∣3amp;−13amp;−5∣= 0 - 0 + 1\cdot\left|\begin{array}{cc}3&-1\\3&-5\end{array}\right|
=1⋅[(3)(−5)−(−1)(3)]= 1\cdot[(3)(-5)-(-1)(3)]
=(−15+3)=−12= (-15+3) = -12

Answer: Δ=−12\Delta = -12
5(ii)Evaluate ∣3amp;−4amp;51amp;1amp;−22amp;3amp;1∣\left|\begin{array}{ccc}3 & -4 & 5\\1 & 1 & -2\\2 & 3 & 1\end{array}\right|Show solution
Expanding along R1R_1:
Δ=3∣1amp;−23amp;1∣−(−4)∣1amp;−22amp;1∣+5∣1amp;12amp;3∣\Delta = 3\left|\begin{array}{cc}1&-2\\3&1\end{array}\right| -(-4)\left|\begin{array}{cc}1&-2\\2&1\end{array}\right| +5\left|\begin{array}{cc}1&1\\2&3\end{array}\right|
=3(1+6)+4(1+4)+5(3−2)= 3(1+6) + 4(1+4) + 5(3-2)
=3(7)+4(5)+5(1)= 3(7) + 4(5) + 5(1)
=21+20+5=46= 21 + 20 + 5 = 46

Answer: Δ=46\Delta = 46
5(iii)Evaluate ∣0amp;1amp;2−1amp;0amp;−3−2amp;3amp;0∣\begin{vmatrix}0 & 1 & 2\\-1 & 0 & -3\\-2 & 3 & 0\end{vmatrix}Show solution
Expanding along R1R_1:
Δ=0∣0amp;−33amp;0∣−1∣−1amp;−3−2amp;0∣+2∣−1amp;0−2amp;3∣\Delta = 0\left|\begin{array}{cc}0&-3\\3&0\end{array}\right| -1\left|\begin{array}{cc}-1&-3\\-2&0\end{array}\right| +2\left|\begin{array}{cc}-1&0\\-2&3\end{array}\right|
=0−1[(−1)(0)−(−3)(−2)]+2[(−1)(3)−(0)(−2)]= 0 - 1[(−1)(0)−(−3)(−2)] + 2[(−1)(3)−(0)(−2)]
=0−1[0−6]+2[−3−0]= 0 - 1[0 - 6] + 2[-3 - 0]
=0+6−6=0= 0 + 6 - 6 = 0

Answer: Δ=0\Delta = 0
5(iv)Evaluate ∣2amp;−1amp;−20amp;2amp;−13amp;−5amp;0∣\begin{vmatrix}2 & -1 & -2\\0 & 2 & -1\\3 & -5 & 0\end{vmatrix}Show solution
Expanding along R1R_1:
Δ=2∣2amp;−1−5amp;0∣−(−1)∣0amp;−13amp;0∣+(−2)∣0amp;23amp;−5∣\Delta = 2\left|\begin{array}{cc}2&-1\\-5&0\end{array}\right| -(-1)\left|\begin{array}{cc}0&-1\\3&0\end{array}\right| +(-2)\left|\begin{array}{cc}0&2\\3&-5\end{array}\right|
=2[(2)(0)−(−1)(−5)]+1[(0)(0)−(−1)(3)]+(−2)[(0)(−5)−(2)(3)]= 2[(2)(0)-(-1)(-5)] + 1[(0)(0)-(-1)(3)] + (-2)[(0)(-5)-(2)(3)]
=2[0−5]+1[0+3]+(−2)[0−6]= 2[0-5] + 1[0+3] + (-2)[0-6]
=2(−5)+3+(−2)(−6)= 2(-5) + 3 + (-2)(-6)
=−10+3+12=5= -10 + 3 + 12 = 5

Answer: Δ=5\Delta = 5
6If A=[1amp;1amp;−22amp;1amp;−35amp;4amp;−9]A = \begin{bmatrix}1 & 1 & -2\\2 & 1 & -3\\5 & 4 & -9\end{bmatrix}, find ∣A∣|A|.Show solution
Expanding along R1R_1:
∣A∣=1∣1amp;−34amp;−9∣−1∣2amp;−35amp;−9∣+(−2)∣2amp;15amp;4∣|A| = 1\left|\begin{array}{cc}1&-3\\4&-9\end{array}\right| -1\left|\begin{array}{cc}2&-3\\5&-9\end{array}\right| +(-2)\left|\begin{array}{cc}2&1\\5&4\end{array}\right|
=1[(−9)−(−12)]−1[(−18)−(−15)]+(−2)[(8)−(5)]= 1[(-9)-(-12)] - 1[(-18)-(-15)] + (-2)[(8)-(5)]
=1[3]−1[−3]+(−2)[3]= 1[3] - 1[-3] + (-2)[3]
=3+3−6=0= 3 + 3 - 6 = 0

Answer: ∣A∣=0|A| = 0
7(i)Find values of xx if ∣2amp;45amp;1∣=∣2xamp;46amp;x∣\begin{vmatrix}2 & 4\\5 & 1\end{vmatrix} = \begin{vmatrix}2x & 4\\6 & x\end{vmatrix}Show solution
LHS:
∣2amp;45amp;1∣=(2)(1)−(4)(5)=2−20=−18\begin{vmatrix}2&4\\5&1\end{vmatrix} = (2)(1)-(4)(5) = 2-20 = -18

RHS:
∣2xamp;46amp;x∣=(2x)(x)−(4)(6)=2x2−24\begin{vmatrix}2x&4\\6&x\end{vmatrix} = (2x)(x)-(4)(6) = 2x^2 - 24

Setting LHS = RHS:
2x2−24=−182x^2 - 24 = -18
2x2=62x^2 = 6
x2=3x^2 = 3
x=±3x = \pm\sqrt{3}

Answer: x=±3x = \pm\sqrt{3}
7(ii)Find values of xx if ∣2amp;34amp;5∣=∣xamp;32xamp;5∣\begin{vmatrix}2 & 3\\4 & 5\end{vmatrix} = \begin{vmatrix}x & 3\\2x & 5\end{vmatrix}Show solution
LHS:
∣2amp;34amp;5∣=(2)(5)−(3)(4)=10−12=−2\begin{vmatrix}2&3\\4&5\end{vmatrix} = (2)(5)-(3)(4) = 10-12 = -2

RHS:
∣xamp;32xamp;5∣=(x)(5)−(3)(2x)=5x−6x=−x\begin{vmatrix}x&3\\2x&5\end{vmatrix} = (x)(5)-(3)(2x) = 5x - 6x = -x

Setting LHS = RHS:
−x=−2-x = -2
x=2x = 2

Answer: x=2x = 2
8If ∣xamp;218amp;x∣=∣6amp;218amp;6∣\begin{vmatrix}x & 2\\18 & x\end{vmatrix} = \begin{vmatrix}6 & 2\\18 & 6\end{vmatrix}, then xx is equal to: (A) 6, (B) ±6\pm 6, (C) −6-6, (D) 0Show solution
Correct Option: (B) ±6\pm 6

LHS: x2−36x^2 - 36

RHS: 36−36=036 - 36 = 0

x2−36=0  ⟹  x2=36  ⟹  x=±6x^2 - 36 = 0 \implies x^2 = 36 \implies x = \pm 6

Hence the correct answer is (B) ±6\pm 6.

Exercise 4.2

1(i)Find area of the triangle with vertices (1,0)(1,0), (6,0)(6,0), (4,3)(4,3).Show solution
Formula:
Δ=12∣x1amp;y1amp;1x2amp;y2amp;1x3amp;y3amp;1∣\Delta = \frac{1}{2}\left|\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{array}\right|

Working:
Δ=12∣1amp;0amp;16amp;0amp;14amp;3amp;1∣\Delta = \frac{1}{2}\left|\begin{array}{ccc}1&0&1\\6&0&1\\4&3&1\end{array}\right|

Expanding along C2C_2 (or R1R_1):
=12[1(0⋅1−1⋅3)−0+1(6⋅3−0⋅4)]= \frac{1}{2}\left[1(0\cdot1 - 1\cdot3) - 0 + 1(6\cdot3 - 0\cdot4)\right]
=12[1(0−3)−0+1(18−0)]= \frac{1}{2}\left[1(0-3) - 0 + 1(18-0)\right]
=12[−3+18]=152= \frac{1}{2}[-3 + 18] = \frac{15}{2}

Answer: Area =152= \dfrac{15}{2} sq. units
1(ii)Find area of the triangle with vertices (2,7)(2,7), (1,1)(1,1), (10,8)(10,8).Show solution
Δ=12∣2amp;7amp;11amp;1amp;110amp;8amp;1∣\Delta = \frac{1}{2}\left|\begin{array}{ccc}2&7&1\\1&1&1\\10&8&1\end{array}\right|

Expanding along R1R_1:
=12[2(1−8)−7(1−10)+1(8−10)]= \frac{1}{2}\left[2(1-8) - 7(1-10) + 1(8-10)\right]
=12[2(−7)−7(−9)+1(−2)]= \frac{1}{2}\left[2(-7) - 7(-9) + 1(-2)\right]
=12[−14+63−2]= \frac{1}{2}[-14 + 63 - 2]
=472= \frac{47}{2}

Answer: Area =472= \dfrac{47}{2} sq. units
1(iii)Find area of the triangle with vertices (−2,−3)(-2,-3), (3,2)(3,2), (−1,−8)(-1,-8).Show solution
Δ=12∣−2amp;−3amp;13amp;2amp;1−1amp;−8amp;1∣\Delta = \frac{1}{2}\left|\begin{array}{ccc}-2&-3&1\\3&2&1\\-1&-8&1\end{array}\right|

Expanding along R1R_1:
=12[(−2)(2⋅1−1⋅(−8))−(−3)(3⋅1−1⋅(−1))+1(3⋅(−8)−2⋅(−1))]= \frac{1}{2}\left[(-2)(2\cdot1 - 1\cdot(-8)) - (-3)(3\cdot1 - 1\cdot(-1)) + 1(3\cdot(-8) - 2\cdot(-1))\right]
=12[(−2)(2+8)+3(3+1)+1(−24+2)]= \frac{1}{2}\left[(-2)(2+8) + 3(3+1) + 1(-24+2)\right]
=12[−20+12−22]= \frac{1}{2}\left[-20 + 12 - 22\right]
=12(−30)=−15= \frac{1}{2}(-30) = -15

Since area is positive:

Answer: Area =∣−15∣=15= |-15| = 15 sq. units
2Show that points A(a,b+c)(a, b+c), B(b,c+a)(b, c+a), C(c,a+b)(c, a+b) are collinear.Show solution
To show: Area of triangle ABC = 0.

Area=12∣aamp;b+camp;1bamp;c+aamp;1camp;a+bamp;1∣\text{Area} = \frac{1}{2}\left|\begin{array}{ccc}a&b+c&1\\b&c+a&1\\c&a+b&1\end{array}\right|

Observe: In each row, sum of first two elements equals a+b+ca+b+c:
- Row 1: a+(b+c)=a+b+ca + (b+c) = a+b+c
- Row 2: b+(c+a)=a+b+cb + (c+a) = a+b+c
- Row 3: c+(a+b)=a+b+cc + (a+b) = a+b+c

Apply C2→C1+C2C_2 \to C_1 + C_2:
=12∣aamp;a+b+camp;1bamp;a+b+camp;1camp;a+b+camp;1∣= \frac{1}{2}\left|\begin{array}{ccc}a&a+b+c&1\\b&a+b+c&1\\c&a+b+c&1\end{array}\right|

Take (a+b+c)(a+b+c) common from C2C_2:
=a+b+c2∣aamp;1amp;1bamp;1amp;1camp;1amp;1∣= \frac{a+b+c}{2}\left|\begin{array}{ccc}a&1&1\\b&1&1\\c&1&1\end{array}\right|

Since C2=C3C_2 = C_3, the determinant =0= 0.

Hence Area =0= 0, so A, B, C are collinear. ■\blacksquare
3(i)Find values of kk if area of triangle is 4 sq. units and vertices are (k,0)(k,0), (4,0)(4,0), (0,2)(0,2).Show solution
Given: Area = 4 sq. units.

12∣kamp;0amp;14amp;0amp;10amp;2amp;1∣=±4\frac{1}{2}\left|\begin{array}{ccc}k&0&1\\4&0&1\\0&2&1\end{array}\right| = \pm 4

Expanding along C2C_2:
12[−0+0−2∣kamp;14amp;1∣]=±4\frac{1}{2}\left[-0 + 0 - 2\left|\begin{array}{cc}k&1\\4&1\end{array}\right|\right] = \pm 4

12⋅(−2)(k−4)=±4\frac{1}{2}\cdot(-2)(k-4) = \pm 4

−(k−4)=±4-(k-4) = \pm 4

Case 1: −(k−4)=4⇒k−4=−4⇒k=0-(k-4) = 4 \Rightarrow k-4 = -4 \Rightarrow k = 0

Case 2: −(k−4)=−4⇒k−4=4⇒k=8-(k-4) = -4 \Rightarrow k-4 = 4 \Rightarrow k = 8

Answer: k=0k = 0 or k=8k = 8
3(ii)Find values of kk if area of triangle is 4 sq. units and vertices are (−2,0)(-2,0), (0,4)(0,4), (0,k)(0,k).Show solution
Given: Area = 4 sq. units.

12∣−2amp;0amp;10amp;4amp;10amp;kamp;1∣=±4\frac{1}{2}\left|\begin{array}{ccc}-2&0&1\\0&4&1\\0&k&1\end{array}\right| = \pm 4

Expanding along C1C_1:
12[(−2)∣4amp;1kamp;1∣]=±4\frac{1}{2}\left[(-2)\left|\begin{array}{cc}4&1\\k&1\end{array}\right|\right] = \pm 4

12⋅(−2)(4−k)=±4\frac{1}{2}\cdot(-2)(4-k) = \pm 4

−(4−k)=±4-(4-k) = \pm 4

Case 1: −(4−k)=4⇒k−4=4⇒k=8-(4-k) = 4 \Rightarrow k-4 = 4 \Rightarrow k = 8

Case 2: −(4−k)=−4⇒k−4=−4⇒k=0-(4-k) = -4 \Rightarrow k-4 = -4 \Rightarrow k = 0

Answer: k=0k = 0 or k=8k = 8
4(i)Find equation of line joining (1,2)(1,2) and (3,6)(3,6) using determinants.Show solution
Let P(x,y)P(x,y) be any point on the line. Then A(1,2)(1,2), B(3,6)(3,6), P(x,y)(x,y) are collinear, so area of triangle ABP = 0.

