Question

If \[ A = \begin{bmatrix} 1 & 2 & -2 \\ 2 & -1 & 2\\ -1 & 1 & -2 \end{bmatrix}, \] then find $A + 2A^{-1}$.

• A. $\begin{bmatrix} 1 & 4 & 0 \\ 4 & -5 & -4 \\ 0 & -2 & -7 \end{bmatrix}$
• B. $\begin{bmatrix} 0 & 2 & 2 \\ 2 & -4 & -6 \\ 2 & -3 & -5 \end{bmatrix}$
• C. $\begin{bmatrix} 0 & 2 & 1 \\ 2 & -4 & -3 \\ 2 & -6 & -5 \end{bmatrix}$
• D. $\begin{bmatrix} 1 & 4 & -1 \\ 4 & -5 & -1 \\ 1 & -5 & -7 \end{bmatrix}$

Hint:

Remember to use the formula $A^{-1} = \frac{1}{|A|} Adj(A)$ and carefully compute matrix multiplication for $A + 2A^{-1}$.

Answer 1

The Correct Option is A

Calculate the inverse of $A$, $A^{-1}$, and then compute $A + 2A^{-1}$. Using determinant and adjoint methods: \[ |A| = 1 \times (-1 \times -2 - 2 \times 1) - 2 \times (2 \times -2 - 2 \times -1) + (-2) \times (2 \times 1 - (-1) \times -1) = 1(2 + 2) - 2(-4 + 2) - 2(2 - 1) = 4 + 4 - 2 = 6. \] The adjoint matrix $Adj(A)$ is computed, then \[ A^{-1} = \frac{1}{|A|} Adj(A). \] Calculate $2 A^{-1}$ and add $A$, resulting in the given matrix in option (1).