Question

Find the inverse of matrix

Answer 1

$\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$

Answer 2

The question asks to find the inverse of a matrix, but the matrix itself is not provided in the question text. Based on the provided similar question, it is assumed that the question intends to ask for the inverse of the matrix $A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$ using elementary row operations.

To find the inverse of matrix $A$, we augment it with the identity matrix $I$ of the same size, forming the augmented matrix $[A | I]$.

$[A | I] = \begin{bmatrix} 1 & 2 & -2 & | & 1 & 0 & 0 \\ -1 & 3 & 0 & | & 0 & 1 & 0 \\ 0 & -2 & 1 & | & 0 & 0 & 1 \end{bmatrix}$

Apply elementary row operations to transform the left side into the identity matrix.

1. $R_2 \rightarrow R_2 + R_1$: $\begin{bmatrix} 1 & 2 & -2 & | & 1 & 0 & 0 \\ 0 & 5 & -2 & | & 1 & 1 & 0 \\ 0 & -2 & 1 & | & 0 & 0 & 1 \end{bmatrix}$

2. $R_2 \rightarrow \frac{1}{5} R_2$: $\begin{bmatrix} 1 & 2 & -2 & | & 1 & 0 & 0 \\ 0 & 1 & -2/5 & | & 1/5 & 1/5 & 0 \\ 0 & -2 & 1 & | & 0 & 0 & 1 \end{bmatrix}$

3. $R_1 \rightarrow R_1 - 2R_2$ and $R_3 \rightarrow R_3 + 2R_2$: $\begin{bmatrix} 1 & 0 & -6/5 & | & 3/5 & -2/5 & 0 \\ 0 & 1 & -2/5 & | & 1/5 & 1/5 & 0 \\ 0 & 0 & 1/5 & | & 2/5 & 2/5 & 1 \end{bmatrix}$

4. $R_3 \rightarrow 5R_3$: $\begin{bmatrix} 1 & 0 & -6/5 & | & 3/5 & -2/5 & 0 \\ 0 & 1 & -2/5 & | & 1/5 & 1/5 & 0 \\ 0 & 0 & 1 & | & 2 & 2 & 5 \end{bmatrix}$

5. $R_1 \rightarrow R_1 + \frac{6}{5}R_3$ and $R_2 \rightarrow R_2 + \frac{2}{5}R_3$: $\begin{bmatrix} 1 & 0 & 0 & | & 3 & 2 & 6 \\ 0 & 1 & 0 & | & 1 & 1 & 2 \\ 0 & 0 & 1 & | & 2 & 2 & 5 \end{bmatrix}$

The left side is now the identity matrix. The matrix on the right side is the inverse of $A$.

$A^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$

Explanation: The inverse of the matrix $A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$ is found by augmenting $A$ with the identity matrix $I$ to form $[A|I]$ and applying elementary row operations to transform $[A|I]$ into $[I|A^{-1}]$. The sequence of row operations used is $R_2 \rightarrow R_2 + R_1$, $R_2 \rightarrow \frac{1}{5}R_2$, $R_1 \rightarrow R_1 - 2R_2$, $R_3 \rightarrow R_3 + 2R_2$, $R_3 \rightarrow 5R_3$, $R_1 \rightarrow R_1 + \frac{6}{5}R_3$, $R_2 \rightarrow R_2 + \frac{2}{5}R_3$. The resulting matrix on the right side is the inverse matrix.