12∣1amp;2amp;13amp;6amp;1xamp;yamp;1∣=0\frac{1}{2}\left|\begin{array}{ccc}1&2&1\\3&6&1\\x&y&1\end{array}\right| = 0

Expanding along R1R_1:
1(6−y)−2(3−x)+1(3y−6x)=01(6-y) - 2(3-x) + 1(3y-6x) = 0
6−y−6+2x+3y−6x=06 - y - 6 + 2x + 3y - 6x = 0
2y−4x=02y - 4x = 0
y=2xy = 2x

Answer: Equation of line is y=2xy = 2x.
4(ii)Find equation of line joining (3,1)(3,1) and (9,3)(9,3) using determinants.Show solution
Let P(x,y)P(x,y) be any point on the line. Then the three points are collinear:

12∣3amp;1amp;19amp;3amp;1xamp;yamp;1∣=0\frac{1}{2}\left|\begin{array}{ccc}3&1&1\\9&3&1\\x&y&1\end{array}\right| = 0

Expanding along R1R_1:
3(3−y)−1(9−x)+1(9y−3x)=03(3-y) - 1(9-x) + 1(9y-3x) = 0
9−3y−9+x+9y−3x=09 - 3y - 9 + x + 9y - 3x = 0
6y−2x=06y - 2x = 0
x−3y=0x - 3y = 0

Answer: Equation of line is x=3yx = 3y (or x−3y=0x - 3y = 0).
5If area of triangle is 35 sq units with vertices (2,−6)(2,-6), (5,4)(5,4) and (k,4)(k,4). Then kk is: (A) 12, (B) −2-2, (C) −12,−2-12,-2, (D) 12,−212,-2Show solution
Correct Option: (D) 12,−212, -2

12∣2amp;−6amp;15amp;4amp;1kamp;4amp;1∣=±35\frac{1}{2}\left|\begin{array}{ccc}2&-6&1\\5&4&1\\k&4&1\end{array}\right| = \pm 35

Expanding along R1R_1:
12[2(4−4)−(−6)(5−k)+1(20−4k)]=±35\frac{1}{2}\left[2(4-4)-(-6)(5-k)+1(20-4k)\right] = \pm 35
12[0+6(5−k)+20−4k]=±35\frac{1}{2}\left[0 + 6(5-k) + 20 - 4k\right] = \pm 35
12[30−6k+20−4k]=±35\frac{1}{2}\left[30 - 6k + 20 - 4k\right] = \pm 35
12(50−10k)=±35\frac{1}{2}(50 - 10k) = \pm 35
50−10k=±7050 - 10k = \pm 70

Case 1: 50−10k=70⇒−10k=20⇒k=−250 - 10k = 70 \Rightarrow -10k = 20 \Rightarrow k = -2

Case 2: 50−10k=−70⇒−10k=−120⇒k=1250 - 10k = -70 \Rightarrow -10k = -120 \Rightarrow k = 12

Answer: k=12k = 12 or k=−2k = -2, i.e., option (D).

Exercise 4.3

1(i)Write Minors and Cofactors of the elements of ∣2amp;−40amp;3∣\left|\begin{array}{cc}2&-4\\0&3\end{array}\right|Show solution
The elements are a11=2, a12=−4, a21=0, a22=3a_{11}=2,\ a_{12}=-4,\ a_{21}=0,\ a_{22}=3.

Minors:
- M11M_{11} = determinant after deleting R1,C1R_1, C_1 =3= 3
- M12M_{12} = determinant after deleting R1,C2R_1, C_2 =0= 0
- M21M_{21} = determinant after deleting R2,C1R_2, C_1 =−4= -4
- M22M_{22} = determinant after deleting R2,C2R_2, C_2 =2= 2

Cofactors (Aij=(−1)i+jMijA_{ij} = (-1)^{i+j}M_{ij}):
- A11=(−1)1+1M11=+3=3A_{11} = (-1)^{1+1}M_{11} = +3 = 3
- A12=(−1)1+2M12=−0=0A_{12} = (-1)^{1+2}M_{12} = -0 = 0
- A21=(−1)2+1M21=−(−4)=4A_{21} = (-1)^{2+1}M_{21} = -(-4) = 4
- A22=(−1)2+2M22=+2=2A_{22} = (-1)^{2+2}M_{22} = +2 = 2
1(ii)Write Minors and Cofactors of the elements of ∣aamp;cbamp;d∣\left|\begin{array}{cc}a&c\\b&d\end{array}\right|Show solution
The elements are a11=a, a12=c, a21=b, a22=da_{11}=a,\ a_{12}=c,\ a_{21}=b,\ a_{22}=d.

Minors:
- M11=dM_{11} = d
- M12=bM_{12} = b
- M21=cM_{21} = c
- M22=aM_{22} = a

Cofactors:
- A11=(−1)2d=dA_{11} = (-1)^{2}d = d
- A12=(−1)3b=−bA_{12} = (-1)^{3}b = -b
- A21=(−1)3c=−cA_{21} = (-1)^{3}c = -c
- A22=(−1)4a=aA_{22} = (-1)^{4}a = a
2(i)Write Minors and Cofactors of the elements of ∣1amp;0amp;00amp;1amp;00amp;0amp;1∣\left|\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right|Show solution
This is the identity matrix I3I_3.

Minors (each MijM_{ij} is the determinant of the 2×22\times2 submatrix after deleting row ii and column jj):

M11=∣1amp;00amp;1∣=1,M12=∣0amp;00amp;1∣=0,M13=∣0amp;10amp;0∣=0M_{11}=\left|\begin{array}{cc}1&0\\0&1\end{array}\right|=1,\quad M_{12}=\left|\begin{array}{cc}0&0\\0&1\end{array}\right|=0,\quad M_{13}=\left|\begin{array}{cc}0&1\\0&0\end{array}\right|=0
M21=∣0amp;00amp;1∣=0,M22=∣1amp;00amp;1∣=1,M23=∣1amp;00amp;0∣=0M_{21}=\left|\begin{array}{cc}0&0\\0&1\end{array}\right|=0,\quad M_{22}=\left|\begin{array}{cc}1&0\\0&1\end{array}\right|=1,\quad M_{23}=\left|\begin{array}{cc}1&0\\0&0\end{array}\right|=0
M31=∣0amp;01amp;0∣=0,M32=∣1amp;00amp;0∣=0,M33=∣1amp;00amp;1∣=1M_{31}=\left|\begin{array}{cc}0&0\\1&0\end{array}\right|=0,\quad M_{32}=\left|\begin{array}{cc}1&0\\0&0\end{array}\right|=0,\quad M_{33}=\left|\begin{array}{cc}1&0\\0&1\end{array}\right|=1

Cofactors (Aij=(−1)i+jMijA_{ij}=(-1)^{i+j}M_{ij}):
A11=1, A12=0, A13=0A_{11}=1,\ A_{12}=0,\ A_{13}=0
A21=0, A22=1, A23=0A_{21}=0,\ A_{22}=1,\ A_{23}=0
A31=0, A32=0, A33=1A_{31}=0,\ A_{32}=0,\ A_{33}=1
2(ii)Write Minors and Cofactors of the elements of ∣1amp;0amp;43amp;5amp;−10amp;1amp;2∣\left|\begin{array}{ccc}1&0&4\\3&5&-1\\0&1&2\end{array}\right|Show solution
Minors:
M11=∣5amp;−11amp;2∣=10+1=11M_{11}=\left|\begin{array}{cc}5&-1\\1&2\end{array}\right|=10+1=11
M12=∣3amp;−10amp;2∣=6−0=6M_{12}=\left|\begin{array}{cc}3&-1\\0&2\end{array}\right|=6-0=6
M13=∣3amp;50amp;1∣=3−0=3M_{13}=\left|\begin{array}{cc}3&5\\0&1\end{array}\right|=3-0=3
M21=∣0amp;41amp;2∣=0−4=−4M_{21}=\left|\begin{array}{cc}0&4\\1&2\end{array}\right|=0-4=-4
M22=∣1amp;40amp;2∣=2−0=2M_{22}=\left|\begin{array}{cc}1&4\\0&2\end{array}\right|=2-0=2
M23=∣1amp;00amp;1∣=1−0=1M_{23}=\left|\begin{array}{cc}1&0\\0&1\end{array}\right|=1-0=1
M31=∣0amp;45amp;−1∣=0−20=−20M_{31}=\left|\begin{array}{cc}0&4\\5&-1\end{array}\right|=0-20=-20
M32=∣1amp;43amp;−1∣=−1−12=−13M_{32}=\left|\begin{array}{cc}1&4\\3&-1\end{array}\right|=-1-12=-13
M33=∣1amp;03amp;5∣=5−0=5M_{33}=\left|\begin{array}{cc}1&0\\3&5\end{array}\right|=5-0=5

Cofactors (Aij=(−1)i+jMijA_{ij}=(-1)^{i+j}M_{ij}):
A11=+11=11,A12=−6,A13=+3=3A_{11}=+11=11,\quad A_{12}=-6,\quad A_{13}=+3=3
A21=−(−4)=4,A22=+2=2,A23=−1A_{21}=-(-4)=4,\quad A_{22}=+2=2,\quad A_{23}=-1
A31=+(−20)=−20,A32=−(−13)=13,A33=+5=5A_{31}=+(-20)=-20,\quad A_{32}=-(-13)=13,\quad A_{33}=+5=5
3Using Cofactors of elements of second row, evaluate Δ=∣5amp;3amp;82amp;0amp;11amp;2amp;3∣\Delta = \left|\begin{array}{ccc}5&3&8\\2&0&1\\1&2&3\end{array}\right|Show solution
Elements of second row: a21=2, a22=0, a23=1a_{21}=2,\ a_{22}=0,\ a_{23}=1

Cofactors of second row elements:
A21=(−1)2+1∣3amp;82amp;3∣=−(9−16)=−(−7)=7A_{21} = (-1)^{2+1}\left|\begin{array}{cc}3&8\\2&3\end{array}\right| = -(9-16) = -(-7) = 7
A22=(−1)2+2∣5amp;81amp;3∣=+(15−8)=7A_{22} = (-1)^{2+2}\left|\begin{array}{cc}5&8\\1&3\end{array}\right| = +(15-8) = 7
A23=(−1)2+3∣5amp;31amp;2∣=−(10−3)=−7A_{23} = (-1)^{2+3}\left|\begin{array}{cc}5&3\\1&2\end{array}\right| = -(10-3) = -7

Expanding along R2R_2:
Δ=a21A21+a22A22+a23A23\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}
=2(7)+0(7)+1(−7)= 2(7) + 0(7) + 1(-7)
=14+0−7=7= 14 + 0 - 7 = 7

Answer: Δ=7\Delta = 7
4Using Cofactors of elements of third column, evaluate Δ=∣1amp;xamp;yz1amp;yamp;zx1amp;zamp;xy∣\Delta = \left|\begin{array}{ccc}1&x&yz\\1&y&zx\\1&z&xy\end{array}\right|Show solution
Elements of third column: a13=yz, a23=zx, a33=xya_{13}=yz,\ a_{23}=zx,\ a_{33}=xy

Cofactors:
A13=(−1)1+3∣1amp;y1amp;z∣=+(z−y)A_{13} = (-1)^{1+3}\left|\begin{array}{cc}1&y\\1&z\end{array}\right| = +(z-y)
A23=(−1)2+3∣1amp;x1amp;z∣=−(z−x)=(x−z)A_{23} = (-1)^{2+3}\left|\begin{array}{cc}1&x\\1&z\end{array}\right| = -(z-x) = (x-z)
A33=(−1)3+3∣1amp;x1amp;y∣=+(y−x)A_{33} = (-1)^{3+3}\left|\begin{array}{cc}1&x\\1&y\end{array}\right| = +(y-x)

Expanding along C3C_3:
Δ=a13A13+a23A23+a33A33\Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}
=yz(z−y)+zx(x−z)+xy(y−x)= yz(z-y) + zx(x-z) + xy(y-x)
=yz2−y2z+x2z−xz2+xy2−x2y= yz^2 - y^2z + x^2z - xz^2 + xy^2 - x^2y
=x2(z−y)+y2(x−z)+z2(y−x)= x^2(z-y) + y^2(x-z) + z^2(y-x)
=(x−y)(y−z)(z−x)= (x-y)(y-z)(z-x)

Answer: Δ=(x−y)(y−z)(z−x)\Delta = (x-y)(y-z)(z-x)
5If Δ=∣a11amp;a12amp;a13a21amp;a22amp;a23a31amp;a32amp;a33∣\Delta = \left|\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right| and AijA_{ij} is Cofactor of aija_{ij}, then value of Δ\Delta is given by: (A) a11A31+a12A32+a13A33a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{33}, (B) a11A11+a12A21+a13A31a_{11}A_{11}+a_{12}A_{21}+a_{13}A_{31}, (C) a21A11+a22A12+a23A13a_{21}A_{11}+a_{22}A_{12}+a_{23}A_{13}, (D) a11A11+a21A21+a31A31a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}Show solution
Correct Option: (D) a11A11+a21A21+a31A31a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}

The value of a determinant is obtained by multiplying elements of a row (or column) with their corresponding cofactors and summing. Option (D) represents expansion along the first column: each element a11,a21,a31a_{11}, a_{21}, a_{31} of column 1 is multiplied by its own cofactor A11,A21,A31A_{11}, A_{21}, A_{31} respectively. This is a valid expansion.

(A) mixes row 1 elements with row 3 cofactors — gives 0, not Δ\Delta.
(B) mixes row 1 elements with column 1 cofactors — incorrect pairing.
(C) mixes row 2 elements with row 1 cofactors — gives 0, not Δ\Delta.

