If `A=[[1, -1, 2], [3, 0, -2], [1, 0, 3]]`, then (adjA)A=

Options

A. `[[2, 0, 0], [0, 2, 0], [0, 0, 2]]`
B. `[[7, 0, 0], [0, 7, 0], [0, 0, 7]]`
C. `[[-7, 0, 0], [0, -7, 0], [0, 0, -7]]`
D. `[[11, 0, 0], [0, 11, 0], [0, 0, 11]]`

Answer

To solve the problem, we need to find (adj A)A for the matrix \( A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \).

### Step 1: Calculate the Determinant of A

To find \( (adj A)A \), we can use the property that \( (adj A)A = \text{det}(A) \cdot I_n \), where \( I_n \) is the identity matrix of the same order as \( A \).

First, we calculate the determinant of matrix \( A \).

\[ A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \]

Using the cofactor expansion along the second column (since it has two zeros):

\[ \text{det}(A) = -(-1) \cdot \text{det} \begin{bmatrix} 3 & -2 \\ 1 & 3 \end{bmatrix} \]

Calculating the determinant of the 2x2 matrix:

\[ \text{det} \begin{bmatrix} 3 & -2 \\ 1 & 3 \end{bmatrix} = (3 \cdot 3) - (-2 \cdot 1) = 9 + 2 = 11 \]

Thus,

\[ \text{det}(A) = 1 \cdot 11 = 11 \]

### Step 2: Formulate (adj A)A

Now, we can express \( (adj A)A \) using the determinant we just calculated:

\[ (adj A)A = \text{det}(A) \cdot I_3 \]

Where \( I_3 \) is the 3x3 identity matrix:

\[ I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

Thus,

\[ (adj A)A = 11 \cdot I_3 = 11 \cdot \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} \]

### Final Answer

Therefore,

\[ (adj A)A = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} \]

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