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: Find the Determinant of A

The determinant of a 3x3 matrix

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

is calculated as:

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

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:

\[ \text{det}(A) = 1(0 - 6) - 4(0) + 5(3) = -6 + 0 + 15 = 9 \]

### Step 2: Find the Matrix of Cofactors

The cofactor \( C_{ij} \) of an element \( a_{ij} \) is given by:

\[ C_{ij} = (-1)^{i+j} \cdot M_{ij} \]

where \( M_{ij} \) is the determinant of the matrix obtained by deleting the \( i \)-th row and \( j \)-th column.

Calculating the cofactors:

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

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

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

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

### Step 3: Find 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 will multiply \( A \) by \( \text{adj}(A) \):

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

Calculating the entries:

1. First row:
- \( 1 \cdot -6 + 4 \cdot 0 + 5 \cdot 3 = -6 + 0 + 15 = 9 \)
- \( 1 \cdot 5 + 4 \cdot 0 + 5 \cdot -1 = 5 + 0 - 5 = 0 \)
- \( 1 \cdot 14 + 4 \cdot 9 + 5 \cdot -10 = 14 + 36 - 50 = 0 \)

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

3. Third row:
- \( 0 \cdot -6 + 1 \cdot 0 + 0 \cdot 3 = 0 + 0 + 0 = 0 \)
- \( 0 \cdot 5 + 1 \cdot 0 + 0 \cdot -1 = 0 + 0 + 0 = 0 \)
- \( 0 \cdot 14 + 1 \cdot 9 + 0 \cdot -10 = 0 + 9 + 0 = 9 \)

Thus, we have:

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

where \( I \) is the identity matrix.

### Final Answer

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