If $A$ is a square matrix of order 3 and $A \cdot (\operatorname{Adj}(A)) = 10I$, then the value of $\frac{1}{25} |\operatorname{adj}(A)|$ is:

Question

For the matrices $ A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} $ and $ B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix} $, if $ (A^{15} + B) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} $, then among the following which one is true?}

Answer 1

Step 1: From the identity $ A \cdot \operatorname{adj}(A) = \det(A) I $, and given $ A \cdot \operatorname{adj}(A) = 10 I $, we deduce that $ \det(A) = 10 $. Step 2: For an $n \times n$ matrix, $ |\operatorname{adj}(A)| = |\det(A)|^{n-1} $. With $n = 3$, this becomes $ |\operatorname{adj}(A)| = |\det(A)|^{2} = 10^2 = 100 $. Step 3: Given $A \cdot \operatorname{adj}(A) = 10 I$, we confirmed $ \det(A) = 10 $. Consequently, $ |\operatorname{adj}(A)| = 10^{2} = 100 $. The expression $ \frac{1}{25} |\operatorname{adj}(A)| $ evaluates to $ \frac{100}{25} = 4 $. As option (d) is 4, it is the correct answer.

If $A$ is a square matrix of order 3 and $A \cdot (\operatorname{Adj}(A)) = 10I$, then the value of $\frac{1}{25} |\operatorname{adj}(A)|$ is:

Show Hint

The Correct Option is A

Solution and Explanation

Top Questions on Matrices

Questions Asked in CUET (UG) exam