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Question: The inverse of the matrix $\begin{bmatrix}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$ is...

The inverse of the matrix [10033052−1]\begin{bmatrix}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix} is

A

−13[−30031092−3]-\frac{1}{3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 2 & -3 \end{bmatrix}

B

−13[−3003−10−9−23]-\frac{1}{3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}

C

−13[3003−10−9−23]-\frac{1}{3} \begin{bmatrix} 3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}

D

−13[−300−3−10−9−23]-\frac{1}{3} \begin{bmatrix} -3 & 0 & 0 \\ -3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}

Answer

−13[−3003−10−9−23]-\frac{1}{3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}

Explanation

Solution

To find the inverse of the matrix A=[10033052−1]A = \begin{bmatrix}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}, we can use the Gauss-Jordan method by augmenting the matrix with the identity matrix and performing row operations:

[A∣I]=[100∣100330∣01052−1∣001][A | I] = \begin{bmatrix} 1 & 0 & 0 & | & 1 & 0 & 0 \\ 3 & 3 & 0 & | & 0 & 1 & 0 \\ 5 & 2 & -1 & | & 0 & 0 & 1 \end{bmatrix}

  1. R2→R2−3R1R_2 \rightarrow R_2 - 3R_1: [100∣100030∣−31052−1∣001]\begin{bmatrix} 1 & 0 & 0 & | & 1 & 0 & 0 \\ 0 & 3 & 0 & | & -3 & 1 & 0 \\ 5 & 2 & -1 & | & 0 & 0 & 1 \end{bmatrix}

  2. R3→R3−5R1R_3 \rightarrow R_3 - 5R_1: [100∣100030∣−31002−1∣−501]\begin{bmatrix} 1 & 0 & 0 & | & 1 & 0 & 0 \\ 0 & 3 & 0 & | & -3 & 1 & 0 \\ 0 & 2 & -1 & | & -5 & 0 & 1 \end{bmatrix}

  3. R2→13R2R_2 \rightarrow \frac{1}{3}R_2: [100∣100010∣−113002−1∣−501]\begin{bmatrix} 1 & 0 & 0 & | & 1 & 0 & 0 \\ 0 & 1 & 0 & | & -1 & \frac{1}{3} & 0 \\ 0 & 2 & -1 & | & -5 & 0 & 1 \end{bmatrix}

  4. R3→R3−2R2R_3 \rightarrow R_3 - 2R_2: [100∣100010∣−113000−1∣−3−231]\begin{bmatrix} 1 & 0 & 0 & | & 1 & 0 & 0 \\ 0 & 1 & 0 & | & -1 & \frac{1}{3} & 0 \\ 0 & 0 & -1 & | & -3 & -\frac{2}{3} & 1 \end{bmatrix}

  5. R3→−1R3R_3 \rightarrow -1R_3: [100∣100010∣−1130001∣323−1]\begin{bmatrix} 1 & 0 & 0 & | & 1 & 0 & 0 \\ 0 & 1 & 0 & | & -1 & \frac{1}{3} & 0 \\ 0 & 0 & 1 & | & 3 & \frac{2}{3} & -1 \end{bmatrix}

The left side is now the identity matrix. The right side is the inverse matrix A−1A^{-1}.

A−1=[100−1130323−1]A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & -1 \end{bmatrix}

To match the form of the options, we can factor out −13-\frac{1}{3}:

A−1=−13[−3×1−3×0−3×0−3×(−1)−3×13−3×0−3×3−3×23−3×(−1)]=−13[−3003−10−9−23]A^{-1} = -\frac{1}{3} \begin{bmatrix} -3 \times 1 & -3 \times 0 & -3 \times 0 \\ -3 \times (-1) & -3 \times \frac{1}{3} & -3 \times 0 \\ -3 \times 3 & -3 \times \frac{2}{3} & -3 \times (-1) \end{bmatrix} = -\frac{1}{3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}

This matches option (b).