Q81 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
If $A=\left[\begin{array}{ccc}-1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$ then the inverse of $(A I)^t$ (where $\mathrm{I}$ is an identity matrix) is
- A$\left[\begin{array}{ccc} 1 & 8 & -5 \\ -1 & 7 & -4 \\ 0 & 5 & 3 \end{array}\right]$
- B$\left[\begin{array}{ccc} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{array}\right]$
- Correct answerC$\left[\begin{array}{ccc} 1 & -8 & 5 \\ -1 & 7 & -4 \\ 1 & -5 & 3 \end{array}\right]$
- D$\left[\begin{array}{ccc} -1 & 1 & -1 \\ 8 & -7 & 5 \\ -5 & 4 & -3 \end{array}\right]$
Solution
Since $AI = A$, we need $(A^t)^{-1}$.
$A^t = \begin{bmatrix} -1 & 1 & 3 \\ 1 & 2 & 1 \\ 2 & 3 & 1 \end{bmatrix}$
$|A^t| = -1(2-3) - 1(1-2) + 3(3-4) = 1 + 1 - 3 = -1$
Cofactors of $A^t$:
$C_{11} = -1,\ C_{12} = 1,\ C_{13} = -1$
$C_{21} = 8,\ C_{22} = -7,\ C_{23} = 5$
$C_{31} = -5,\ C_{32} = 4,\ C_{33} = -3$
$\text{adj}(A^t) = C^t = \begin{bmatrix} -1 & 8 & -5 \\ 1 & -7 & 4 \\ -1 & 5 & -3 \end{bmatrix}$
$(A^t)^{-1} = \dfrac{1}{-1}\begin{bmatrix} -1 & 8 & -5 \\ 1 & -7 & 4 \\ -1 & 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & -8 & 5 \\ -1 & 7 & -4 \\ 1 & -5 & 3 \end{bmatrix}$
$A^t = \begin{bmatrix} -1 & 1 & 3 \\ 1 & 2 & 1 \\ 2 & 3 & 1 \end{bmatrix}$
$|A^t| = -1(2-3) - 1(1-2) + 3(3-4) = 1 + 1 - 3 = -1$
Cofactors of $A^t$:
$C_{11} = -1,\ C_{12} = 1,\ C_{13} = -1$
$C_{21} = 8,\ C_{22} = -7,\ C_{23} = 5$
$C_{31} = -5,\ C_{32} = 4,\ C_{33} = -3$
$\text{adj}(A^t) = C^t = \begin{bmatrix} -1 & 8 & -5 \\ 1 & -7 & 4 \\ -1 & 5 & -3 \end{bmatrix}$
$(A^t)^{-1} = \dfrac{1}{-1}\begin{bmatrix} -1 & 8 & -5 \\ 1 & -7 & 4 \\ -1 & 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & -8 & 5 \\ -1 & 7 & -4 \\ 1 & -5 & 3 \end{bmatrix}$
Hence, the correct answer is option: 'C'.