Answer: (D)

Exercise 4.4

1Find adjoint of the matrix [1amp;23amp;4]\begin{bmatrix}1&2\\3&4\end{bmatrix}Show solution
Given: A=[1amp;23amp;4]A = \begin{bmatrix}1&2\\3&4\end{bmatrix}

Cofactors:
A11=+4,A12=−3,A21=−2,A22=+1A_{11}=+4,\quad A_{12}=-3,\quad A_{21}=-2,\quad A_{22}=+1

Adjoint = Transpose of cofactor matrix:
adj(A)=[A11amp;A21A12amp;A22]=[4amp;−2−3amp;1]\text{adj}(A) = \begin{bmatrix}A_{11}&A_{21}\\A_{12}&A_{22}\end{bmatrix} = \begin{bmatrix}4&-2\\-3&1\end{bmatrix}

Answer: adj(A)=[4amp;−2−3amp;1]\text{adj}(A) = \begin{bmatrix}4&-2\\-3&1\end{bmatrix}
2Find adjoint of the matrix [1amp;−1amp;22amp;3amp;5−2amp;0amp;1]\begin{bmatrix}1&-1&2\\2&3&5\\-2&0&1\end{bmatrix}Show solution
Given: A=[1amp;−1amp;22amp;3amp;5−2amp;0amp;1]A = \begin{bmatrix}1&-1&2\\2&3&5\\-2&0&1\end{bmatrix}

Cofactors:
A11=+∣3amp;50amp;1∣=3,A12=−∣2amp;5−2amp;1∣=−(2+10)=−12A_{11}=+\left|\begin{array}{cc}3&5\\0&1\end{array}\right|=3,\quad A_{12}=-\left|\begin{array}{cc}2&5\\-2&1\end{array}\right|=-(2+10)=-12
A13=+∣2amp;3−2amp;0∣=(0+6)=6A_{13}=+\left|\begin{array}{cc}2&3\\-2&0\end{array}\right|=(0+6)=6
A21=−∣−1amp;20amp;1∣=−(−1−0)=1,A22=+∣1amp;2−2amp;1∣=(1+4)=5A_{21}=-\left|\begin{array}{cc}-1&2\\0&1\end{array}\right|=-(-1-0)=1,\quad A_{22}=+\left|\begin{array}{cc}1&2\\-2&1\end{array}\right|=(1+4)=5
A23=−∣1amp;−1−2amp;0∣=−(0−2)=2A_{23}=-\left|\begin{array}{cc}1&-1\\-2&0\end{array}\right|=-(0-2)=2
A31=+∣−1amp;23amp;5∣=(−5−6)=−11,A32=−∣1amp;22amp;5∣=−(5−4)=−1A_{31}=+\left|\begin{array}{cc}-1&2\\3&5\end{array}\right|=(-5-6)=-11,\quad A_{32}=-\left|\begin{array}{cc}1&2\\2&5\end{array}\right|=-(5-4)=-1
A33=+∣1amp;−12amp;3∣=(3+2)=5A_{33}=+\left|\begin{array}{cc}1&-1\\2&3\end{array}\right|=(3+2)=5

Adjoint = Transpose of cofactor matrix:
adj(A)=[A11amp;A21amp;A31A12amp;A22amp;A32A13amp;A23amp;A33]=[3amp;1amp;−11−12amp;5amp;−16amp;2amp;5]\text{adj}(A) = \begin{bmatrix}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{bmatrix} = \begin{bmatrix}3&1&-11\\-12&5&-1\\6&2&5\end{bmatrix}
3Verify A(adjA)=(adjA)A=∣A∣IA(\text{adj}A) = (\text{adj}A)A = |A|I for A=[2amp;3−4amp;−6]A = \begin{bmatrix}2&3\\-4&-6\end{bmatrix}Show solution
Step 1: ∣A∣=(2)(−6)−(3)(−4)=−12+12=0|A| = (2)(-6)-(3)(-4) = -12+12 = 0

So ∣A∣I=0⋅I=[0amp;00amp;0]|A|I = 0\cdot I = \begin{bmatrix}0&0\\0&0\end{bmatrix}

Step 2: Cofactors: A11=−6, A12=4, A21=−3, A22=2A_{11}=-6,\ A_{12}=4,\ A_{21}=-3,\ A_{22}=2

adj(A)=[−6amp;−34amp;2]\text{adj}(A) = \begin{bmatrix}-6&-3\\4&2\end{bmatrix}

Step 3: Compute A(adjA)A(\text{adj}A):
[2amp;3−4amp;−6][−6amp;−34amp;2]=[−12+12amp;−6+624−24amp;12−12]=[0amp;00amp;0]\begin{bmatrix}2&3\\-4&-6\end{bmatrix}\begin{bmatrix}-6&-3\\4&2\end{bmatrix} = \begin{bmatrix}-12+12&-6+6\\24-24&12-12\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}

Step 4: Compute (adjA)A(\text{adj}A)A:
[−6amp;−34amp;2][2amp;3−4amp;−6]=[−12+12amp;−18+188−8amp;12−12]=[0amp;00amp;0]\begin{bmatrix}-6&-3\\4&2\end{bmatrix}\begin{bmatrix}2&3\\-4&-6\end{bmatrix} = \begin{bmatrix}-12+12&-18+18\\8-8&12-12\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}

Hence A(adjA)=(adjA)A=∣A∣I=[0amp;00amp;0]A(\text{adj}A) = (\text{adj}A)A = |A|I = \begin{bmatrix}0&0\\0&0\end{bmatrix}. ✓\checkmark
4Verify A(adjA)=(adjA)A=∣A∣IA(\text{adj}A) = (\text{adj}A)A = |A|I for A=[1amp;−1amp;23amp;0amp;−21amp;0amp;3]A = \begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}Show solution
Step 1: Find ∣A∣|A| (expanding along R1R_1):
∣A∣=1(0−0)−(−1)(9+2)+2(0−0)=0+11+0=11|A| = 1(0-0)-(-1)(9+2)+2(0-0) = 0+11+0 = 11

So ∣A∣I=11I=[11amp;0amp;00amp;11amp;00amp;0amp;11]|A|I = 11I = \begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}

Step 2: Find cofactors:
A11=∣0amp;−20amp;3∣=0,A12=−∣3amp;−21amp;3∣=−(9+2)=−11A_{11}=\left|\begin{array}{cc}0&-2\\0&3\end{array}\right|=0,\quad A_{12}=-\left|\begin{array}{cc}3&-2\\1&3\end{array}\right|=-(9+2)=-11
A13=∣3amp;01amp;0∣=0A_{13}=\left|\begin{array}{cc}3&0\\1&0\end{array}\right|=0
A21=−∣−1amp;20amp;3∣=−(−3−0)=3,A22=∣1amp;21amp;3∣=3−2=1A_{21}=-\left|\begin{array}{cc}-1&2\\0&3\end{array}\right|=-(-3-0)=3,\quad A_{22}=\left|\begin{array}{cc}1&2\\1&3\end{array}\right|=3-2=1
A23=−∣1amp;−11amp;0∣=−(0+1)=−1A_{23}=-\left|\begin{array}{cc}1&-1\\1&0\end{array}\right|=-(0+1)=-1
A31=∣−1amp;20amp;−2∣=2−0=2,A32=−∣1amp;23amp;−2∣=−(−2−6)=8A_{31}=\left|\begin{array}{cc}-1&2\\0&-2\end{array}\right|=2-0=2,\quad A_{32}=-\left|\begin{array}{cc}1&2\\3&-2\end{array}\right|=-(-2-6)=8
A33=∣1amp;−13amp;0∣=0+3=3A_{33}=\left|\begin{array}{cc}1&-1\\3&0\end{array}\right|=0+3=3

adj(A)=[0amp;3amp;2−11amp;1amp;80amp;−1amp;3]\text{adj}(A) = \begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}

Step 3: Compute A(adjA)A(\text{adj}A):
[1amp;−1amp;23amp;0amp;−21amp;0amp;3][0amp;3amp;2−11amp;1amp;80amp;−1amp;3]\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}\begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}
=[0+11+0amp;3−1−2amp;2−8+60+0+0amp;9+0+2amp;6+0−60+0+0amp;3+0−3amp;2+0+9]=[11amp;0amp;00amp;11amp;00amp;0amp;11]=11I✓= \begin{bmatrix}0+11+0&3-1-2&2-8+6\\0+0+0&9+0+2&6+0-6\\0+0+0&3+0-3&2+0+9\end{bmatrix} = \begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix} = 11I\checkmark

Step 4: Compute (adjA)A(\text{adj}A)A:
[0amp;3amp;2−11amp;1amp;80amp;−1amp;3][1amp;−1amp;23amp;0amp;−21amp;0amp;3]\begin{bmatrix}0&3&2\\-11&1&8\\0&-1&3\end{bmatrix}\begin{bmatrix}1&-1&2\\3&0&-2\\1&0&3\end{bmatrix}
=[0+9+2amp;0+0+0amp;0−6+6−11+3+8amp;11+0+0amp;−22−2+240−3+3amp;0+0+0amp;0+2+9]=[11amp;0amp;00amp;11amp;00amp;0amp;11]=11I✓= \begin{bmatrix}0+9+2&0+0+0&0-6+6\\-11+3+8&11+0+0&-22-2+24\\0-3+3&0+0+0&0+2+9\end{bmatrix} = \begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix} = 11I\checkmark

Hence verified.
5Find the inverse of [2amp;−24amp;3]\begin{bmatrix}2&-2\\4&3\end{bmatrix} (if it exists).Show solution
Given: A=[2amp;−24amp;3]A = \begin{bmatrix}2&-2\\4&3\end{bmatrix}

Step 1: ∣A∣=(2)(3)−(−2)(4)=6+8=14≠0|A| = (2)(3)-(-2)(4) = 6+8 = 14 \neq 0, so inverse exists.

Step 2: adj(A)=[3amp;2−4amp;2]\text{adj}(A) = \begin{bmatrix}3&2\\-4&2\end{bmatrix}

Step 3:
A−1=1∣A∣adj(A)=114[3amp;2−4amp;2]A^{-1} = \frac{1}{|A|}\text{adj}(A) = \frac{1}{14}\begin{bmatrix}3&2\\-4&2\end{bmatrix}

Answer: A−1=114[3amp;2−4amp;2]A^{-1} = \dfrac{1}{14}\begin{bmatrix}3&2\\-4&2\end{bmatrix}
6Find the inverse of [−1amp;5−3amp;2]\begin{bmatrix}-1&5\\-3&2\end{bmatrix} (if it exists).Show solution
Given: A=[−1amp;5−3amp;2]A = \begin{bmatrix}-1&5\\-3&2\end{bmatrix}

Step 1: ∣A∣=(−1)(2)−(5)(−3)=−2+15=13≠0|A| = (-1)(2)-(5)(-3) = -2+15 = 13 \neq 0

Step 2: adj(A)=[2amp;−53amp;−1]\text{adj}(A) = \begin{bmatrix}2&-5\\3&-1\end{bmatrix}

Step 3:
A−1=113[2amp;−53amp;−1]A^{-1} = \frac{1}{13}\begin{bmatrix}2&-5\\3&-1\end{bmatrix}
7Find the inverse of [1amp;2amp;30amp;2amp;40amp;0amp;5]\begin{bmatrix}1&2&3\\0&2&4\\0&0&5\end{bmatrix} (if it exists).Show solution
Given: A=[1amp;2amp;30amp;2amp;40amp;0amp;5]A = \begin{bmatrix}1&2&3\\0&2&4\\0&0&5\end{bmatrix} (upper triangular)

Step 1: ∣A∣=1⋅2⋅5=10≠0|A| = 1\cdot2\cdot5 = 10 \neq 0

Step 2: Cofactors:
A11=∣2amp;40amp;5∣=10,A12=−∣0amp;40amp;5∣=0,A13=∣0amp;20amp;0∣=0A_{11}=\left|\begin{array}{cc}2&4\\0&5\end{array}\right|=10,\quad A_{12}=-\left|\begin{array}{cc}0&4\\0&5\end{array}\right|=0,\quad A_{13}=\left|\begin{array}{cc}0&2\\0&0\end{array}\right|=0
A21=−∣2amp;30amp;5∣=−10,A22=∣1amp;30amp;5∣=5,A23=−∣1amp;20amp;0∣=0A_{21}=-\left|\begin{array}{cc}2&3\\0&5\end{array}\right|=-10,\quad A_{22}=\left|\begin{array}{cc}1&3\\0&5\end{array}\right|=5,\quad A_{23}=-\left|\begin{array}{cc}1&2\\0&0\end{array}\right|=0
A31=∣2amp;32amp;4∣=8−6=2,A32=−∣1amp;30amp;4∣=−4,A33=∣1amp;20amp;2∣=2A_{31}=\left|\begin{array}{cc}2&3\\2&4\end{array}\right|=8-6=2,\quad A_{32}=-\left|\begin{array}{cc}1&3\\0&4\end{array}\right|=-4,\quad A_{33}=\left|\begin{array}{cc}1&2\\0&2\end{array}\right|=2

adj(A)=[10amp;−10amp;20amp;5amp;−40amp;0amp;2]\text{adj}(A) = \begin{bmatrix}10&-10&2\\0&5&-4\\0&0&2\end{bmatrix}

Step 3:
A−1=110[10amp;−10amp;20amp;5amp;−40amp;0amp;2]A^{-1} = \frac{1}{10}\begin{bmatrix}10&-10&2\\0&5&-4\\0&0&2\end{bmatrix}
8Find the inverse of [1amp;0amp;03amp;3amp;05amp;2amp;−1]\begin{bmatrix}1&0&0\\3&3&0\\5&2&-1\end{bmatrix} (if it exists).Show solution
Given: A=[1amp;0amp;03amp;3amp;05amp;2amp;−1]A = \begin{bmatrix}1&0&0\\3&3&0\\5&2&-1\end{bmatrix}

Step 1: ∣A∣=1(3⋅(−1)−0⋅2)−0+0=−3≠0|A| = 1(3\cdot(-1)-0\cdot2)-0+0 = -3 \neq 0

Step 2: Cofactors:
A11=∣3amp;02amp;−1∣=−3,A12=−∣3amp;05amp;−1∣=−(−3)=3,A13=∣3amp;35amp;2∣=6−15=−9A_{11}=\left|\begin{array}{cc}3&0\\2&-1\end{array}\right|=-3,\quad A_{12}=-\left|\begin{array}{cc}3&0\\5&-1\end{array}\right|=-(-3)=3,\quad A_{13}=\left|\begin{array}{cc}3&3\\5&2\end{array}\right|=6-15=-9
A21=−∣0amp;02amp;−1∣=0,A22=∣1amp;05amp;−1∣=−1,A23=−∣1amp;05amp;2∣=−2A_{21}=-\left|\begin{array}{cc}0&0\\2&-1\end{array}\right|=0,\quad A_{22}=\left|\begin{array}{cc}1&0\\5&-1\end{array}\right|=-1,\quad A_{23}=-\left|\begin{array}{cc}1&0\\5&2\end{array}\right|=-2
A31=∣0amp;03amp;0∣=0,A32=−∣1amp;03amp;0∣=0,A33=∣1amp;03amp;3∣=3A_{31}=\left|\begin{array}{cc}0&0\\3&0\end{array}\right|=0,\quad A_{32}=-\left|\begin{array}{cc}1&0\\3&0\end{array}\right|=0,\quad A_{33}=\left|\begin{array}{cc}1&0\\3&3\end{array}\right|=3

adj(A)=[−3amp;0amp;03amp;−1amp;0−9amp;−2amp;3]\text{adj}(A) = \begin{bmatrix}-3&0&0\\3&-1&0\\-9&-2&3\end{bmatrix}

