If `A=|{:(1,0,0),(0,1,0),(0,0,1):}|" and A"=[{:(0,-3,4),(1,2,3),(0,5,5):}]," then find "(I-A)^(-1)`

Answer

To solve the problem of finding \((I - A)^{-1}\), where \(I\) is the identity matrix and \(A\) is given as:

\[ A = \begin{pmatrix} 0 & -3 & 4 \\ 1 & 2 & 3 \\ 0 & 5 & 5 \end{pmatrix} \]

we will follow these steps:

### Step 1: Define the Identity Matrix \(I\)

The identity matrix \(I\) for a \(3 \times 3\) matrix is:

\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 2: Calculate \(I - A\)

Now we will subtract matrix \(A\) from \(I\):

\[ I - A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} - \begin{pmatrix} 0 & -3 & 4 \\ 1 & 2 & 3 \\ 0 & 5 & 5 \end{pmatrix} \]

Performing the subtraction element-wise:

\[ I - A = \begin{pmatrix} 1 - 0 & 0 - (-3) & 0 - 4 \\ 0 - 1 & 1 - 2 & 0 - 3 \\ 0 - 0 & 0 - 5 & 1 - 5 \end{pmatrix} = \begin{pmatrix} 1 & 3 & -4 \\ -1 & -1 & -3 \\ 0 & -5 & -4 \end{pmatrix} \]

### Step 3: Calculate the Determinant of \(I - A\)

Let \(B = I - A\):

\[ B = \begin{pmatrix} 1 & 3 & -4 \\ -1 & -1 & -3 \\ 0 & -5 & -4 \end{pmatrix} \]

We will calculate the determinant of matrix \(B\):

\[ \text{det}(B) = 1 \cdot \begin{vmatrix} -1 & -3 \\ -5 & -4 \end{vmatrix} - 3 \cdot \begin{vmatrix} -1 & -3 \\ 0 & -4 \end{vmatrix} + (-4) \cdot \begin{vmatrix} -1 & -1 \\ 0 & -5 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} -1 & -3 \\ -5 & -4 \end{vmatrix} = (-1)(-4) - (-3)(-5) = 4 - 15 = -11\)

2. \(\begin{vmatrix} -1 & -3 \\ 0 & -4 \end{vmatrix} = (-1)(-4) - (-3)(0) = 4\)

3. \(\begin{vmatrix} -1 & -1 \\ 0 & -5 \end{vmatrix} = (-1)(-5) - (-1)(0) = 5\)

Now substituting back into the determinant formula:

\[ \text{det}(B) = 1 \cdot (-11) - 3 \cdot 4 + (-4) \cdot 5 = -11 - 12 - 20 = -43 \]

### Step 4: Find the Cofactor Matrix of \(B\)

Next, we will find the cofactor matrix of \(B\). The cofactor \(C_{ij}\) is given by:

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

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

Calculating the cofactors:

1. \(C_{11} = \begin{vmatrix} -1 & -3 \\ -5 & -4 \end{vmatrix} = -11\)

2. \(C_{12} = -\begin{vmatrix} -1 & -3 \\ 0 & -4 \end{vmatrix} = -4\)

3. \(C_{13} = \begin{vmatrix} -1 & -1 \\ 0 & -5 \end{vmatrix} = 5\)

4. \(C_{21} = -\begin{vmatrix} 3 & -4 \\ -5 & -4 \end{vmatrix} = -(-12 + 20) = 8\)

5. \(C_{22} = \begin{vmatrix} 1 & -4 \\ 0 & -4 \end{vmatrix} = -4\)

6. \(C_{23} = -\begin{vmatrix} 1 & 3 \\ 0 & -5 \end{vmatrix} = 5\)

7. \(C_{31} = \begin{vmatrix} 3 & -4 \\ -1 & -3 \end{vmatrix} = -9 + 4 = -5\)

8. \(C_{32} = -\begin{vmatrix} 1 & -4 \\ -1 & -3 \end{vmatrix} = 1\)

9. \(C_{33} = \begin{vmatrix} 1 & 3 \\ -1 & -1 \end{vmatrix} = -1 - (-3) = 2\)

Thus, the cofactor matrix \(C\) is:

\[ C = \begin{pmatrix} -11 & -4 & 5 \\ 8 & -4 & 5 \\ -5 & 1 & 2 \end{pmatrix} \]

### Step 5: Find the Adjoint of \(B\)

The adjoint of \(B\) is the transpose of the cofactor matrix:

\[ \text{adj}(B) = C^T = \begin{pmatrix} -11 & 8 & -5 \\ -4 & -4 & 1 \\ 5 & 5 & 2 \end{pmatrix} \]

### Step 6: Calculate the Inverse of \(B\)

The inverse of \(B\) is given by:

\[ B^{-1} = \frac{1}{\text{det}(B)} \cdot \text{adj}(B) \]

Substituting the determinant and adjoint:

\[ B^{-1} = \frac{1}{-43} \cdot \begin{pmatrix} -11 & 8 & -5 \\ -4 & -4 & 1 \\ 5 & 5 & 2 \end{pmatrix} \]

Thus, the final result for \((I - A)^{-1}\) is:

\[ (I - A)^{-1} = \begin{pmatrix} \frac{11}{43} & -\frac{8}{43} & \frac{5}{43} \\ \frac{4}{43} & \frac{4}{43} & -\frac{1}{43} \\ -\frac{5}{43} & -\frac{5}{43} & -\frac{2}{43} \end{pmatrix} \]