The inverse of the matrix \(A = \begin{bmatrix}1 & 2 & 2\\ 1 & 3 & 2\\ 1 & 2 & 3\end{bmatrix}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix}5 & -2 & -2\\ -1 & 1 & 0\\ -1 & 0 & 1\end{bmatrix}\)
Cofactors of the first row:
\(C_{11} = 9 - 4 = 5\)
\(C_{12} = -(3 - 2) = -1\)
\(C_{13} = 2 - 3 = -1\)
So
\(|A| = 1(5) + 2(-1) + 2(-1) = 1\)
The other cofactors:
\(C_{21} = -(6 - 4) = -2\)
\(C_{22} = 3 - 2 = 1\)
\(C_{23} = -(2 - 2) = 0\)
\(C_{31} = 4 - 6 = -2\)
\(C_{32} = -(2 - 2) = 0\)
\(C_{33} = 3 - 2 = 1\)
The adjoint is the transpose of the cofactor matrix, and with \(|A| = 1\) the inverse equals the adjoint:
\( \displaystyle A^{-1} = \frac{1}{1}\begin{bmatrix}5 & -2 & -2\\ -1 & 1 & 0\\ -1 & 0 & 1\end{bmatrix} \)
Check: the first row of A times the first column of \(A^{-1}\) is \(5 - 2 - 2 = 1\), and times the second column is \(-2 + 2 + 0 = 0\).
Why the other options are incorrect
Option (1) \(\begin{bmatrix}5 & -1 & -1\\ -2 & 1 & 0\\ -2 & 0 & 1\end{bmatrix}\) — Incorrect. This is the cofactor matrix itself; it has not been transposed.
Option (2) \(\begin{bmatrix}-5 & 2 & 2\\ 1 & -1 & 0\\ 1 & 0 & -1\end{bmatrix}\) — Incorrect. This is \(-A^{-1}\); it would need \(|A| = -1\), but \(|A| = 1\).
Option (4) \(\begin{bmatrix}5 & 2 & 2\\ 1 & 1 & 0\\ 1 & 0 & 1\end{bmatrix}\) — Incorrect. These are the minors without the cofactor signs \((-1)^{i+j}\).