Question

The inverse of matrix \[ \begin{pmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\3 & -3 & 4 \end{pmatrix} \] is

• A. \[ \begin{pmatrix} 4 & 1 & -1\\ 3 & -1 & 3 \\ 4 & -3 & -3 \end{pmatrix} \]
• B. \(\begin{pmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{pmatrix}\)
• C. \[ \begin{pmatrix} 3 & -4 & 4\\ 4 & 1 & -3\\ 0 & 1 & 1 \end{pmatrix} \]
• D. \[ \begin{pmatrix} -4 & 3 & 4\\ 3 & -3 & 4 \end{pmatrix} \]

Hint:

To find the inverse of a 3x3 matrix, calculate the determinant and the adjoint matrix. Then apply the formula for the inverse matrix.

Answer 1

The Correct Option is B

Given Matrix:

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

Step 1: Check Invertibility (Calculate Determinant)

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

\[ = 0 \cdot (0) - 1 \cdot (4) + (-1) \cdot (-3) = -4 + 3 = -1 \neq 0 \]

The matrix is invertible since \(\det(A) = -1 \neq 0\).

Step 2: Find the Adjugate Matrix

Cofactor matrix \( C \):

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

Adjugate matrix (transpose of cofactor matrix):

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

Step 3: Compute the Inverse

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

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

Final Answer:

\[ \boxed{ \begin{pmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{pmatrix} } \]