Step 3:
A−1=1−3[−3amp;0amp;03amp;−1amp;0−9amp;−2amp;3]=[1amp;0amp;0−1amp;13amp;03amp;23amp;−1]A^{-1} = \frac{1}{-3}\begin{bmatrix}-3&0&0\\3&-1&0\\-9&-2&3\end{bmatrix} = \begin{bmatrix}1&0&0\\-1&\frac{1}{3}&0\\3&\frac{2}{3}&-1\end{bmatrix}
9Find the inverse of [2amp;1amp;34amp;−1amp;0−7amp;2amp;1]\begin{bmatrix}2&1&3\\4&-1&0\\-7&2&1\end{bmatrix} (if it exists).Show solution
Given: A=[2amp;1amp;34amp;−1amp;0−7amp;2amp;1]A = \begin{bmatrix}2&1&3\\4&-1&0\\-7&2&1\end{bmatrix}

Step 1: ∣A∣|A| expanding along R1R_1:
=2[(−1)(1)−(0)(2)]−1[(4)(1)−(0)(−7)]+3[(4)(2)−(−1)(−7)]= 2[(-1)(1)-(0)(2)] - 1[(4)(1)-(0)(-7)] + 3[(4)(2)-(-1)(-7)]
=2(−1)−1(4)+3(8−7)= 2(-1) - 1(4) + 3(8-7)
=−2−4+3=−3= -2 - 4 + 3 = -3

Step 2: Cofactors:
A11=(−1−0)=−1,A12=−(4−0)=−4,A13=(8−7)=1A_{11}=(-1-0)=-1,\quad A_{12}=-(4-0)=-4,\quad A_{13}=(8-7)=1
A21=−(1−6)=5,A22=(2+21)=23,A23=−(4+7)=−11A_{21}=-(1-6)=5,\quad A_{22}=(2+21)=23,\quad A_{23}=-(4+7)=-11
A31=(0+3)=3,A32=−(0−12)=12,A33=(−2−4)=−6A_{31}=(0+3)=3,\quad A_{32}=-(0-12)=12,\quad A_{33}=(-2-4)=-6

adj(A)=[−1amp;5amp;3−4amp;23amp;121amp;−11amp;−6]\text{adj}(A) = \begin{bmatrix}-1&5&3\\-4&23&12\\1&-11&-6\end{bmatrix}

Step 3:
A−1=1−3[−1amp;5amp;3−4amp;23amp;121amp;−11amp;−6]=[13amp;−53amp;−143amp;−233amp;−4−13amp;113amp;2]A^{-1} = \frac{1}{-3}\begin{bmatrix}-1&5&3\\-4&23&12\\1&-11&-6\end{bmatrix} = \begin{bmatrix}\frac{1}{3}&-\frac{5}{3}&-1\\\frac{4}{3}&-\frac{23}{3}&-4\\-\frac{1}{3}&\frac{11}{3}&2\end{bmatrix}
10Find the inverse of [1amp;−1amp;20amp;2amp;−33amp;−2amp;4]\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix} (if it exists).Show solution
Given: A=[1amp;−1amp;20amp;2amp;−33amp;−2amp;4]A = \begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}

Step 1: ∣A∣|A| expanding along R1R_1:
=1(8−6)−(−1)(0+9)+2(0−6)= 1(8-6)-(-1)(0+9)+2(0-6)
=1(2)+1(9)+2(−6)=2+9−12=−1= 1(2)+1(9)+2(-6) = 2+9-12 = -1

Step 2: Cofactors:
A11=(8−6)=2,A12=−(0+9)=−9,A13=(0−6)=−6A_{11}=(8-6)=2,\quad A_{12}=-(0+9)=-9,\quad A_{13}=(0-6)=-6
A21=−(−4+4)=0,A22=(4−6)=−2,A23=−(−2+3)=−1A_{21}=-(-4+4)=0,\quad A_{22}=(4-6)=-2,\quad A_{23}=-(−2+3)=-1
A31=(3−4)=−1,A32=−(−3−0)=3,A33=(2−0)=2A_{31}=(3-4)=-1,\quad A_{32}=-(-3-0)=3,\quad A_{33}=(2-0)=2

adj(A)=[2amp;0amp;−1−9amp;−2amp;3−6amp;−1amp;2]\text{adj}(A) = \begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix}

Step 3:
A−1=1−1[2amp;0amp;−1−9amp;−2amp;3−6amp;−1amp;2]=[−2amp;0amp;19amp;2amp;−36amp;1amp;−2]A^{-1} = \frac{1}{-1}\begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix} = \begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}
11Find the inverse of [1amp;0amp;00amp;cos⁡αamp;sin⁡α0amp;sin⁡αamp;−cos⁡α]\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix} (if it exists).Show solution
Given: A=[1amp;0amp;00amp;cos⁡αamp;sin⁡α0amp;sin⁡αamp;−cos⁡α]A = \begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}

Step 1: ∣A∣|A| expanding along C1C_1:
=1⋅∣cos⁡αamp;sin⁡αsin⁡αamp;−cos⁡α∣=(−cos⁡2α−sin⁡2α)=−1= 1\cdot\left|\begin{array}{cc}\cos\alpha&\sin\alpha\\\sin\alpha&-\cos\alpha\end{array}\right| = (-\cos^2\alpha - \sin^2\alpha) = -1

Since ∣A∣=−1≠0|A| = -1 \neq 0, inverse exists.

Step 2: Cofactors:
A11=(−cos⁡2α−sin⁡2α)=−1,A12=0,A13=0A_{11}=(-\cos^2\alpha-\sin^2\alpha)=-1,\quad A_{12}=0,\quad A_{13}=0
A21=−(0−0)=0,A22=(−cos⁡α−0)=−cos⁡α,A23=−(sin⁡α−0)=−sin⁡αA_{21}=-(0-0)=0,\quad A_{22}=(-\cos\alpha-0)=-\cos\alpha,\quad A_{23}=-(\sin\alpha-0)=-\sin\alpha
A31=(0−0)=0,A32=−(sin⁡α−0)=−sin⁡α,A33=(cos⁡α−0)=cos⁡αA_{31}=(0-0)=0,\quad A_{32}=-(\sin\alpha-0)=-\sin\alpha,\quad A_{33}=(\cos\alpha-0)=\cos\alpha

adj(A)=[−1amp;0amp;00amp;−cos⁡αamp;−sin⁡α0amp;−sin⁡αamp;cos⁡α]\text{adj}(A) = \begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}

Step 3:
A−1=1−1[−1amp;0amp;00amp;−cos⁡αamp;−sin⁡α0amp;−sin⁡αamp;cos⁡α]=[1amp;0amp;00amp;cos⁡αamp;sin⁡α0amp;sin⁡αamp;−cos⁡α]A^{-1} = \frac{1}{-1}\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix} = \begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}

(Note: A−1=AA^{-1} = A in this case.)
12Let A=[3amp;72amp;5]A = \begin{bmatrix}3&7\\2&5\end{bmatrix} and B=[6amp;87amp;9]B = \begin{bmatrix}6&8\\7&9\end{bmatrix}. Verify that (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}.Show solution
Step 1: Find ABAB:
AB=[3amp;72amp;5][6amp;87amp;9]=[18+49amp;24+6312+35amp;16+45]=[67amp;8747amp;61]AB = \begin{bmatrix}3&7\\2&5\end{bmatrix}\begin{bmatrix}6&8\\7&9\end{bmatrix} = \begin{bmatrix}18+49&24+63\\12+35&16+45\end{bmatrix} = \begin{bmatrix}67&87\\47&61\end{bmatrix}

Step 2: Find (AB)−1(AB)^{-1}:
∣AB∣=67×61−87×47=4087−4089=−2|AB| = 67\times61 - 87\times47 = 4087 - 4089 = -2
adj(AB)=[61amp;−87−47amp;67]\text{adj}(AB) = \begin{bmatrix}61&-87\\-47&67\end{bmatrix}
(AB)−1=1−2[61amp;−87−47amp;67](AB)^{-1} = \frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix}

Step 3: Find A−1A^{-1}:
∣A∣=15−14=1,A−1=[5amp;−7−2amp;3]|A| = 15-14=1,\quad A^{-1} = \begin{bmatrix}5&-7\\-2&3\end{bmatrix}

Step 4: Find B−1B^{-1}:
∣B∣=54−56=−2,B−1=1−2[9amp;−8−7amp;6]|B| = 54-56=-2,\quad B^{-1} = \frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}

Step 5: Find B−1A−1B^{-1}A^{-1}:
B−1A−1=1−2[9amp;−8−7amp;6][5amp;−7−2amp;3]B^{-1}A^{-1} = \frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}\begin{bmatrix}5&-7\\-2&3\end{bmatrix}
=1−2[45+16amp;−63−24−35−12amp;49+18]=1−2[61amp;−87−47amp;67]= \frac{1}{-2}\begin{bmatrix}45+16&-63-24\\-35-12&49+18\end{bmatrix} = \frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix}

Since (AB)−1=1−2[61amp;−87−47amp;67]=B−1A−1(AB)^{-1} = \frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix} = B^{-1}A^{-1}, the result is verified. ✓\checkmark
13If A=[3amp;1−1amp;2]A = \begin{bmatrix}3&1\\-1&2\end{bmatrix}, show that A2−5A+7I=OA^2 - 5A + 7I = O. Hence find A−1A^{-1}.Show solution
Step 1: Compute A2A^2:
A2=[3amp;1−1amp;2][3amp;1−1amp;2]=[9−1amp;3+2−3−2amp;−1+4]=[8amp;5−5amp;3]A^2 = \begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix} = \begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix} = \begin{bmatrix}8&5\\-5&3\end{bmatrix}

Step 2: Compute A2−5A+7IA^2 - 5A + 7I:
=[8amp;5−5amp;3]−[15amp;5−5amp;10]+[7amp;00amp;7]=[0amp;00amp;0]=O✓= \begin{bmatrix}8&5\\-5&3\end{bmatrix} - \begin{bmatrix}15&5\\-5&10\end{bmatrix} + \begin{bmatrix}7&0\\0&7\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix} = O \quad\checkmark

Step 3: Find A−1A^{-1}:

From A2−5A+7I=OA^2 - 5A + 7I = O:
A2−5A=−7IA^2 - 5A = -7I
A(A−5I)=−7IA(A - 5I) = -7I

Multiplying both sides by A−1A^{-1} on the right:
A−5I=−7A−1A - 5I = -7A^{-1}
A−1=17(5I−A)=17([5amp;00amp;5]−[3amp;1−1amp;2])=17[2amp;−11amp;3]A^{-1} = \frac{1}{7}(5I - A) = \frac{1}{7}\left(\begin{bmatrix}5&0\\0&5\end{bmatrix} - \begin{bmatrix}3&1\\-1&2\end{bmatrix}\right) = \frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}

Answer: A−1=17[2amp;−11amp;3]A^{-1} = \dfrac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}
14For the matrix A=[3amp;21amp;1]A = \begin{bmatrix}3&2\\1&1\end{bmatrix}, find the numbers aa and bb such that A2+aA+bI=OA^2 + aA + bI = O.Show solution
Step 1: Compute A2A^2:
A2=[3amp;21amp;1][3amp;21amp;1]=[9+2amp;6+23+1amp;2+1]=[11amp;84amp;3]A^2 = \begin{bmatrix}3&2\\1&1\end{bmatrix}\begin{bmatrix}3&2\\1&1\end{bmatrix} = \begin{bmatrix}9+2&6+2\\3+1&2+1\end{bmatrix} = \begin{bmatrix}11&8\\4&3\end{bmatrix}

Step 2: Set up A2+aA+bI=OA^2 + aA + bI = O:
[11amp;84amp;3]+a[3amp;21amp;1]+b[1amp;00amp;1]=[0amp;00amp;0]\begin{bmatrix}11&8\\4&3\end{bmatrix} + a\begin{bmatrix}3&2\\1&1\end{bmatrix} + b\begin{bmatrix}1&0\\0&1\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}

Step 3: Equating elements:
- (1,1)(1,1): 11+3a+b=011 + 3a + b = 0
- (1,2)(1,2): 8+2a=0⇒a=−48 + 2a = 0 \Rightarrow a = -4
- (2,1)(2,1): 4+a=0⇒a=−44 + a = 0 \Rightarrow a = -4 ✓
- (2,2)(2,2): 3+a+b=0⇒3−4+b=0⇒b=13 + a + b = 0 \Rightarrow 3 - 4 + b = 0 \Rightarrow b = 1

Verification with (1,1): 11+3(−4)+1=11−12+1=011 + 3(-4) + 1 = 11 - 12 + 1 = 0 ✓

Answer: a=−4, b=1a = -4,\ b = 1
15For the matrix A=[1amp;1amp;11amp;2amp;−32amp;−1amp;3]A = \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}, show that A3−6A2+5A+11I=OA^3 - 6A^2 + 5A + 11I = O. Hence find A−1A^{-1}.Show solution
Step 1: Compute A2A^2:
A2=A⋅A=[1amp;1amp;11amp;2amp;−32amp;−1amp;3][1amp;1amp;11amp;2amp;−32amp;−1amp;3]A^2 = A\cdot A = \begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}
=[1+1+2amp;1+2−1amp;1−3+31+2−6amp;1+4+3amp;1−6−92−1+6amp;2−2−3amp;2+3+9]=[4amp;2amp;1−3amp;8amp;−147amp;−3amp;14]= \begin{bmatrix}1+1+2&1+2-1&1-3+3\\1+2-6&1+4+3&1-6-9\\2-1+6&2-2-3&2+3+9\end{bmatrix} = \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}

