If `A=[(3,-3,4),(2,-3,4),(0,-1,1)]`, then `A^(-1)=`

Options

A. A
B. `A^(2)`
C. `A^(3)`
D. `A^(4)`

Answer

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

### Step 1: Calculate the Determinant of \( A \)

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 \( A \):

- \( a = 3, b = -3, c = 4 \)
- \( d = 2, e = -3, f = 4 \)
- \( g = 0, h = -1, i = 1 \)

Calculating the determinant:

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

Calculating each term:

1. \( (-3)(1) - (4)(-1) = -3 + 4 = 1 \)
2. \( (2)(1) - (4)(0) = 2 - 0 = 2 \)
3. \( (2)(-1) - (-3)(0) = -2 - 0 = -2 \)

Now substituting back:

\[ \text{det}(A) = 3(1) + 3(2) + 4(-2) = 3 + 6 - 8 = 1 \]

### Step 2: Calculate the Adjoint of \( A \)

The adjoint of a matrix \( A \) is the transpose of the cofactor matrix. We need to find the cofactors for each element of \( A \).

**Cofactor Calculation:**

1. **Cofactor \( C_{11} \)** (remove row 1, column 1):
\[ C_{11} = \text{det}\begin{pmatrix} -3 & 4 \\ -1 & 1 \end{pmatrix} = (-3)(1) - (4)(-1) = -3 + 4 = 1 \]

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

3. **Cofactor \( C_{13} \)** (remove row 1, column 3):
\[ C_{13} = \text{det}\begin{pmatrix} 2 & -3 \\ 0 & -1 \end{pmatrix} = (2)(-1) - (-3)(0) = -2 \]

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

5. **Cofactor \( C_{22} \)** (remove row 2, column 2):
\[ C_{22} = \text{det}\begin{pmatrix} 3 & 4 \\ 0 & 1 \end{pmatrix} = (3)(1) - (4)(0) = 3 \]

6. **Cofactor \( C_{23} \)** (remove row 2, column 3):
\[ C_{23} = -\text{det}\begin{pmatrix} 3 & -3 \\ 0 & -1 \end{pmatrix} = -((3)(-1) - (-3)(0)) = 3 \]

7. **Cofactor \( C_{31} \)** (remove row 3, column 1):
\[ C_{31} = \text{det}\begin{pmatrix} -3 & 4 \\ -3 & 4 \end{pmatrix} = (-3)(4) - (4)(-3) = -12 + 12 = 0 \]

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

9. **Cofactor \( C_{33} \)** (remove row 3, column 3):
\[ C_{33} = \text{det}\begin{pmatrix} 3 & -3 \\ 2 & -3 \end{pmatrix} = (3)(-3) - (-3)(2) = -9 + 6 = -3 \]

Now, the cofactor matrix is:

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

Taking the transpose gives us the adjoint:

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

### Step 3: Calculate the Inverse of \( A \)

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

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

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

\[ A^{-1} = \text{adj}(A) = \begin{pmatrix} 1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3 \end{pmatrix} \]

### Final Answer

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