If matrix `A=[[1, 0, -1], [3, 4, 5], [0, 6, 7]]` and its inverse is denoted by `A^(-1)=[[a_(11), a_(12), a_(13)], [a_(21), a_(22), a_(23)], [a_(31), a_(32), a_(33)]],` then the value of `a_(23)`=

Options

A. `(-2)/(3)`
B. `(1)/(5)`
C. `(2)/(5)`
D. `(21)/(20)`

Answer

To find the value of \( a_{23} \) in the inverse of the matrix \( A \), we will follow these steps:

Given:
\[ A = \begin{bmatrix} 1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & 6 & 7 \end{bmatrix} \]

### Step 1: Calculate the determinant of matrix \( A \)

The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is represented as:
\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

For our matrix \( A \):
- \( a = 1, b = 0, c = -1 \)
- \( d = 3, e = 4, f = 5 \)
- \( g = 0, h = 6, i = 7 \)

Calculating the determinant:
\[ \text{det}(A) = 1(4 \cdot 7 - 5 \cdot 6) - 0 + (-1)(3 \cdot 6 - 4 \cdot 0) \]
\[ = 1(28 - 30) - 0 - (18 - 0) \]
\[ = 1(-2) - 18 = -2 - 18 = -20 \]

### Step 2: Calculate the cofactor matrix of \( A \)

The cofactor matrix \( C \) is calculated by finding the determinant of the \( 2 \times 2 \) matrices formed by removing the row and column of each element, with alternating signs.

Calculating the cofactors:
- \( C_{11} = \text{det}\begin{bmatrix} 4 & 5 \\ 6 & 7 \end{bmatrix} = (4 \cdot 7 - 5 \cdot 6) = -2 \)
- \( C_{12} = -\text{det}\begin{bmatrix} 3 & 5 \\ 0 & 7 \end{bmatrix} = -(3 \cdot 7 - 5 \cdot 0) = -21 \)
- \( C_{13} = \text{det}\begin{bmatrix} 3 & 4 \\ 0 & 6 \end{bmatrix} = (3 \cdot 6 - 4 \cdot 0) = 18 \)

- \( C_{21} = -\text{det}\begin{bmatrix} 0 & -1 \\ 6 & 7 \end{bmatrix} = -(-1 \cdot 6 - 0 \cdot 7) = 6 \)
- \( C_{22} = \text{det}\begin{bmatrix} 1 & -1 \\ 0 & 7 \end{bmatrix} = (1 \cdot 7 - (-1) \cdot 0) = 7 \)
- \( C_{23} = -\text{det}\begin{bmatrix} 1 & 0 \\ 0 & 6 \end{bmatrix} = -(1 \cdot 6 - 0 \cdot 0) = -6 \)

- \( C_{31} = \text{det}\begin{bmatrix} 0 & -1 \\ 4 & 5 \end{bmatrix} = (0 \cdot 5 - (-1) \cdot 4) = 4 \)
- \( C_{32} = -\text{det}\begin{bmatrix} 1 & -1 \\ 3 & 5 \end{bmatrix} = -(1 \cdot 5 - (-1) \cdot 3) = -8 \)
- \( C_{33} = \text{det}\begin{bmatrix} 1 & 0 \\ 3 & 4 \end{bmatrix} = (1 \cdot 4 - 0 \cdot 3) = 4 \)

Thus, the cofactor matrix \( C \) is:
\[ C = \begin{bmatrix} -2 & -21 & 18 \\ 6 & 7 & -6 \\ 4 & -8 & 4 \end{bmatrix} \]

### Step 3: Transpose the cofactor matrix to get the adjoint

The adjoint \( \text{adj}(A) \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = C^T = \begin{bmatrix} -2 & 6 & 4 \\ -21 & 7 & -8 \\ 18 & -6 & 4 \end{bmatrix} \]

### Step 4: Calculate the inverse of matrix \( A \)

The inverse of matrix \( A \) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Substituting the values:
\[ A^{-1} = \frac{1}{-20} \cdot \begin{bmatrix} -2 & 6 & 4 \\ -21 & 7 & -8 \\ 18 & -6 & 4 \end{bmatrix} \]
\[ = \begin{bmatrix} \frac{1}{10} & -\frac{3}{10} & -\frac{1}{5} \\ \frac{21}{20} & -\frac{7}{20} & \frac{2}{5} \\ -\frac{9}{10} & \frac{3}{10} & -\frac{1}{5} \end{bmatrix} \]

### Step 5: Identify \( a_{23} \)

From the inverse matrix \( A^{-1} \), we can see that:
\[ a_{23} = \frac{2}{5} \]

### Final Answer
Thus, the value of \( a_{23} \) is:
\[ \boxed{\frac{2}{5}} \]