Step 2: Compute A3=A2⋅AA^3 = A^2\cdot A:
=[4amp;2amp;1−3amp;8amp;−147amp;−3amp;14][1amp;1amp;11amp;2amp;−32amp;−1amp;3]= \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}
=[4+2+2amp;4+4−1amp;4−6+3−3+8−28amp;−3+16+14amp;−3−24−427−3+28amp;7−6−14amp;7+9+42]=[8amp;7amp;1−23amp;27amp;−6932amp;−13amp;58]= \begin{bmatrix}4+2+2&4+4-1&4-6+3\\-3+8-28&-3+16+14&-3-24-42\\7-3+28&7-6-14&7+9+42\end{bmatrix} = \begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix}

Step 3: Compute A3−6A2+5A+11IA^3 - 6A^2 + 5A + 11I:
=[8amp;7amp;1−23amp;27amp;−6932amp;−13amp;58]−6[4amp;2amp;1−3amp;8amp;−147amp;−3amp;14]+5[1amp;1amp;11amp;2amp;−32amp;−1amp;3]+11[1amp;0amp;00amp;1amp;00amp;0amp;1]= \begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix} - 6\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix} + 5\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix} + 11\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
=[8−24+5+11amp;7−12+5+0amp;1−6+5+0−23+18+5+0amp;27−48+10+11amp;−69+84−15+032−42+10+0amp;−13+18−5+0amp;58−84+15+11]= \begin{bmatrix}8-24+5+11&7-12+5+0&1-6+5+0\\-23+18+5+0&27-48+10+11&-69+84-15+0\\32-42+10+0&-13+18-5+0&58-84+15+11\end{bmatrix}
=[0amp;0amp;00amp;0amp;00amp;0amp;0]=O✓= \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix} = O \quad\checkmark

Step 4: Find A−1A^{-1}:

From A3−6A2+5A+11I=OA^3 - 6A^2 + 5A + 11I = O, multiply by A−1A^{-1}:
A2−6A+5I+11A−1=OA^2 - 6A + 5I + 11A^{-1} = O
A−1=111(6A−A2−5I)A^{-1} = \frac{1}{11}(6A - A^2 - 5I)
=111(6[1amp;1amp;11amp;2amp;−32amp;−1amp;3]−[4amp;2amp;1−3amp;8amp;−147amp;−3amp;14]−[5amp;0amp;00amp;5amp;00amp;0amp;5])= \frac{1}{11}\left(6\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix} - \begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix} - \begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}\right)
=111[6−4−5amp;6−2−0amp;6−1−06+3−0amp;12−8−5amp;−18+14−012−7−0amp;−6+3−0amp;18−14−5]= \frac{1}{11}\begin{bmatrix}6-4-5&6-2-0&6-1-0\\6+3-0&12-8-5&-18+14-0\\12-7-0&-6+3-0&18-14-5\end{bmatrix}
=111[−3amp;4amp;59amp;−1amp;−45amp;−3amp;−1]= \frac{1}{11}\begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}

Answer: A−1=111[−3amp;4amp;59amp;−1amp;−45amp;−3amp;−1]A^{-1} = \dfrac{1}{11}\begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}
16If A=[2amp;−1amp;1−1amp;2amp;−11amp;−1amp;2]A = \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}, verify that A3−6A2+9A−4I=OA^3 - 6A^2 + 9A - 4I = O and hence find A−1A^{-1}.Show solution
Step 1: Compute A2A^2:
A2=[2amp;−1amp;1−1amp;2amp;−11amp;−1amp;2]2=[4+1+1amp;−2−2−1amp;2+1+2−2−2−1amp;1+4+1amp;−1−2−22+1+2amp;−1−2−2amp;1+1+4]=[6amp;−5amp;5−5amp;6amp;−55amp;−5amp;6]A^2 = \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}^2 = \begin{bmatrix}4+1+1&-2-2-1&2+1+2\\-2-2-1&1+4+1&-1-2-2\\2+1+2&-1-2-2&1+1+4\end{bmatrix} = \begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}

Step 2: Compute A3=A2⋅AA^3 = A^2\cdot A:
=[6amp;−5amp;5−5amp;6amp;−55amp;−5amp;6][2amp;−1amp;1−1amp;2amp;−11amp;−1amp;2]= \begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}
=[12+5+5amp;−6−10−5amp;6+5+10−10−6−5amp;5+12+5amp;−5−6−1010+5+6amp;−5−10−6amp;5+5+12]=[22amp;−21amp;21−21amp;22amp;−2121amp;−21amp;22]= \begin{bmatrix}12+5+5&-6-10-5&6+5+10\\-10-6-5&5+12+5&-5-6-10\\10+5+6&-5-10-6&5+5+12\end{bmatrix} = \begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}

Step 3: Verify A3−6A2+9A−4I=OA^3 - 6A^2 + 9A - 4I = O:
=[22amp;−21amp;21−21amp;22amp;−2121amp;−21amp;22]−6[6amp;−5amp;5−5amp;6amp;−55amp;−5amp;6]+9[2amp;−1amp;1−1amp;2amp;−11amp;−1amp;2]−4[1amp;0amp;00amp;1amp;00amp;0amp;1]= \begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix} - 6\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix} + 9\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix} - 4\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
=[22−36+18−4amp;−21+30−9−0amp;21−30+9−0−21+30−9−0amp;22−36+18−4amp;−21+30−9−021−30+9−0amp;−21+30−9−0amp;22−36+18−4]=[0amp;0amp;00amp;0amp;00amp;0amp;0]=O✓= \begin{bmatrix}22-36+18-4&-21+30-9-0&21-30+9-0\\-21+30-9-0&22-36+18-4&-21+30-9-0\\21-30+9-0&-21+30-9-0&22-36+18-4\end{bmatrix} = \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix} = O\checkmark

Step 4: Find A−1A^{-1}:

From A3−6A2+9A−4I=OA^3 - 6A^2 + 9A - 4I = O, multiply by A−1A^{-1}:
A2−6A+9I−4A−1=OA^2 - 6A + 9I - 4A^{-1} = O
A−1=14(A2−6A+9I)A^{-1} = \frac{1}{4}(A^2 - 6A + 9I)
=14([6amp;−5amp;5−5amp;6amp;−55amp;−5amp;6]−[12amp;−6amp;6−6amp;12amp;−66amp;−6amp;12]+[9amp;0amp;00amp;9amp;00amp;0amp;9])= \frac{1}{4}\left(\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix} - \begin{bmatrix}12&-6&6\\-6&12&-6\\6&-6&12\end{bmatrix} + \begin{bmatrix}9&0&0\\0&9&0\\0&0&9\end{bmatrix}\right)
=14[3amp;1amp;−11amp;3amp;1−1amp;1amp;3]= \frac{1}{4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}

Answer: A−1=14[3amp;1amp;−11amp;3amp;1−1amp;1amp;3]A^{-1} = \dfrac{1}{4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}
17Let AA be a nonsingular square matrix of order 3×33\times3. Then ∣adjA∣|\text{adj}A| is equal to: (A) ∣A∣|A|, (B) ∣A∣2|A|^2, (C) ∣A∣3|A|^3, (D) 3∣A∣3|A|Show solution
Correct Option: (B) ∣A∣2|A|^2

For a square matrix of order nn, ∣adjA∣=∣A∣n−1|\text{adj}A| = |A|^{n-1}.

Here n=3n = 3, so ∣adjA∣=∣A∣3−1=∣A∣2|\text{adj}A| = |A|^{3-1} = |A|^2.

Answer: (B)
18If AA is an invertible matrix of order 2, then det⁡(A−1)\det(A^{-1}) is equal to: (A) det⁡(A)\det(A), (B) 1det⁡(A)\frac{1}{\det(A)}, (C) 1, (D) 0Show solution
Correct Option: (B) 1det⁡(A)\dfrac{1}{\det(A)}

Since AA−1=IAA^{-1} = I, taking determinants: det⁡(A)⋅det⁡(A−1)=det⁡(I)=1\det(A)\cdot\det(A^{-1}) = \det(I) = 1.

Therefore det⁡(A−1)=1det⁡(A)\det(A^{-1}) = \dfrac{1}{\det(A)}.

Answer: (B)

Exercise 4.5

1Examine the consistency of the system: x+2y=2x + 2y = 2, 2x+3y=32x + 3y = 3.Show solution
Matrix form: AX=BAX = B where A=[1amp;22amp;3]A = \begin{bmatrix}1&2\\2&3\end{bmatrix}, X=[xy]X = \begin{bmatrix}x\\y\end{bmatrix}, B=[23]B = \begin{bmatrix}2\\3\end{bmatrix}

Step 1: ∣A∣=3−4=−1≠0|A| = 3 - 4 = -1 \neq 0

Since ∣A∣≠0|A| \neq 0, the system has a unique solution and is therefore consistent.
2Examine the consistency of the system: 2x−y=52x - y = 5, x+y=4x + y = 4.Show solution
Matrix form: A=[2amp;−11amp;1]A = \begin{bmatrix}2&-1\\1&1\end{bmatrix}, B=[54]B = \begin{bmatrix}5\\4\end{bmatrix}

Step 1: ∣A∣=2+1=3≠0|A| = 2+1 = 3 \neq 0

Since ∣A∣≠0|A| \neq 0, the system is consistent (has a unique solution).
3Examine the consistency of the system: x+3y=5x + 3y = 5, 2x+6y=82x + 6y = 8.Show solution
Matrix form: A=[1amp;32amp;6]A = \begin{bmatrix}1&3\\2&6\end{bmatrix}, B=[58]B = \begin{bmatrix}5\\8\end{bmatrix}

Step 1: ∣A∣=6−6=0|A| = 6 - 6 = 0

Step 2: Find adj(A)\text{adj}(A):
adj(A)=[6amp;−3−2amp;1]\text{adj}(A) = \begin{bmatrix}6&-3\\-2&1\end{bmatrix}

Step 3: Compute (adjA)B(\text{adj}A)B:
[6amp;−3−2amp;1][58]=[30−24−10+8]=[6−2]≠[00]\begin{bmatrix}6&-3\\-2&1\end{bmatrix}\begin{bmatrix}5\\8\end{bmatrix} = \begin{bmatrix}30-24\\-10+8\end{bmatrix} = \begin{bmatrix}6\\-2\end{bmatrix} \neq \begin{bmatrix}0\\0\end{bmatrix}

Since ∣A∣=0|A|=0 and (adjA)B≠O(\text{adj}A)B \neq O, the system is inconsistent (no solution).
4Examine the consistency of the system: x+y+z=1x+y+z=1, 2x+3y+2z=22x+3y+2z=2, ax+ay+2az=4ax+ay+2az=4.Show solution
Matrix form: A=[1amp;1amp;12amp;3amp;2aamp;aamp;2a]A = \begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}, B=[124]B = \begin{bmatrix}1\\2\\4\end{bmatrix}

Step 1: ∣A∣|A| expanding along R1R_1:
=1(6a−2a)−1(4a−2a)+1(2a−3a)= 1(6a-2a) - 1(4a-2a) + 1(2a-3a)
=4a−2a−a=a= 4a - 2a - a = a

Case 1: If a≠0a \neq 0, then ∣A∣≠0|A| \neq 0, so the system is consistent (unique solution).

Case 2: If a=0a = 0, then A=[1amp;1amp;12amp;3amp;20amp;0amp;0]A = \begin{bmatrix}1&1&1\\2&3&2\\0&0&0\end{bmatrix}, ∣A∣=0|A|=0.

adj(A)B\text{adj}(A)B needs to be checked. With a=0a=0, the third equation becomes 0=40=4, which is impossible. So the system is inconsistent when a=0a=0.
5Examine the consistency of the system: 3x−y−2z=23x-y-2z=2, 2y−z=−12y-z=-1, 3x−5y=33x-5y=3.Show solution
Matrix form: A=[3amp;−1amp;−20amp;2amp;−13amp;−5amp;0]A = \begin{bmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{bmatrix}, B=[2−13]B = \begin{bmatrix}2\\-1\\3\end{bmatrix}

Step 1: ∣A∣|A| expanding along R1R_1:
=3(0−5)−(−1)(0+3)+(−2)(0−6)= 3(0-5)-(-1)(0+3)+(-2)(0-6)
=3(−5)+1(3)+(−2)(−6)= 3(-5)+1(3)+(-2)(-6)
=−15+3+12=0= -15+3+12 = 0

Step 2: Find (adjA)B(\text{adj}A)B. Cofactors of AA:
A11=(0−5)=−5,A12=−(0+3)=−3,A13=(0−6)=−6A_{11}=(0-5)=-5,\quad A_{12}=-(0+3)=-3,\quad A_{13}=(0-6)=-6
A21=−(0−10)=10,A22=(0+6)=6,A23=−(−15+3)=12A_{21}=-(0-10)=10,\quad A_{22}=(0+6)=6,\quad A_{23}=-(−15+3)=12
A31=(1+4)=5,A32=−(−3−0)=3,A33=(6−0)=6A_{31}=(1+4)=5,\quad A_{32}=-(−3-0)=3,\quad A_{33}=(6-0)=6

adj(A)=[−5amp;10amp;5−3amp;6amp;3−6amp;12amp;6]\text{adj}(A) = \begin{bmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{bmatrix}

(adjA)B=[−5amp;10amp;5−3amp;6amp;3−6amp;12amp;6][2−13]=[−10−10+15−6−6+9−12−12+18]=[−5−3−6]≠O(\text{adj}A)B = \begin{bmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{bmatrix}\begin{bmatrix}2\\-1\\3\end{bmatrix} = \begin{bmatrix}-10-10+15\\-6-6+9\\-12-12+18\end{bmatrix} = \begin{bmatrix}-5\\-3\\-6\end{bmatrix} \neq O

Since ∣A∣=0|A|=0 and (adjA)B≠O(\text{adj}A)B \neq O, the system is inconsistent.
6Examine the consistency of the system: 5x−y+4z=55x-y+4z=5, 2x+3y+5z=22x+3y+5z=2, 5x−2y+6z=−15x-2y+6z=-1.Show solution
Matrix form: A=[5amp;−1amp;42amp;3amp;55amp;−2amp;6]A = \begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6\end{bmatrix}, B=[52−1]B = \begin{bmatrix}5\\2\\-1\end{bmatrix}

Step 1: ∣A∣|A| expanding along R1R_1:
=5(18+10)−(−1)(12−25)+4(−4−15)= 5(18+10)-(-1)(12-25)+4(-4-15)
=5(28)+1(−13)+4(−19)= 5(28)+1(-13)+4(-19)
=140−13−76=51= 140 - 13 - 76 = 51

