## Related questions with answers

Reduce $[A | I]$ to find $A^{−1}$. In each case, check your calculations by multiplying the given matrix by the derived inverse.

$\left[\begin{array}{lll} 1 & 4 & 2 \\ 0 & 2 & 1 \\ 3 & 5 & 3 \end{array}\right]$

Solution

VerifiedWe have

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

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

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

to

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

$\begin{align*} \begin{bmatrix}\begin{array}{ccc|ccc} 1&4&2&1&0&0\\ 0&2&1&0&1&0\\ 3&5&3&0&0&1 \end{array}\end{bmatrix}{\small\begin{matrix} R_{1}\longleftrightarrow R_{3}\\ R_{3}-\dfrac{1}{3}R_{1} \rightarrow R_{3} \end{matrix}}&\sim \begin{bmatrix}\begin{array}{ccc|ccc} 3&5&3&0&0&1\\ 0&2&1&0&1&0\\ 0&\dfrac{7}{3}&1&1&0&-\dfrac{1}{3} \end{array}\end{bmatrix}\\ \begin{bmatrix}\begin{array}{ccc|ccc} 3&5&3&0&0&1\\ 0&2&1&0&1&0\\ 0&\dfrac{7}{3}&1&1&0&-\dfrac{1}{3} \end{array}\end{bmatrix}{\small\begin{matrix} R_{2}\longleftrightarrow R_{3}\\ R_{3}-\dfrac{6}{7}R_{2} \rightarrow R_{3} \end{matrix}}&\sim \begin{bmatrix}\begin{array}{ccc|ccc} 3&5&3&0&0&1\\ 0&\dfrac{7}{3}&1&1&0&-\dfrac{1}{3}\\ 0&0&\dfrac{1}{7}&-\dfrac{6}{7}&1&\dfrac{2}{7} \end{array}\end{bmatrix}\\ \begin{bmatrix}\begin{array}{ccc|ccc} 3&5&3&0&0&1\\ 0&\dfrac{7}{3}&1&1&0&-\dfrac{1}{3}\\ 0&0&\dfrac{1}{7}&-\dfrac{6}{7}&1&\dfrac{2}{7} \end{array}\end{bmatrix}{\small\begin{matrix} 7R_{3}\rightarrow R_{3}\\ R_{2}-R_{3}\rightarrow R_{3} \end{matrix}}&\sim \begin{bmatrix}\begin{array}{ccc|ccc} 3&5&3&0&0&1\\ 0&\dfrac{7}{3}&0&7&-7&-\dfrac{7}{3}\\ 0&0&1&-6&7&2 \end{array}\end{bmatrix}\\ \begin{bmatrix}\begin{array}{ccc|ccc} 3&5&3&0&0&1\\ 0&\dfrac{7}{3}&0&7&-7&-\dfrac{7}{3}\\ 0&0&1&-6&7&2 \end{array}\end{bmatrix}{\small\begin{matrix} R_{1}-3R_{3}\rightarrow R_{1}\\ \dfrac{3}{7}R_{2}\rightarrow R_{2} \end{matrix}}&\sim \begin{bmatrix}\begin{array}{ccc|ccc} 3&5&0&18&-21&-5\\ 0&1&0&3&-3&-1\\ 0&0&1&-6&7&2 \end{array}\end{bmatrix}\\ \begin{bmatrix}\begin{array}{ccc|ccc} 3&5&0&18&-21&-5\\ 0&1&0&3&-3&-1\\ 0&0&1&-6&7&2 \end{array}\end{bmatrix}{\small\begin{matrix} R_{1}-5R_{2}\rightarrow R_{1}\\ \dfrac{1}{3}R_{1}\rightarrow R_{1} \end{matrix}}&\sim \begin{bmatrix}\begin{array}{ccc|ccc} 1&0&0&1&-2&0\\ 0&1&0&3&-3&-1\\ 0&0&1&-6&7&2 \end{array}\end{bmatrix} \end{align*}$

## Create an account to view solutions

## Create an account to view solutions

## Recommended textbook solutions

#### Introduction to Linear Algebra

5th Edition•ISBN: 9780201658590 (5 more)Jimmy T Arnold, Lee W. Johnson, R Dean Riess#### Linear Algebra with Applications

5th Edition•ISBN: 9780321796974 (4 more)Otto Bretscher#### Linear Algebra and Its Applications

5th Edition•ISBN: 9780321982384David C. Lay, Judi J. McDonald, Steven R. Lay## More related questions

1/4

1/7