Question

The inverse of a matrix $A = \begin{bmatrix} 0 & 2 & 4 \\ 2 & 4 & 2 \\ 3 & 3 & 1 \end{bmatrix}$ is}

• A. $\begin{bmatrix} -1 & \frac{1}{2} & 0 \\ 1 & 0 & 0\\ \frac{2}{3} & -\frac{3}{8} & \frac{1}{8} \end{bmatrix}$
• B. $\begin{bmatrix} -\frac{1}{8} & \frac{5}{8} & -\frac{3}{4} \\ \frac{1}{4} & -\frac{3}{4} & \frac{1}{2} \\ -\frac{3}{8} & \frac{3}{8} & -\frac{1}{4} \end{bmatrix}$
• C. $\begin{bmatrix} \frac{1}{8} & -\frac{5}{8} & \frac{3}{4} \\ -\frac{1}{4} & \frac{3}{4} & -\frac{1}{2} \\ \frac{3}{8} & -\frac{3}{8} & \frac{1}{4} \end{bmatrix}$
• D. $\begin{bmatrix} -1 & 1 & 0 \\ \frac{1}{2} & 0 & 0 \\ 0 & -\frac{3}{8} & \frac{1}{8} \end{bmatrix}$

Hint:

To quickly verify an inverse matrix in a multiple-choice exam, multiply the first row of original matrix $A$ by the first column of the candidate inverse matrix:
\[ \text{Row 1} \cdot \text{Col 1} = [0, 2, 4] \cdot \begin{bmatrix} 1/8 -1/4 3/8 \end{bmatrix} = 0 - \frac{2}{4} + \frac{12}{8} = -\frac{1}{2} + \frac{3}{2} = 1 \]
This quick check confirms that Option C is correct.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept

The inverse of a square matrix \(A\) is a matrix \(A^{-1}\) such that

\[ AA^{-1}=A^{-1}A=I, \]

where \(I\) is the identity matrix of the same order.

For the given matrix

\[ A= \begin{bmatrix} 0 & 2 & 4\\ 2 & 4 & 2\\ 3 & 3 & 1 \end{bmatrix}, \]

the inverse can be calculated using the adjugate formula:

\[ A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A). \]

A matrix has an inverse only when

\[ \det(A)\neq 0. \]

Step 2: Detailed Calculation

1. Calculate the determinant of \(A\)

Expanding along the first row:

\[ \begin{aligned} \det(A) &= 0 \begin{vmatrix} 4 & 2\\ 3 & 1 \end{vmatrix} - 2 \begin{vmatrix} 2 & 2\\ 3 & 1 \end{vmatrix} + 4 \begin{vmatrix} 2 & 4\\ 3 & 3 \end{vmatrix}. \end{aligned} \]

Now calculate the \(2\times2\) determinants:

\[ \begin{aligned} \det(A) &= 0 - 2\left[(2)(1)-(2)(3)\right] + 4\left[(2)(3)-(4)(3)\right]\\ &= -2(2-6)+4(6-12)\\ &= -2(-4)+4(-6)\\ &= 8-24\\ &=-16. \end{aligned} \]

Since

\[ \det(A)=-16\neq 0, \]

the inverse of \(A\) exists.

2. Find the cofactor matrix

The cofactors are:

\[ \begin{aligned} C_{11} &= \begin{vmatrix} 4 & 2\\ 3 & 1 \end{vmatrix} =4-6=-2,\\[4pt] C_{12} &= - \begin{vmatrix} 2 & 2\\ 3 & 1 \end{vmatrix} =-(2-6)=4,\\[4pt] C_{13} &= \begin{vmatrix} 2 & 4\\ 3 & 3 \end{vmatrix} =6-12=-6, \end{aligned} \] \[ \begin{aligned} C_{21} &= - \begin{vmatrix} 2 & 4\\ 3 & 1 \end{vmatrix} =-(2-12)=10,\\[4pt] C_{22} &= \begin{vmatrix} 0 & 4\\ 3 & 1 \end{vmatrix} =0-12=-12,\\[4pt] C_{23} &= - \begin{vmatrix} 0 & 2\\ 3 & 3 \end{vmatrix} =-(0-6)=6, \end{aligned} \] \[ \begin{aligned} C_{31} &= \begin{vmatrix} 2 & 4\\ 4 & 2 \end{vmatrix} =4-16=-12,\\[4pt] C_{32} &= - \begin{vmatrix} 0 & 4\\ 2 & 2 \end{vmatrix} =-(0-8)=8,\\[4pt] C_{33} &= \begin{vmatrix} 0 & 2\\ 2 & 4 \end{vmatrix} =0-4=-4. \end{aligned} \]

Therefore, the cofactor matrix is

\[ C= \begin{bmatrix} -2 & 4 & -6\\ 10 & -12 & 6\\ -12 & 8 & -4 \end{bmatrix}. \]

3. Find the adjugate matrix

The adjugate is the transpose of the cofactor matrix:

\[ \operatorname{adj}(A)=C^{T} = \begin{bmatrix} -2 & 10 & -12\\ 4 & -12 & 8\\ -6 & 6 & -4 \end{bmatrix}. \]

4. Calculate \(A^{-1}\)

\[ \begin{aligned} A^{-1} &= \frac{1}{\det(A)}\operatorname{adj}(A)\\[4pt] &= -\frac{1}{16} \begin{bmatrix} -2 & 10 & -12\\ 4 & -12 & 8\\ -6 & 6 & -4 \end{bmatrix}. \end{aligned} \]

Multiplying every entry by \(-\frac{1}{16}\), we get

\[ A^{-1} = \begin{bmatrix} \frac{1}{8} & -\frac{5}{8} & \frac{3}{4}\\ -\frac{1}{4} & \frac{3}{4} & -\frac{1}{2}\\ \frac{3}{8} & -\frac{3}{8} & \frac{1}{4} \end{bmatrix}. \]

Step 3: Final Answer

\[ \boxed{ A^{-1} = \begin{bmatrix} \frac{1}{8} & -\frac{5}{8} & \frac{3}{4}\\ -\frac{1}{4} & \frac{3}{4} & -\frac{1}{2}\\ \frac{3}{8} & -\frac{3}{8} & \frac{1}{4} \end{bmatrix} } \]

Hence, the correct answer is Option (C).