Matrices — Practice Quiz

5

Step 1: Compute A² = A·A = [[2,1],[1,3]] · [[2,1],[1,3]]. Step 2: A² = [[(4+1),(2+3)],[(2+3),(1+9)]] = [[5,5],[5,10]]. Step 3: Now 5A = [[10,5],[5,15]]. So A² - 5A = [[5-10, 5-5],[5-5, 10-15]] = [[-5,0],[0,-5]] = -5I. Step 4: We need A² - 5A + kI = O, so -5I + kI = O, giving (k-5)I = O, so k = 5. Final Answer: k = 5. This is the Cayley-Hamilton theorem in action - the characteristic polynomial of A evaluated at A gives zero matrix.

81

Step 1: A 3×3 symmetric matrix has the form where a_ij = a_ji. So only elements on and above the diagonal are free choices. Step 2: The independent entries are: 3 diagonal elements (a₁₁, a₂₂, a₃₃) and 3 upper triangular elements (a₁₂, a₁₃, a₂₃). Total = 6 independent entries. Step 3: Each independent entry can take values from {0, 1, 2} - that is 3 choices each. Step 4: Total symmetric matrices = 3⁶ = 729. Wait - but we need entries from {0,1,2} only. With 6 free choices each with 3 options: 3⁶ = 729. However re-examining: diagonal has 3 entries, upper triangle has 3 entries = 6 total free en

A

Step 1: Compute powers of A. A = [[0,1],[-1,0]]. Step 2: A² = [[0,1],[-1,0]]·[[0,1],[-1,0]] = [[(0)(0)+(1)(-1),(0)(1)+(1)(0)],[(-1)(0)+(0)(-1),(-1)(1)+(0)(0)]] = [[-1,0],[0,-1]] = -I. Step 3: A³ = A²·A = (-I)·A = -A. A⁴ = A³·A = (-A)·A = -A² = -(-I) = I. Step 4: The cycle repeats every 4: A¹=A, A²=-I, A³=-A, A⁴=I. Now 2023 = 4×505 + 3, so A²⁰²³ = A³ = -A. Final Answer: A²⁰²³ = -A. Common mistake: Students may divide 2023 by 4 incorrectly. 2023 ÷ 4 = 505 remainder 3, so the answer matches A³ = -A.

x = -1

Step 1: Since A is symmetric, a₁₃ = a₃₁, which means the matrix is already symmetric for any x. Step 2: For A² = 5A - 4I, compute A². With general x, A² entry (1,1) = 1+4+x² ; (1,2)=2+2+0=4 but actual = 1·2+2·1+x·0=4. Step 3: From A² = 5A - 4I: Entry (1,1): 1+4+x² = 5(1)-4 = 1, so x² + 5 = 1 is impossible... Let me use entry (3,3): x²+0+1 = 5(1)-4=1, so x²=0, x=0. Step 4: But checking entry (1,3): x+0+x = 5x, so 2x=5x giving 3x=0, x=0. Yet option says x=-1. Re-examining with x=-1: A=[[1,2,-1],[2,1,0],[-1,0,1]]; (1,3) entry of A²= 1(-1)+2(0)+(-1)(1)=-2; 5A-4I entry(1,3)=5(-1)=-5. Not equal. Fin