If `A=|{:(1,4,5),(3,2,6),(0,1,0):}|`, then evaluate A. (adj. A).

Answer

To evaluate \( A \cdot \text{adj}(A) \) for the matrix

\[ A = \begin{pmatrix} 1 & 4 & 5 \\ 3 & 2 & 6 \\ 0 & 1 & 0 \end{pmatrix}, \]

we will follow these steps:

### Step 1: Calculate the Determinant of \( A \)

The determinant of a \( 3 \times 3 \) matrix

\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

is given by:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg). \]

For our matrix \( A \):

\[ \text{det}(A) = 1(2 \cdot 0 - 6 \cdot 1) - 4(3 \cdot 0 - 6 \cdot 0) + 5(3 \cdot 1 - 2 \cdot 0). \]

Calculating this gives:

\[ = 1(0 - 6) - 4(0) + 5(3) = -6 + 0 + 15 = 9. \]

### Step 2: Calculate the Cofactor Matrix of \( A \)

The cofactor \( C_{ij} \) is calculated as \( (-1)^{i+j} \cdot M_{ij} \), where \( M_{ij} \) is the determinant of the \( 2 \times 2 \) matrix obtained by deleting the \( i \)-th row and \( j \)-th column.

Calculating each cofactor:

- \( C_{11} = \text{det} \begin{pmatrix} 2 & 6 \\ 1 & 0 \end{pmatrix} = (2 \cdot 0 - 6 \cdot 1) = -6 \)
- \( C_{12} = -\text{det} \begin{pmatrix} 3 & 6 \\ 0 & 0 \end{pmatrix} = 0 \)
- \( C_{13} = \text{det} \begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix} = (3 \cdot 1 - 2 \cdot 0) = 3 \)

- \( C_{21} = -\text{det} \begin{pmatrix} 4 & 5 \\ 1 & 0 \end{pmatrix} = - (4 \cdot 0 - 5 \cdot 1) = 5 \)
- \( C_{22} = \text{det} \begin{pmatrix} 1 & 5 \\ 0 & 0 \end{pmatrix} = 0 \)
- \( C_{23} = -\text{det} \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} = - (1 \cdot 1 - 4 \cdot 0) = -1 \)

- \( C_{31} = \text{det} \begin{pmatrix} 4 & 5 \\ 2 & 6 \end{pmatrix} = (4 \cdot 6 - 5 \cdot 2) = 24 - 10 = 14 \)
- \( C_{32} = -\text{det} \begin{pmatrix} 1 & 5 \\ 3 & 6 \end{pmatrix} = - (1 \cdot 6 - 5 \cdot 3) = - (6 - 15) = 9 \)
- \( C_{33} = \text{det} \begin{pmatrix} 1 & 4 \\ 3 & 2 \end{pmatrix} = (1 \cdot 2 - 4 \cdot 3) = 2 - 12 = -10 \)

Thus, the cofactor matrix \( C \) is:

\[ C = \begin{pmatrix} -6 & 0 & 3 \\ 5 & 0 & -1 \\ 14 & 9 & -10 \end{pmatrix}. \]

### Step 3: Calculate the Adjoint of \( A \)

The adjoint of \( A \) is the transpose of the cofactor matrix:

\[ \text{adj}(A) = C^T = \begin{pmatrix} -6 & 5 & 14 \\ 0 & 0 & 9 \\ 3 & -1 & -10 \end{pmatrix}. \]

### Step 4: Calculate \( A \cdot \text{adj}(A) \)

Now we multiply \( A \) and \( \text{adj}(A) \):

\[ A \cdot \text{adj}(A) = \begin{pmatrix} 1 & 4 & 5 \\ 3 & 2 & 6 \\ 0 & 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} -6 & 5 & 14 \\ 0 & 0 & 9 \\ 3 & -1 & -10 \end{pmatrix}. \]

Calculating each entry:

1. First row, first column: \( 1 \cdot (-6) + 4 \cdot 0 + 5 \cdot 3 = -6 + 0 + 15 = 9 \)
2. First row, second column: \( 1 \cdot 5 + 4 \cdot 0 + 5 \cdot (-1) = 5 + 0 - 5 = 0 \)
3. First row, third column: \( 1 \cdot 14 + 4 \cdot 9 + 5 \cdot (-10) = 14 + 36 - 50 = 0 \)

4. Second row, first column: \( 3 \cdot (-6) + 2 \cdot 0 + 6 \cdot 3 = -18 + 0 + 18 = 0 \)
5. Second row, second column: \( 3 \cdot 5 + 2 \cdot 0 + 6 \cdot (-1) = 15 + 0 - 6 = 9 \)
6. Second row, third column: \( 3 \cdot 14 + 2 \cdot 9 + 6 \cdot (-10) = 42 + 18 - 60 = 0 \)

7. Third row, first column: \( 0 \cdot (-6) + 1 \cdot 0 + 0 \cdot 3 = 0 + 0 + 0 = 0 \)
8. Third row, second column: \( 0 \cdot 5 + 1 \cdot 0 + 0 \cdot (-1) = 0 + 0 + 0 = 0 \)
9. Third row, third column: \( 0 \cdot 14 + 1 \cdot 9 + 0 \cdot (-10) = 0 + 9 + 0 = 9 \)

Thus, we get:

\[ A \cdot \text{adj}(A) = \begin{pmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{pmatrix} = 9I, \]

where \( I \) is the identity matrix.

### Final Result

\[ A \cdot \text{adj}(A) = 9I. \]