Determinants — Practice Quiz

[[−2,0],[0,−2]]

Step 1: By Theorem 1, A·(adj A) = |A|·I for any square matrix A. Step 2: Calculate |A| = (1)(4) - (2)(3) = 4 - 6 = -2. Step 3: Therefore A·(adj A) = |A|·I = -2 · [[1,0],[0,1]] = [[-2,0],[0,-2]]. Step 4: Verification - adj A for a 2×2 matrix [[a,b],[c,d]] is [[d,-b],[-c,a]]. So adj A = [[4,-2],[-3,1]]. Step 5: A·(adj A) = [[1,2],[3,4]]·[[4,-2],[-3,1]] = [[4-6,-2+2],[12-12,-6+4]] = [[-2,0],[0,-2]] ✓. Common mistake: confusing A·adj(A) with adj(A)·A or computing the wrong determinant.

A^(-1) = adj(A)/|A| and |A^(-1)| = -1/3

Step 1: From the theory of inverses, for a non-singular matrix A, the inverse is A^(-1) = (1/|A|) · adj(A). Step 2: Now we need |A^(-1)|. Using the property |AB| = |A||B|, and knowing A·A^(-1) = I: |A·A^(-1)| = |I| = 1, so |A|·|A^(-1)| = 1. Step 3: Therefore |A^(-1)| = 1/|A|. Step 4: Given |A| = -3, we get |A^(-1)| = 1/(-3) = -1/3. Step 5: Note: det is a real number that can be negative, so |A^(-1)| = -1/3 is correct. Common mistake: students sometimes think |A^(-1)| must be positive or compute it as 1/|A|² = 1/9.

Because R2 - R1 = R3 - R2, making R1, R2, R3 linearly dependent

Step 1: Since a, b, c are in A.P., we have b-a = c-b, i.e., 2b = a+c. Step 2: Apply R2→R2-R1 and R3→R3-R2: New R2 = [1,1,b-a] and New R3 = [1,1,c-b]. Step 3: Since a,b,c are in A.P., b-a = c-b (common difference d). So new R2 = [1,1,d] and new R3 = [1,1,d]. Step 4: Now R2 = R3 (two identical rows). By a fundamental property of determinants, if two rows are identical, the determinant is zero. Step 5: This means the three original rows are linearly dependent whenever a, b, c are in A.P. The determinant is always 0 under this condition.