Wait, let me recompute:
=5(3⋅6−5⋅(−2))−(−1)(2⋅6−5⋅5)+4(2⋅(−2)−3⋅5)= 5(3\cdot6-5\cdot(-2))-(-1)(2\cdot6-5\cdot5)+4(2\cdot(-2)-3\cdot5)
=5(18+10)+1(12−25)+4(−4−15)= 5(18+10)+1(12-25)+4(-4-15)
=5(28)+1(−13)+4(−19)= 5(28)+1(-13)+4(-19)
=140−13−76=51= 140 - 13 - 76 = 51

Hmm, let me redo:
∣A∣=5(18−(−10))−(−1)(12−25)+4(−4−15)|A| = 5(18-(-10))-(-1)(12-25)+4(-4-15)
=5(28)+1(−13)+4(−19)=140−13−76=51= 5(28)+1(-13)+4(-19) = 140-13-76 = 51

Actually: 3×6=183\times6 = 18, 5×(−2)=−105\times(-2)=-10, so 18−(−10)=2818-(-10)=28. ✓
2×6=122\times6=12, 5×5=255\times5=25, so 12−25=−1312-25=-13. ✓
2×(−2)=−42\times(-2)=-4, 3×5=153\times5=15, so −4−15=−19-4-15=-19. ✓

∣A∣=140−13−76=51≠0|A| = 140 - 13 - 76 = 51 \neq 0

Since ∣A∣≠0|A| \neq 0, the system is consistent (has a unique solution).
7Solve using matrix method: 5x+2y=45x+2y=4, 7x+3y=57x+3y=5.Show solution
Matrix form: AX=BAX = B, where A=[5amp;27amp;3]A = \begin{bmatrix}5&2\\7&3\end{bmatrix}, X=[xy]X = \begin{bmatrix}x\\y\end{bmatrix}, B=[45]B = \begin{bmatrix}4\\5\end{bmatrix}

Step 1: ∣A∣=15−14=1≠0|A| = 15-14 = 1 \neq 0

Step 2: A−1=11[3amp;−2−7amp;5]=[3amp;−2−7amp;5]A^{-1} = \frac{1}{1}\begin{bmatrix}3&-2\\-7&5\end{bmatrix} = \begin{bmatrix}3&-2\\-7&5\end{bmatrix}

Step 3: X=A−1BX = A^{-1}B:
[xy]=[3amp;−2−7amp;5][45]=[12−10−28+25]=[2−3]\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}3&-2\\-7&5\end{bmatrix}\begin{bmatrix}4\\5\end{bmatrix} = \begin{bmatrix}12-10\\-28+25\end{bmatrix} = \begin{bmatrix}2\\-3\end{bmatrix}

Answer: x=2, y=−3x = 2,\ y = -3
8Solve using matrix method: 2x−y=−22x-y=-2, 3x+4y=33x+4y=3.Show solution
Matrix form: A=[2amp;−13amp;4]A = \begin{bmatrix}2&-1\\3&4\end{bmatrix}, B=[−23]B = \begin{bmatrix}-2\\3\end{bmatrix}

Step 1: ∣A∣=8+3=11|A| = 8+3 = 11

Step 2: A−1=111[4amp;1−3amp;2]A^{-1} = \frac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}

Step 3: X=A−1BX = A^{-1}B:
[xy]=111[4amp;1−3amp;2][−23]=111[−8+36+6]=111[−512]\begin{bmatrix}x\\y\end{bmatrix} = \frac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}\begin{bmatrix}-2\\3\end{bmatrix} = \frac{1}{11}\begin{bmatrix}-8+3\\6+6\end{bmatrix} = \frac{1}{11}\begin{bmatrix}-5\\12\end{bmatrix}

Answer: x=−511, y=1211x = -\dfrac{5}{11},\ y = \dfrac{12}{11}
9Solve using matrix method: 4x−3y=34x-3y=3, 3x−5y=73x-5y=7.Show solution
Matrix form: A=[4amp;−33amp;−5]A = \begin{bmatrix}4&-3\\3&-5\end{bmatrix}, B=[37]B = \begin{bmatrix}3\\7\end{bmatrix}

Step 1: ∣A∣=−20+9=−11|A| = -20+9 = -11

Step 2: A−1=1−11[−5amp;3−3amp;4]A^{-1} = \frac{1}{-11}\begin{bmatrix}-5&3\\-3&4\end{bmatrix}

Step 3: X=A−1BX = A^{-1}B:
[xy]=1−11[−5amp;3−3amp;4][37]=1−11[−15+21−9+28]=1−11[619]\begin{bmatrix}x\\y\end{bmatrix} = \frac{1}{-11}\begin{bmatrix}-5&3\\-3&4\end{bmatrix}\begin{bmatrix}3\\7\end{bmatrix} = \frac{1}{-11}\begin{bmatrix}-15+21\\-9+28\end{bmatrix} = \frac{1}{-11}\begin{bmatrix}6\\19\end{bmatrix}

Answer: x=−611, y=−1911x = -\dfrac{6}{11},\ y = -\dfrac{19}{11}
10Solve using matrix method: 5x+2y=35x+2y=3, 3x+2y=53x+2y=5.Show solution
Matrix form: A=[5amp;23amp;2]A = \begin{bmatrix}5&2\\3&2\end{bmatrix}, B=[35]B = \begin{bmatrix}3\\5\end{bmatrix}

Step 1: ∣A∣=10−6=4|A| = 10-6 = 4

Step 2: A−1=14[2amp;−2−3amp;5]A^{-1} = \frac{1}{4}\begin{bmatrix}2&-2\\-3&5\end{bmatrix}

Step 3: X=A−1BX = A^{-1}B:
[xy]=14[2amp;−2−3amp;5][35]=14[6−10−9+25]=14[−416]=[−14]\begin{bmatrix}x\\y\end{bmatrix} = \frac{1}{4}\begin{bmatrix}2&-2\\-3&5\end{bmatrix}\begin{bmatrix}3\\5\end{bmatrix} = \frac{1}{4}\begin{bmatrix}6-10\\-9+25\end{bmatrix} = \frac{1}{4}\begin{bmatrix}-4\\16\end{bmatrix} = \begin{bmatrix}-1\\4\end{bmatrix}

Answer: x=−1, y=4x = -1,\ y = 4
11Solve using matrix method: 2x+y+z=12x+y+z=1, x−2y−z=32x-2y-z=\frac{3}{2}, 3y−5z=93y-5z=9.Show solution
Matrix form: A=[2amp;1amp;11amp;−2amp;−10amp;3amp;−5]A = \begin{bmatrix}2&1&1\\1&-2&-1\\0&3&-5\end{bmatrix}, B=[13/29]B = \begin{bmatrix}1\\3/2\\9\end{bmatrix}

Step 1: ∣A∣|A| expanding along R1R_1:
=2(10+3)−1(−5−0)+1(3−0)= 2(10+3) - 1(-5-0) + 1(3-0)
=2(13)+5+3=26+5+3=34= 2(13)+5+3 = 26+5+3 = 34

Step 2: Cofactors:
A11=(10+3)=13,A12=−(−5−0)=5,A13=(3−0)=3A_{11}=(10+3)=13,\quad A_{12}=-(-5-0)=5,\quad A_{13}=(3-0)=3
A21=−(−5−3)=8,A22=(−10−0)=−10,A23=−(6−0)=−6A_{21}=-(−5-3)=8,\quad A_{22}=(-10-0)=-10,\quad A_{23}=-(6-0)=-6
A31=(−1+2)=1,A32=−(−2−1)=3,A33=(−4−1)=−5A_{31}=(-1+2)=1,\quad A_{32}=-(-2-1)=3,\quad A_{33}=(-4-1)=-5

A−1=134[13amp;8amp;15amp;−10amp;33amp;−6amp;−5]A^{-1} = \frac{1}{34}\begin{bmatrix}13&8&1\\5&-10&3\\3&-6&-5\end{bmatrix}

Step 3: X=A−1BX = A^{-1}B:
[xyz]=134[13amp;8amp;15amp;−10amp;33amp;−6amp;−5][13/29]\begin{bmatrix}x\\y\\z\end{bmatrix} = \frac{1}{34}\begin{bmatrix}13&8&1\\5&-10&3\\3&-6&-5\end{bmatrix}\begin{bmatrix}1\\3/2\\9\end{bmatrix}
=134[13+12+95−15+273−9−45]=134[3417−51]=[11/2−3/2]= \frac{1}{34}\begin{bmatrix}13+12+9\\5-15+27\\3-9-45\end{bmatrix} = \frac{1}{34}\begin{bmatrix}34\\17\\-51\end{bmatrix} = \begin{bmatrix}1\\1/2\\-3/2\end{bmatrix}

Answer: x=1, y=12, z=−32x = 1,\ y = \dfrac{1}{2},\ z = -\dfrac{3}{2}
12Solve using matrix method: x−y+z=4x-y+z=4, 2x+y−3z=02x+y-3z=0, x+y+z=2x+y+z=2.Show solution
Matrix form: A=[1amp;−1amp;12amp;1amp;−31amp;1amp;1]A = \begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix}, B=[402]B = \begin{bmatrix}4\\0\\2\end{bmatrix}

Step 1: ∣A∣|A| expanding along R1R_1:
=1(1+3)−(−1)(2+3)+1(2−1)= 1(1+3)-(-1)(2+3)+1(2-1)
=4+5+1=10= 4+5+1 = 10

Step 2: Cofactors:
A11=4,A12=−(2+3)=−5,A13=(2−1)=1A_{11}=4,\quad A_{12}=-(2+3)=-5,\quad A_{13}=(2-1)=1
A21=−(−1−1)=2,A22=(1−1)=0,A23=−(1+1)=−2A_{21}=-(-1-1)=2,\quad A_{22}=(1-1)=0,\quad A_{23}=-(1+1)=-2
A31=(3−1)=2,A32=−(−3−2)=5,A33=(1+2)=3A_{31}=(3-1)=2,\quad A_{32}=-(-3-2)=5,\quad A_{33}=(1+2)=3

A−1=110[4amp;2amp;2−5amp;0amp;51amp;−2amp;3]A^{-1} = \frac{1}{10}\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}

Step 3: X=A−1BX = A^{-1}B:
=110[4amp;2amp;2−5amp;0amp;51amp;−2amp;3][402]=110[16+0+4−20+0+104+0+6]=110[20−1010]=[2−11]= \frac{1}{10}\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}\begin{bmatrix}4\\0\\2\end{bmatrix} = \frac{1}{10}\begin{bmatrix}16+0+4\\-20+0+10\\4+0+6\end{bmatrix} = \frac{1}{10}\begin{bmatrix}20\\-10\\10\end{bmatrix} = \begin{bmatrix}2\\-1\\1\end{bmatrix}

Answer: x=2, y=−1, z=1x = 2,\ y = -1,\ z = 1
13Solve using matrix method: 2x+3y+3z=52x+3y+3z=5, x−2y+z=−4x-2y+z=-4, 3x−y−2z=33x-y-2z=3.Show solution
Matrix form: A=[2amp;3amp;31amp;−2amp;13amp;−1amp;−2]A = \begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix}, B=[5−43]B = \begin{bmatrix}5\\-4\\3\end{bmatrix}

Step 1: ∣A∣|A| expanding along R1R_1:
=2(4+1)−3(−2−3)+3(−1+6)= 2(4+1)-3(-2-3)+3(-1+6)
=2(5)−3(−5)+3(5)=10+15+15=40= 2(5)-3(-5)+3(5) = 10+15+15 = 40

Step 2: Cofactors:
A11=(4+1)=5,A12=−(−2−3)=5,A13=(−1+6)=5A_{11}=(4+1)=5,\quad A_{12}=-(-2-3)=5,\quad A_{13}=(-1+6)=5
A21=−(−6+3)=3,A22=(−4−9)=−13,A23=−(−2−9)=11A_{21}=-(−6+3)=3,\quad A_{22}=(-4-9)=-13,\quad A_{23}=-(-2-9)=11
A31=(3+6)=9,A32=−(2−3)=1,A33=(−4−3)=−7A_{31}=(3+6)=9,\quad A_{32}=-(2-3)=1,\quad A_{33}=(-4-3)=-7

A−1=140[5amp;3amp;95amp;−13amp;15amp;11amp;−7]A^{-1} = \frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}

Step 3: X=A−1BX = A^{-1}B:
=140[5amp;3amp;95amp;−13amp;15amp;11amp;−7][5−43]=140[25−12+2725+52+325−44−21]=140[4080−40]=[12−1]= \frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\begin{bmatrix}5\\-4\\3\end{bmatrix} = \frac{1}{40}\begin{bmatrix}25-12+27\\25+52+3\\25-44-21\end{bmatrix} = \frac{1}{40}\begin{bmatrix}40\\80\\-40\end{bmatrix} = \begin{bmatrix}1\\2\\-1\end{bmatrix}

Answer: x=1, y=2, z=−1x = 1,\ y = 2,\ z = -1
14Solve using matrix method: x−y+2z=7x-y+2z=7, 3x+4y−5z=−53x+4y-5z=-5, 2x−y+3z=122x-y+3z=12.Show solution
Matrix form: A=[1amp;−1amp;23amp;4amp;−52amp;−1amp;3]A = \begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix}, B=[7−512]B = \begin{bmatrix}7\\-5\\12\end{bmatrix}

Step 1: ∣A∣|A| expanding along R1R_1:
=1(12−5)−(−1)(9+10)+2(−3−8)= 1(12-5)-(-1)(9+10)+2(-3-8)
=7+19−22=4= 7+19-22 = 4

Step 2: Cofactors:
A11=(12−5)=7,A12=−(9+10)=−19,A13=(−3−8)=−11A_{11}=(12-5)=7,\quad A_{12}=-(9+10)=-19,\quad A_{13}=(-3-8)=-11
A21=−(−3+2)=1,A22=(3−4)=−1,A23=−(−1+2)=−1A_{21}=-(-3+2)=1,\quad A_{22}=(3-4)=-1,\quad A_{23}=-(-1+2)=-1
A31=(5−8)=−3,A32=−(−5−6)=11,A33=(4+3)=7A_{31}=(5-8)=-3,\quad A_{32}=-(-5-6)=11,\quad A_{33}=(4+3)=7

A−1=14[7amp;1amp;−3−19amp;−1amp;11−11amp;−1amp;7]A^{-1} = \frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}

