The inverse of `[(3,5,7),(2,-3,1),(1,1,2)]` is ……………

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 3 & 5 & 7 \\ 2 & -3 & 1 \\ 1 & 1 & 2 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the Determinant of Matrix A

The determinant of a 3x3 matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) is given by:

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

For our matrix:

\[ \text{det}(A) = 3((-3)(2) - (1)(1)) - 5((2)(2) - (1)(1)) + 7((2)(1) - (-3)(1)) \]

Calculating each term:

- \( 3((-6) - 1) = 3(-7) = -21 \)
- \( -5(4 - 1) = -5(3) = -15 \)
- \( 7(2 + 3) = 7(5) = 35 \)

So,

\[ \text{det}(A) = -21 + 15 + 35 = 29 \]

### Step 2: Find the Matrix of Minors

Next, we calculate the minors for each element of the matrix.

- Minor of \( a_{11} = 3 \):

\[ \begin{vmatrix} -3 & 1 \\ 1 & 2 \end{vmatrix} = (-3)(2) - (1)(1) = -6 - 1 = -7 \]

- Minor of \( a_{12} = 5 \):

\[ \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = (2)(2) - (1)(1) = 4 - 1 = 3 \]

- Minor of \( a_{13} = 7 \):

\[ \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} = (2)(1) - (-3)(1) = 2 + 3 = 5 \]

- Minor of \( a_{21} = 2 \):

\[ \begin{vmatrix} 5 & 7 \\ 1 & 2 \end{vmatrix} = (5)(2) - (7)(1) = 10 - 7 = 3 \]

- Minor of \( a_{22} = -3 \):

\[ \begin{vmatrix} 3 & 7 \\ 1 & 2 \end{vmatrix} = (3)(2) - (7)(1) = 6 - 7 = -1 \]

- Minor of \( a_{23} = 1 \):

\[ \begin{vmatrix} 3 & 5 \\ 1 & 1 \end{vmatrix} = (3)(1) - (5)(1) = 3 - 5 = -2 \]

- Minor of \( a_{31} = 1 \):

\[ \begin{vmatrix} 5 & 7 \\ -3 & 1 \end{vmatrix} = (5)(1) - (7)(-3) = 5 + 21 = 26 \]

- Minor of \( a_{32} = 1 \):

\[ \begin{vmatrix} 3 & 7 \\ 2 & 1 \end{vmatrix} = (3)(1) - (7)(2) = 3 - 14 = -11 \]

- Minor of \( a_{33} = 2 \):

\[ \begin{vmatrix} 3 & 5 \\ 2 & -3 \end{vmatrix} = (3)(-3) - (5)(2) = -9 - 10 = -19 \]

### Step 3: Calculate the Cofactor Matrix

Now we apply the checkerboard pattern of signs to the minors:

\[ C = \begin{pmatrix} -7 & -3 & 5 \\ -3 & -1 & 2 \\ 26 & 11 & -19 \end{pmatrix} \]

### Step 4: Transpose the Cofactor Matrix to get the Adjugate

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

\[ \text{adj}(A) = C^T = \begin{pmatrix} -7 & -3 & 26 \\ -3 & -1 & 11 \\ 5 & 2 & -19 \end{pmatrix} \]

### Step 5: 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 we found:

\[ A^{-1} = \frac{1}{29} \begin{pmatrix} -7 & -3 & 26 \\ -3 & -1 & 11 \\ 5 & 2 & -19 \end{pmatrix} \]

Thus, the inverse of the matrix \( A \) is:

\[ A^{-1} = \begin{pmatrix} -\frac{7}{29} & -\frac{3}{29} & \frac{26}{29} \\ -\frac{3}{29} & -\frac{1}{29} & \frac{11}{29} \\ \frac{5}{29} & \frac{2}{29} & -\frac{19}{29} \end{pmatrix} \]