Find the inverse of each of the following :
`[{:(2,1,3),(4,-1,0),(-7,2,1):}]`

Answer

To find the inverse of the given matrix \( A = \begin{pmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the Determinant of Matrix A
The determinant of a \( 3 \times 3 \) matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) is calculated using the formula:

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

For our matrix \( A \):

\[ \text{det}(A) = 2((-1)(1) - (0)(2)) - 1((4)(1) - (0)(-7)) + 3((4)(2) - (-1)(-7)) \]

Calculating each term:

1. \( 2((-1)(1) - (0)(2)) = 2(-1 - 0) = -2 \)
2. \( -1((4)(1) - (0)(-7)) = -1(4 - 0) = -4 \)
3. \( 3((4)(2) - (-1)(-7)) = 3(8 - 7) = 3(1) = 3 \)

Now, summing these up:

\[ \text{det}(A) = -2 - 4 + 3 = -3 \]

### Step 2: Find the Adjoint of Matrix A
The adjoint of a matrix is the transpose of its cofactor matrix. We need to calculate the cofactors for each element in the matrix.

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = \text{det}\begin{pmatrix} -1 & 0 \\ 2 & 1 \end{pmatrix} = (-1)(1) - (0)(2) = -1 \]

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = -\text{det}\begin{pmatrix} 4 & 0 \\ -7 & 1 \end{pmatrix} = -((4)(1) - (0)(-7)) = -4 \]

3. **Cofactor \( C_{13} \)**:
\[ C_{13} = \text{det}\begin{pmatrix} 4 & -1 \\ -7 & 2 \end{pmatrix} = (4)(2) - (-1)(-7) = 8 - 7 = 1 \]

4. **Cofactor \( C_{21} \)**:
\[ C_{21} = -\text{det}\begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix} = -((1)(1) - (3)(2)) = -1 + 6 = 5 \]

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = \text{det}\begin{pmatrix} 2 & 3 \\ -7 & 1 \end{pmatrix} = (2)(1) - (3)(-7) = 2 + 21 = 23 \]

6. **Cofactor \( C_{23} \)**:
\[ C_{23} = -\text{det}\begin{pmatrix} 2 & 1 \\ -7 & 2 \end{pmatrix} = -((2)(2) - (1)(-7)) = -4 + 7 = 3 \]

7. **Cofactor \( C_{31} \)**:
\[ C_{31} = \text{det}\begin{pmatrix} 1 & 3 \\ -1 & 0 \end{pmatrix} = (1)(0) - (3)(-1) = 0 + 3 = 3 \]

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = -\text{det}\begin{pmatrix} 2 & 3 \\ 4 & 0 \end{pmatrix} = -((2)(0) - (3)(4)) = 12 \]

9. **Cofactor \( C_{33} \)**:
\[ C_{33} = \text{det}\begin{pmatrix} 2 & 1 \\ 4 & -1 \end{pmatrix} = (2)(-1) - (1)(4) = -2 - 4 = -6 \]

Now we can form the cofactor matrix:

\[ \text{Cofactor Matrix} = \begin{pmatrix} -1 & -4 & 1 \\ 5 & 23 & 3 \\ 3 & 12 & -6 \end{pmatrix} \]

Taking the transpose gives us the adjoint:

\[ \text{Adjoint}(A) = \begin{pmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & 3 & -6 \end{pmatrix} \]

### Step 3: Calculate the Inverse of Matrix A
Using the formula for the inverse:

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{Adjoint}(A) \]

Substituting the values we found:

\[ A^{-1} = \frac{1}{-3} \cdot \begin{pmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & 3 & -6 \end{pmatrix} \]

Calculating the inverse:

\[ A^{-1} = \begin{pmatrix} \frac{1}{3} & -\frac{5}{3} & -1 \\ \frac{4}{3} & -\frac{23}{3} & -4 \\ -\frac{1}{3} & -1 & 2 \end{pmatrix} \]

### Final Answer:
\[ A^{-1} = \begin{pmatrix} \frac{1}{3} & -\frac{5}{3} & -1 \\ \frac{4}{3} & -\frac{23}{3} & -4 \\ -\frac{1}{3} & -1 & 2 \end{pmatrix} \]