For a $3 \times 3$ matrix $A$, if $A(\operatorname{adj} A) = \begin{bmatrix} 99 & 0 & 0 \\0 & 99 & 0 \\0 & 0 & 99 \end{bmatrix}$, then $\det(A)$ is equal to:

Question

Let the foot of the perpendicular from the point $ (1, 2, 4) $ on the line $ \frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} $ be $ P $. Then the distance of $ P $ from the plane $ 3x + 4y + 12z + 23 = 0 $ is:

Answer 1

For any square matrix $A$, the matrix identity is $A \cdot \operatorname{adj}(A) = \det(A) I$, where $I$ is the identity matrix of the same order. We are given that $A \cdot \operatorname{adj}(A) = \begin{bmatrix} 99 & 0 & 0 \\ 0 & 99 & 0 \\ 0 & 0 & 99 \end{bmatrix} = 99 I$. Comparing this with the identity $A \cdot \operatorname{adj}(A) = \det(A) I$, we deduce that $\det(A) = 99$.

For a $3 \times 3$ matrix $A$, if $A(\operatorname{adj} A) = \begin{bmatrix} 99 & 0 & 0 \\0 & 99 & 0 \\0 & 0 & 99 \end{bmatrix}$, then $\det(A)$ is equal to:

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The Correct Option is D

Solution and Explanation

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