Step 3: X=A−1BX = A^{-1}B:
=14[7amp;1amp;−3−19amp;−1amp;11−11amp;−1amp;7][7−512]=14[49−5−36−133+5+132−77+5+84]=14[8412]=[213]= \frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}\begin{bmatrix}7\\-5\\12\end{bmatrix} = \frac{1}{4}\begin{bmatrix}49-5-36\\-133+5+132\\-77+5+84\end{bmatrix} = \frac{1}{4}\begin{bmatrix}8\\4\\12\end{bmatrix} = \begin{bmatrix}2\\1\\3\end{bmatrix}

Answer: x=2, y=1, z=3x = 2,\ y = 1,\ z = 3
15If A=[2amp;−3amp;53amp;2amp;−41amp;1amp;−2]A = \begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}, find A−1A^{-1}. Using A−1A^{-1} solve: 2x−3y+5z=112x-3y+5z=11, 3x+2y−4z=−53x+2y-4z=-5, x+y−2z=−3x+y-2z=-3.Show solution
Step 1: Find ∣A∣|A|:
∣A∣=2(2⋅(−2)−(−4)⋅1)−(−3)(3⋅(−2)−(−4)⋅1)+5(3⋅1−2⋅1)|A| = 2(2\cdot(-2)-(-4)\cdot1)-(-3)(3\cdot(-2)-(-4)\cdot1)+5(3\cdot1-2\cdot1)
=2(−4+4)+3(−6+4)+5(3−2)= 2(-4+4)+3(-6+4)+5(3-2)
=0+3(−2)+5(1)=0−6+5=−1= 0+3(-2)+5(1) = 0-6+5 = -1

Step 2: Cofactors:
A11=(−4+4)=0,A12=−(−6+4)=2,A13=(3−2)=1A_{11}=(-4+4)=0,\quad A_{12}=-(-6+4)=2,\quad A_{13}=(3-2)=1
A21=−(6−5)=−1,A22=(−4−5)=−9,A23=−(2+3)=−5A_{21}=-(6-5)=-1,\quad A_{22}=(-4-5)=-9,\quad A_{23}=-(2+3)=-5
A31=(12−10)=2,A32=−(−8−15)=23,A33=(4+9)=13A_{31}=(12-10)=2,\quad A_{32}=-(-8-15)=23,\quad A_{33}=(4+9)=13

A−1=1−1[0amp;−1amp;22amp;−9amp;231amp;−5amp;13]=[0amp;1amp;−2−2amp;9amp;−23−1amp;5amp;−13]A^{-1} = \frac{1}{-1}\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix} = \begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}

Step 3: The system is AX=BAX = B where B=[11−5−3]B = \begin{bmatrix}11\\-5\\-3\end{bmatrix}.

X=A−1B=[0amp;1amp;−2−2amp;9amp;−23−1amp;5amp;−13][11−5−3]X = A^{-1}B = \begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}\begin{bmatrix}11\\-5\\-3\end{bmatrix}
=[0−5+6−22−45+69−11−25+39]=[123]= \begin{bmatrix}0-5+6\\-22-45+69\\-11-25+39\end{bmatrix} = \begin{bmatrix}1\\2\\3\end{bmatrix}

Answer: x=1, y=2, z=3x = 1,\ y = 2,\ z = 3
16The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method.Show solution
Let cost per kg of onion = xx, wheat = yy, rice = zz (in ₹).

System of equations:
4x+3y+2z=604x + 3y + 2z = 60
2x+4y+6z=902x + 4y + 6z = 90
6x+2y+3z=706x + 2y + 3z = 70

Matrix form: AX=BAX = B, where A=[4amp;3amp;22amp;4amp;66amp;2amp;3]A = \begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix}, B=[609070]B = \begin{bmatrix}60\\90\\70\end{bmatrix}

Step 1: ∣A∣|A| expanding along R1R_1:
=4(12−12)−3(6−36)+2(4−24)= 4(12-12)-3(6-36)+2(4-24)
=4(0)−3(−30)+2(−20)= 4(0)-3(-30)+2(-20)
=0+90−40=50= 0+90-40 = 50

Step 2: Cofactors:
A11=(12−12)=0,A12=−(6−36)=30,A13=(4−24)=−20A_{11}=(12-12)=0,\quad A_{12}=-(6-36)=30,\quad A_{13}=(4-24)=-20
A21=−(9−4)=−5,A22=(12−12)=0,A23=−(8−18)=10A_{21}=-(9-4)=-5,\quad A_{22}=(12-12)=0,\quad A_{23}=-(8-18)=10
A31=(18−8)=10,A32=−(24−4)=−20,A33=(16−6)=10A_{31}=(18-8)=10,\quad A_{32}=-(24-4)=-20,\quad A_{33}=(16-6)=10

A−1=150[0amp;−5amp;1030amp;0amp;−20−20amp;10amp;10]A^{-1} = \frac{1}{50}\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}

Step 3: X=A−1BX = A^{-1}B:
=150[0amp;−5amp;1030amp;0amp;−20−20amp;10amp;10][609070]= \frac{1}{50}\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}\begin{bmatrix}60\\90\\70\end{bmatrix}
=150[0−450+7001800+0−1400−1200+900+700]=150[250400400]=[588]= \frac{1}{50}\begin{bmatrix}0-450+700\\1800+0-1400\\-1200+900+700\end{bmatrix} = \frac{1}{50}\begin{bmatrix}250\\400\\400\end{bmatrix} = \begin{bmatrix}5\\8\\8\end{bmatrix}

Answer: Cost of onion = ₹5/kg, wheat = ₹8/kg, rice = ₹8/kg.

Miscellaneous Exercises on Chapter 4

1Prove that the determinant ∣xamp;sin⁡θamp;cos⁡θ−sin⁡θamp;−xamp;1cos⁡θamp;1amp;x∣\left|\begin{array}{ccc}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{array}\right| is independent of θ\theta.Show solution
Expanding along R1R_1:
Δ=x∣−xamp;11amp;x∣−sin⁡θ∣−sin⁡θamp;1cos⁡θamp;x∣+cos⁡θ∣−sin⁡θamp;−xcos⁡θamp;1∣\Delta = x\left|\begin{array}{cc}-x&1\\1&x\end{array}\right| - \sin\theta\left|\begin{array}{cc}-\sin\theta&1\\\cos\theta&x\end{array}\right| + \cos\theta\left|\begin{array}{cc}-\sin\theta&-x\\\cos\theta&1\end{array}\right|

=x(−x2−1)−sin⁡θ(−xsin⁡θ−cos⁡θ)+cos⁡θ(−sin⁡θ+xcos⁡θ)= x(-x^2-1) - \sin\theta(-x\sin\theta - \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta)

=−x3−x+xsin⁡2θ+sin⁡θcos⁡θ−cos⁡θsin⁡θ+xcos⁡2θ= -x^3 - x + x\sin^2\theta + \sin\theta\cos\theta - \cos\theta\sin\theta + x\cos^2\theta

=−x3−x+x(sin⁡2θ+cos⁡2θ)= -x^3 - x + x(\sin^2\theta + \cos^2\theta)

=−x3−x+x=−x3= -x^3 - x + x = -x^3

Since Δ=−x3\Delta = -x^3, which does not contain θ\theta, the determinant is independent of θ\theta. ■\blacksquare
2Evaluate ∣cos⁡αcos⁡βamp;cos⁡αsin⁡βamp;−sin⁡α−sin⁡βamp;cos⁡βamp;0sin⁡αcos⁡βamp;sin⁡αsin⁡βamp;cos⁡α∣\left|\begin{array}{ccc}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha\end{array}\right|Show solution
Expanding along R2R_2 (or C3C_3):

Expanding along C3C_3:
Δ=−sin⁡α∣−sin⁡βamp;cos⁡βsin⁡αsin⁡βamp;sin⁡αcos⁡β∣⋅(−1)1+3+0+cos⁡α∣cos⁡αcos⁡βamp;cos⁡αsin⁡β−sin⁡βamp;cos⁡β∣⋅(−1)3+3\Delta = -\sin\alpha\left|\begin{array}{cc}-\sin\beta&\cos\beta\\\sin\alpha\sin\beta&\sin\alpha\cos\beta\end{array}\right|\cdot(-1)^{1+3} + 0 + \cos\alpha\left|\begin{array}{cc}\cos\alpha\cos\beta&\cos\alpha\sin\beta\\-\sin\beta&\cos\beta\end{array}\right|\cdot(-1)^{3+3}

Wait, let me expand along C3C_3:
Δ=(−sin⁡α)⋅(−1)1+3∣−sin⁡βamp;cos⁡βsin⁡αsin⁡βamp;sin⁡αcos⁡β∣+0⋅(…)+cos⁡α⋅(−1)3+3∣cos⁡αcos⁡βamp;cos⁡αsin⁡β−sin⁡βamp;cos⁡β∣\Delta = (-\sin\alpha)\cdot(-1)^{1+3}\left|\begin{array}{cc}-\sin\beta&\cos\beta\\\sin\alpha\sin\beta&\sin\alpha\cos\beta\end{array}\right| + 0\cdot(\ldots) + \cos\alpha\cdot(-1)^{3+3}\left|\begin{array}{cc}\cos\alpha\cos\beta&\cos\alpha\sin\beta\\-\sin\beta&\cos\beta\end{array}\right|

=(−sin⁡α)(1)(−sin⁡β⋅sin⁡αcos⁡β−cos⁡β⋅sin⁡αsin⁡β)+cos⁡α(1)(cos⁡αcos⁡β⋅cos⁡β−cos⁡αsin⁡β⋅(−sin⁡β))= (-\sin\alpha)(1)(-\sin\beta\cdot\sin\alpha\cos\beta - \cos\beta\cdot\sin\alpha\sin\beta) + \cos\alpha(1)(\cos\alpha\cos\beta\cdot\cos\beta - \cos\alpha\sin\beta\cdot(-\sin\beta))

=(−sin⁡α)(−sin⁡αsin⁡βcos⁡β−sin⁡αcos⁡βsin⁡β)+cos⁡α(cos⁡αcos⁡2β+cos⁡αsin⁡2β)= (-\sin\alpha)(-\sin\alpha\sin\beta\cos\beta - \sin\alpha\cos\beta\sin\beta) + \cos\alpha(\cos\alpha\cos^2\beta + \cos\alpha\sin^2\beta)

=(−sin⁡α)(−2sin⁡αsin⁡βcos⁡β)+cos⁡α(cos⁡α)= (-\sin\alpha)(-2\sin\alpha\sin\beta\cos\beta) + \cos\alpha(\cos\alpha)

Hmm, let me redo more carefully.

M13=∣−sin⁡βamp;cos⁡βsin⁡αsin⁡βamp;sin⁡αcos⁡β∣=−sin⁡β⋅sin⁡αcos⁡β−cos⁡β⋅sin⁡αsin⁡β=−sin⁡αsin⁡βcos⁡β−sin⁡αsin⁡βcos⁡β=−2sin⁡αsin⁡βcos⁡βM_{13} = \left|\begin{array}{cc}-\sin\beta&\cos\beta\\\sin\alpha\sin\beta&\sin\alpha\cos\beta\end{array}\right| = -\sin\beta\cdot\sin\alpha\cos\beta - \cos\beta\cdot\sin\alpha\sin\beta = -\sin\alpha\sin\beta\cos\beta - \sin\alpha\sin\beta\cos\beta = -2\sin\alpha\sin\beta\cos\beta

Wait: (−sin⁡β)(sin⁡αcos⁡β)−(cos⁡β)(sin⁡αsin⁡β)=−sin⁡αsin⁡βcos⁡β−sin⁡αcos⁡βsin⁡β=−2sin⁡αsin⁡βcos⁡β(-\sin\beta)(\sin\alpha\cos\beta) - (\cos\beta)(\sin\alpha\sin\beta) = -\sin\alpha\sin\beta\cos\beta - \sin\alpha\cos\beta\sin\beta = -2\sin\alpha\sin\beta\cos\beta

M33=∣cos⁡αcos⁡βamp;cos⁡αsin⁡β−sin⁡βamp;cos⁡β∣=cos⁡αcos⁡2β+cos⁡αsin⁡2β=cos⁡αM_{33} = \left|\begin{array}{cc}\cos\alpha\cos\beta&\cos\alpha\sin\beta\\-\sin\beta&\cos\beta\end{array}\right| = \cos\alpha\cos^2\beta + \cos\alpha\sin^2\beta = \cos\alpha

So:
Δ=(−sin⁡α)(+1)(−2sin⁡αsin⁡βcos⁡β)+cos⁡α(+1)(cos⁡α)\Delta = (-\sin\alpha)(+1)(-2\sin\alpha\sin\beta\cos\beta) + \cos\alpha(+1)(\cos\alpha)
=2sin⁡2αsin⁡βcos⁡β+cos⁡2α= 2\sin^2\alpha\sin\beta\cos\beta + \cos^2\alpha
=sin⁡2αsin⁡2β+cos⁡2α= \sin^2\alpha\sin2\beta + \cos^2\alpha

This doesn't simplify to a clean answer. Let me try expanding along R2R_2 instead.

Expanding along R2R_2:
Δ=−(−sin⁡β)∣cos⁡αsin⁡βamp;−sin⁡αsin⁡αsin⁡βamp;cos⁡α∣+cos⁡β∣cos⁡αcos⁡βamp;−sin⁡αsin⁡αcos⁡βamp;cos⁡α∣−0\Delta = -(-\sin\beta)\left|\begin{array}{cc}\cos\alpha\sin\beta&-\sin\alpha\\\sin\alpha\sin\beta&\cos\alpha\end{array}\right| + \cos\beta\left|\begin{array}{cc}\cos\alpha\cos\beta&-\sin\alpha\\\sin\alpha\cos\beta&\cos\alpha\end{array}\right| - 0

=sin⁡β(cos⁡αsin⁡βcos⁡α+sin⁡α⋅sin⁡αsin⁡β)+cos⁡β(cos⁡αcos⁡βcos⁡α+sin⁡αsin⁡αcos⁡β)= \sin\beta(\cos\alpha\sin\beta\cos\alpha + \sin\alpha\cdot\sin\alpha\sin\beta) + \cos\beta(\cos\alpha\cos\beta\cos\alpha + \sin\alpha\sin\alpha\cos\beta)

=sin⁡β⋅sin⁡β(cos⁡2α+sin⁡2α)+cos⁡β⋅cos⁡β(cos⁡2α+sin⁡2α)= \sin\beta\cdot\sin\beta(\cos^2\alpha + \sin^2\alpha) + \cos\beta\cdot\cos\beta(\cos^2\alpha + \sin^2\alpha)

=sin⁡2β(1)+cos⁡2β(1)=1= \sin^2\beta(1) + \cos^2\beta(1) = 1

Answer: Δ=1\Delta = 1
3If A−1=[3amp;−1amp;1−15amp;6amp;−55amp;−2amp;2]A^{-1} = \begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix} and B=[1amp;2amp;−2−1amp;3amp;00amp;−2amp;1]B = \begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix}, find (AB)−1(AB)^{-1}.Show solution
Using the property: (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}

Step 1: Find B−1B^{-1}.

