Q87 / 150 · MHT CET 2025 (5 May Shift 2) · Mathematics

If $A=\left[\begin{array}{rrr}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$ then $A(I+\operatorname{adj} A)=$

  • A
    $\left[\begin{array}{rrr}9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 11\end{array}\right]$
  • B
    $\left[\begin{array}{ccr}8 & -2 & 2 \\ 0 & 9 & -3 \\ 3 & -2 & 10\end{array}\right]$
  • Correct answer
    C
    $\left[\begin{array}{rrr}9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 12\end{array}\right]$
  • D
    $\left[\begin{array}{crr}3 & 2 & -2 \\ 0 & 10 & 3 \\ -3 & 2 & 12\end{array}\right]$

Solution

Given $A(I + \operatorname{adj} A)$, expand to $AI + A(\operatorname{adj} A)$. Since $AI = A$ and $A(\operatorname{adj} A) = |A|I$, the expression simplifies to $A + |A|I$.

Compute the determinant of $A = \begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$:

$|A| = 1 \cdot \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} - (-2) \cdot \begin{vmatrix} 0 & -3 \\ 3 & 4 \end{vmatrix} + 2 \cdot \begin{vmatrix} 0 & 2 \\ 3 & -2 \end{vmatrix}$

$|A| = 1(8 - 6) + 2(0 - (-9)) + 2(0 - 6) = 2 + 18 - 12 = 8$

Substitute to get $A + 8I = \begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} + \begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix}$

Add the matrices element-wise:

$\begin{bmatrix} 9 & -2 & 2 \\ 0 & 10 & -3 \\ 3 & -2 & 12 \end{bmatrix}$

This matches option C.

Hence, the correct answer is option: 'C'.

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