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Question

# Reduce $[A | I]$ to find $A^{−1}$. In each case, check your calculations by multiplying the given matrix by the derived inverse.$\left[\begin{array}{lll} 1 & 3 & 5 \\ 0 & 1 & 4 \\ 0 & 2 & 7 \end{array}\right]$

Solution

Verified
Step 1
1 of 3

We have

\begin{align*} A&=\begin{bmatrix} 1&3&5\\ 0&1&4\\ 0&2&7 \end{bmatrix} \end{align*}

To find inverse of $A$, convert matrix $A$ in $2 \times 4$ form as matrix

$\begin{bmatrix} \begin{array}{c|c} A&I \end{array}\end{bmatrix}$

. Reduce matrix

$\begin{bmatrix} \begin{array}{c|c} A&I \end{array}\end{bmatrix}=\begin{bmatrix}\begin{array}{ccc|ccc} 1&3&5&1&0&0\\ 0&1&4&0&1&0\\ 0&2&7&0&0&1 \end{array}\end{bmatrix}$

to

$\begin{bmatrix} \begin{array}{c|c} I&A^{-1} \end{array}\end{bmatrix}$

\begin{align*} \begin{bmatrix}\begin{array}{ccc|ccc} 1&3&5&1&0&0\\ 0&1&4&0&1&0\\ 0&2&7&0&0&1 \end{array}\end{bmatrix}{\small\begin{matrix} R_{2}\longleftrightarrow R_{3}\\ R_{3}-\dfrac{1}{2}R_{2} \rightarrow R_{3} \end{matrix}}&\sim \begin{bmatrix}\begin{array}{ccc|ccc} 1&3&5&1&0&0\\ 0&2&7&0&0&1\\ 0&0&\dfrac{1}{2}&0&1&-\dfrac{1}{2} \end{array}\end{bmatrix}\\ \begin{bmatrix}\begin{array}{ccc|ccc} 1&3&5&1&0&0\\ 0&2&7&0&0&1\\ 0&0&\dfrac{1}{2}&0&1&-\dfrac{1}{2} \end{array}\end{bmatrix}{\small\begin{matrix} 2R_{3}\rightarrow R_{3}\\ R_{2}-7R_{3}\rightarrow R_{3} \end{matrix}}&\sim \begin{bmatrix}\begin{array}{ccc|ccc} 1&3&5&1&0&0\\ 0&2&0&0&-14&8\\ 0&0&1&0&2&-1 \end{array}\end{bmatrix}\\ \begin{bmatrix}\begin{array}{ccc|ccc} 1&3&5&1&0&0\\ 0&2&0&0&-14&8\\ 0&0&1&0&2&-1 \end{array}\end{bmatrix}{\small\begin{matrix} R_{1}-5R_{3}\rightarrow R_{1}\\ \dfrac{1}{2}R_{2}\rightarrow R_{2} \end{matrix}}&\sim \begin{bmatrix}\begin{array}{ccc|ccc} 1&3&0&1&-10&5\\ 0&1&0&0&-7&4\\ 0&0&1&0&2&-1 \end{array}\end{bmatrix}\\ \begin{bmatrix}\begin{array}{ccc|ccc} 1&3&0&1&-10&5\\ 0&1&0&0&-7&4\\ 0&0&1&0&2&-1 \end{array}\end{bmatrix}{\small\begin{matrix} R_{1}-3R_{2}\rightarrow R_{1} \end{matrix}}&\sim \begin{bmatrix}\begin{array}{ccc|ccc} 1&0&0&1&11&-7\\ 0&1&0&0&-7&4\\ 0&0&1&0&2&-1 \end{array}\end{bmatrix} \end{align*}

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