The inverset of `A=[(-4,-3,-3),(1,0,1),(4,4,3)]` is ………

Answer

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

### Step 1: Find 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 calculated using the formula:

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

For our matrix \( A \):

- \( a = -4, b = -3, c = -3 \)
- \( d = 1, e = 0, f = 1 \)
- \( g = 4, h = 4, i = 3 \)

Calculating the determinant:

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

Calculating each term:

1. \( -4(0 - 4) = -4 \cdot -4 = 16 \)
2. \( -(-3)(3 - 4) = 3 \cdot -1 = -3 \)
3. \( -3(4 - 0) = -3 \cdot 4 = -12 \)

Now, summing these values:

\[ \text{det}(A) = 16 - 3 - 12 = 1 \]

### Step 2: Find the Cofactor Matrix

The cofactor \( C_{ij} \) is calculated using the formula:

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

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

Calculating the cofactors:

1. \( C_{11} = M_{11} = \text{det} \begin{pmatrix} 0 & 1 \\ 4 & 3 \end{pmatrix} = 0 \cdot 3 - 1 \cdot 4 = -4 \)
2. \( C_{12} = -M_{12} = -\text{det} \begin{pmatrix} 1 & 1 \\ 4 & 3 \end{pmatrix} = - (1 \cdot 3 - 1 \cdot 4) = -(-1) = 1 \)
3. \( C_{13} = M_{13} = \text{det} \begin{pmatrix} 1 & 0 \\ 4 & 4 \end{pmatrix} = 1 \cdot 4 - 0 \cdot 4 = 4 \)

Continuing with the second row:

4. \( C_{21} = -M_{21} = -\text{det} \begin{pmatrix} -3 & -3 \\ 4 & 3 \end{pmatrix} = -((-3) \cdot 3 - (-3) \cdot 4) = -(-9 + 12) = -3 \)
5. \( C_{22} = M_{22} = \text{det} \begin{pmatrix} -4 & -3 \\ 4 & 3 \end{pmatrix} = -4 \cdot 3 - (-3) \cdot 4 = -12 + 12 = 0 \)
6. \( C_{23} = -M_{23} = -\text{det} \begin{pmatrix} -4 & -3 \\ 1 & 4 \end{pmatrix} = -((-4) \cdot 4 - (-3) \cdot 1) = -(-16 + 3) = -(-13) = 13 \)

Continuing with the third row:

7. \( C_{31} = M_{31} = \text{det} \begin{pmatrix} -3 & -3 \\ 0 & 1 \end{pmatrix} = -3 \cdot 1 - (-3) \cdot 0 = -3 \)
8. \( C_{32} = -M_{32} = -\text{det} \begin{pmatrix} -4 & -3 \\ 1 & 1 \end{pmatrix} = -((-4) \cdot 1 - (-3) \cdot 1) = -(-4 + 3) = -(-1) = 1 \)
9. \( C_{33} = M_{33} = \text{det} \begin{pmatrix} -4 & -3 \\ 1 & 0 \end{pmatrix} = -4 \cdot 0 - (-3) \cdot 1 = 3 \)

Thus, the cofactor matrix is:

\[ \text{Cofactor}(A) = \begin{pmatrix} -4 & 1 & 4 \\ 3 & 0 & 13 \\ -3 & 1 & 3 \end{pmatrix} \]

### Step 3: Find the Adjoint of A

The adjoint of a matrix is the transpose of the cofactor matrix:

\[ \text{Adj}(A) = \begin{pmatrix} -4 & 3 & -3 \\ 1 & 0 & 1 \\ 4 & 13 & 3 \end{pmatrix} \]

### Step 4: Calculate the Inverse of A

The inverse of a matrix is given by:

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

Since \( \text{det}(A) = 1 \):

\[ A^{-1} = 1 \cdot \text{Adj}(A) = \text{Adj}(A) \]

Thus, the inverse is:

\[ A^{-1} = \begin{pmatrix} -4 & 3 & -3 \\ 1 & 0 & 1 \\ 4 & 13 & 3 \end{pmatrix} \]

### Final Answer

The inverse of \( A \) is:

\[ A^{-1} = \begin{pmatrix} -4 & 3 & -3 \\ 1 & 0 & 1 \\ 4 & 13 & 3 \end{pmatrix} \]