∣B∣|B| expanding along R1R_1:
=1(3−0)−2(−1−0)+(−2)(2−0)=3+2−4=1= 1(3-0)-2(-1-0)+(-2)(2-0) = 3+2-4 = 1

Cofactors of BB:
B11=(3−0)=3,B12=−(−1−0)=1,B13=(2−0)=2B_{11}=(3-0)=3,\quad B_{12}=-(-1-0)=1,\quad B_{13}=(2-0)=2
B21=−(2−4)=2,B22=(1−0)=1,B23=−(−2−0)=2B_{21}=-(2-4)=2,\quad B_{22}=(1-0)=1,\quad B_{23}=-(-2-0)=2
B31=(0+6)=6,B32=−(0−2)=2,B33=(3−2)=5B_{31}=(0+6)=6,\quad B_{32}=-(0-2)=2,\quad B_{33}=(3-2)=5

B−1=11[3amp;2amp;61amp;1amp;22amp;2amp;5]B^{-1} = \frac{1}{1}\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}

Step 2: (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}:
=[3amp;2amp;61amp;1amp;22amp;2amp;5][3amp;−1amp;1−15amp;6amp;−55amp;−2amp;2]= \begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}
=[9−30+30amp;−3+12−12amp;3−10+123−15+10amp;−1+6−4amp;1−5+46−30+25amp;−2+12−10amp;2−10+10]= \begin{bmatrix}9-30+30&-3+12-12&3-10+12\\3-15+10&-1+6-4&1-5+4\\6-30+25&-2+12-10&2-10+10\end{bmatrix}
=[9amp;−3amp;5−2amp;1amp;01amp;0amp;2]= \begin{bmatrix}9&-3&5\\-2&1&0\\1&0&2\end{bmatrix}

Answer: (AB)−1=[9amp;−3amp;5−2amp;1amp;01amp;0amp;2](AB)^{-1} = \begin{bmatrix}9&-3&5\\-2&1&0\\1&0&2\end{bmatrix}
4Let A=[1amp;2amp;12amp;3amp;11amp;1amp;5]A = \begin{bmatrix}1&2&1\\2&3&1\\1&1&5\end{bmatrix}. Verify that (i) [adjA]−1=adj(A−1)[\text{adj}A]^{-1} = \text{adj}(A^{-1}), (ii) (A−1)−1=A(A^{-1})^{-1} = A.Show solution
Step 1: Find ∣A∣|A|:
∣A∣=1(15−1)−2(10−1)+1(2−3)=14−18−1=−5|A| = 1(15-1)-2(10-1)+1(2-3) = 14-18-1 = -5

Step 2: Find cofactors of AA:
A11=14,A12=−9,A13=−1A_{11}=14,\quad A_{12}=-9,\quad A_{13}=-1
A21=−9,A22=4,A23=1A_{21}=-9,\quad A_{22}=4,\quad A_{23}=1
A31=−1,A32=1,A33=−1A_{31}=-1,\quad A_{32}=1,\quad A_{33}=-1

adj(A)=[14amp;−9amp;−1−9amp;4amp;1−1amp;1amp;−1]\text{adj}(A) = \begin{bmatrix}14&-9&-1\\-9&4&1\\-1&1&-1\end{bmatrix}

A−1=1−5[14amp;−9amp;−1−9amp;4amp;1−1amp;1amp;−1]A^{-1} = \frac{1}{-5}\begin{bmatrix}14&-9&-1\\-9&4&1\\-1&1&-1\end{bmatrix}

Verification of (ii): (A−1)−1=A(A^{-1})^{-1} = A

Since A−1=1∣A∣adj(A)A^{-1} = \frac{1}{|A|}\text{adj}(A), we have (A−1)−1=A(A^{-1})^{-1} = A by the standard property of invertible matrices (inverting twice returns the original matrix). This follows from A⋅A−1=I⇒(A−1)−1=AA\cdot A^{-1} = I \Rightarrow (A^{-1})^{-1} = A. ✓

Verification of (i): [adjA]−1=adj(A−1)[\text{adj}A]^{-1} = \text{adj}(A^{-1})

We know adj(A)=∣A∣⋅A−1\text{adj}(A) = |A|\cdot A^{-1}, so [adjA]−1=1∣A∣A[\text{adj}A]^{-1} = \frac{1}{|A|}A.

Also, adj(A−1)=∣A−1∣⋅(A−1)−1=1∣A∣⋅A\text{adj}(A^{-1}) = |A^{-1}|\cdot(A^{-1})^{-1} = \frac{1}{|A|}\cdot A.

Hence [adjA]−1=adj(A−1)[\text{adj}A]^{-1} = \text{adj}(A^{-1}). ✓
5Evaluate ∣xamp;yamp;x+yyamp;x+yamp;xx+yamp;xamp;y∣\left|\begin{array}{ccc}x&y&x+y\\y&x+y&x\\x+y&x&y\end{array}\right|Show solution
Apply C1→C1+C2+C3C_1 \to C_1 + C_2 + C_3:

Each row sum: x+y+(x+y)=2(x+y)x+y+(x+y) = 2(x+y)

Δ=∣2(x+y)amp;yamp;x+y2(x+y)amp;x+yamp;x2(x+y)amp;xamp;y∣\Delta = \left|\begin{array}{ccc}2(x+y)&y&x+y\\2(x+y)&x+y&x\\2(x+y)&x&y\end{array}\right|

Take 2(x+y)2(x+y) common from C1C_1:
=2(x+y)∣1amp;yamp;x+y1amp;x+yamp;x1amp;xamp;y∣= 2(x+y)\left|\begin{array}{ccc}1&y&x+y\\1&x+y&x\\1&x&y\end{array}\right|

Apply R2→R2−R1R_2 \to R_2 - R_1 and R3→R3−R1R_3 \to R_3 - R_1:
=2(x+y)∣1amp;yamp;x+y0amp;xamp;−y0amp;x−yamp;−x∣= 2(x+y)\left|\begin{array}{ccc}1&y&x+y\\0&x&-y\\0&x-y&-x\end{array}\right|

Expanding along C1C_1:
=2(x+y)⋅1⋅∣xamp;−yx−yamp;−x∣= 2(x+y)\cdot 1\cdot\left|\begin{array}{cc}x&-y\\x-y&-x\end{array}\right|
=2(x+y)(−x2+y(x−y))= 2(x+y)(-x^2 + y(x-y))
=2(x+y)(−x2+xy−y2)= 2(x+y)(-x^2 + xy - y^2)
=−2(x+y)(x2−xy+y2)= -2(x+y)(x^2 - xy + y^2)
=−2(x3+y3)= -2(x^3 + y^3)

Answer: Δ=−2(x3+y3)\Delta = -2(x^3+y^3)
6Evaluate ∣1amp;xamp;y1amp;x+yamp;y1amp;xamp;x+y∣\left|\begin{array}{ccc}1&x&y\\1&x+y&y\\1&x&x+y\end{array}\right|Show solution
Apply R2→R2−R1R_2 \to R_2 - R_1 and R3→R3−R1R_3 \to R_3 - R_1:
Δ=∣1amp;xamp;y0amp;yamp;00amp;0amp;x∣\Delta = \left|\begin{array}{ccc}1&x&y\\0&y&0\\0&0&x\end{array}\right|

Expanding along C1C_1:
=1⋅∣yamp;00amp;x∣=xy= 1\cdot\left|\begin{array}{cc}y&0\\0&x\end{array}\right| = xy

Answer: Δ=xy\Delta = xy
7Solve the system of equations: 2x+3y+10z=4\dfrac{2}{x}+\dfrac{3}{y}+\dfrac{10}{z}=4, 4x−6y+5z=1\dfrac{4}{x}-\dfrac{6}{y}+\dfrac{5}{z}=1, 6x+9y−20z=2\dfrac{6}{x}+\dfrac{9}{y}-\dfrac{20}{z}=2.Show solution
Substitution: Let u=1x, v=1y, w=1zu = \dfrac{1}{x},\ v = \dfrac{1}{y},\ w = \dfrac{1}{z}.

System becomes:
2u+3v+10w=42u + 3v + 10w = 4
4u−6v+5w=14u - 6v + 5w = 1
6u+9v−20w=26u + 9v - 20w = 2

Matrix form: A=[2amp;3amp;104amp;−6amp;56amp;9amp;−20]A = \begin{bmatrix}2&3&10\\4&-6&5\\6&9&-20\end{bmatrix}, B=[412]B = \begin{bmatrix}4\\1\\2\end{bmatrix}

Step 1: ∣A∣|A| expanding along R1R_1:
=2(120−45)−3(−80−30)+10(36+36)= 2(120-45)-3(-80-30)+10(36+36)
=2(75)−3(−110)+10(72)= 2(75)-3(-110)+10(72)
=150+330+720=1200= 150+330+720 = 1200

Step 2: Cofactors:
A11=(120−45)=75,A12=−(−80−30)=110,A13=(36+36)=72A_{11}=(120-45)=75,\quad A_{12}=-(-80-30)=110,\quad A_{13}=(36+36)=72
A21=−(−60−90)=150,A22=(−40−60)=−100,A23=−(18−18)=0A_{21}=-(−60-90)=150,\quad A_{22}=(-40-60)=-100,\quad A_{23}=-(18-18)=0
A31=(15+60)=75,A32=−(10−40)=30,A33=(−12−12)=−24A_{31}=(15+60)=75,\quad A_{32}=-(10-40)=30,\quad A_{33}=(-12-12)=-24

A−1=11200[75amp;150amp;75110amp;−100amp;3072amp;0amp;−24]A^{-1} = \frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}

Step 3: X=A−1BX = A^{-1}B:
[uvw]=11200[75amp;150amp;75110amp;−100amp;3072amp;0amp;−24][412]\begin{bmatrix}u\\v\\w\end{bmatrix} = \frac{1}{1200}\begin{bmatrix}75&150&75\\110&-100&30\\72&0&-24\end{bmatrix}\begin{bmatrix}4\\1\\2\end{bmatrix}
=11200[300+150+150440−100+60288+0−48]=11200[600400240]=[1/21/31/5]= \frac{1}{1200}\begin{bmatrix}300+150+150\\440-100+60\\288+0-48\end{bmatrix} = \frac{1}{1200}\begin{bmatrix}600\\400\\240\end{bmatrix} = \begin{bmatrix}1/2\\1/3\\1/5\end{bmatrix}

So u=12, v=13, w=15u = \dfrac{1}{2},\ v = \dfrac{1}{3},\ w = \dfrac{1}{5}.

Answer: x=2, y=3, z=5x = 2,\ y = 3,\ z = 5
8If x,y,zx, y, z are nonzero real numbers, then the inverse of matrix A=[xamp;0amp;00amp;yamp;00amp;0amp;z]A = \begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix} is: (A) [x−1amp;0amp;00amp;y−1amp;00amp;0amp;z−1]\begin{bmatrix}x^{-1}&0&0\\0&y^{-1}&0\\0&0&z^{-1}\end{bmatrix}, (B) xyz[x−1amp;0amp;00amp;y−1amp;00amp;0amp;z−1]xyz\begin{bmatrix}x^{-1}&0&0\\0&y^{-1}&0\\0&0&z^{-1}\end{bmatrix}, (C) 1xyz[xamp;0amp;00amp;yamp;00amp;0amp;z]\frac{1}{xyz}\begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix}, (D) 1xyz[1amp;0amp;00amp;1amp;00amp;0amp;1]\frac{1}{xyz}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}Show solution
Correct Option: (A)

∣A∣=xyz|A| = xyz. The cofactor matrix of a diagonal matrix is also diagonal with entries yz,xz,xyyz, xz, xy.

adj(A)=[yzamp;0amp;00amp;xzamp;00amp;0amp;xy]\text{adj}(A) = \begin{bmatrix}yz&0&0\\0&xz&0\\0&0&xy\end{bmatrix}

A−1=1xyz[yzamp;0amp;00amp;xzamp;00amp;0amp;xy]=[1/xamp;0amp;00amp;1/yamp;00amp;0amp;1/z]=[x−1amp;0amp;00amp;y−1amp;00amp;0amp;z−1]A^{-1} = \frac{1}{xyz}\begin{bmatrix}yz&0&0\\0&xz&0\\0&0&xy\end{bmatrix} = \begin{bmatrix}1/x&0&0\\0&1/y&0\\0&0&1/z\end{bmatrix} = \begin{bmatrix}x^{-1}&0&0\\0&y^{-1}&0\\0&0&z^{-1}\end{bmatrix}

Answer: (A)
9Let A=[1amp;sin⁡θamp;1−sin⁡θamp;1amp;sin⁡θ−1amp;−sin⁡θamp;1]A = \begin{bmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{bmatrix}, where 0≤θ≤2π0\leq\theta\leq2\pi. Then: (A) det⁡(A)=0\det(A)=0, (B) det⁡(A)∈(2,∞)\det(A)\in(2,\infty), (C) det⁡(A)∈(2,4)\det(A)\in(2,4), (D) det⁡(A)∈[2,4]\det(A)\in[2,4]Show solution
Correct Option: (D) det⁡(A)∈[2,4]\det(A)\in[2,4]

Expanding along R1R_1:
det⁡(A)=1(1+sin⁡2θ)−sin⁡θ(−sin⁡θ+sin⁡θ)+1(sin⁡2θ+1)\det(A) = 1(1+\sin^2\theta) - \sin\theta(-\sin\theta+\sin\theta) + 1(\sin^2\theta+1)
=(1+sin⁡2θ)−sin⁡θ(0)+(1+sin⁡2θ)= (1+\sin^2\theta) - \sin\theta(0) + (1+\sin^2\theta)
=2(1+sin⁡2θ)=2+2sin⁡2θ= 2(1+\sin^2\theta) = 2 + 2\sin^2\theta

Since 0≤sin⁡2θ≤10 \leq \sin^2\theta \leq 1:
2≤2+2sin⁡2θ≤42 \leq 2+2\sin^2\theta \leq 4

So det⁡(A)∈[2,4]\det(A) \in [2, 4].

Answer: (D)

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