# 7.7 Solving systems with inverses  (Page 3/8)

 Page 3 / 8

Use the formula to find the inverse of matrix $\text{\hspace{0.17em}}A.\text{\hspace{0.17em}}$ Verify your answer by augmenting with the identity matrix.

$A=\left[\begin{array}{cc}1& -1\\ 2& \text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]$

${A}^{-1}=\left[\begin{array}{cc}\frac{3}{5}& \frac{1}{5}\\ -\frac{2}{5}& \frac{1}{5}\end{array}\right]$

## Finding the inverse of the matrix, if it exists

Find the inverse, if it exists, of the given matrix.

$A=\left[\begin{array}{cc}3& 6\\ 1& 2\end{array}\right]$

We will use the method of augmenting with the identity.

$\left[\begin{array}{cc}3& 6\\ 1& 3\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
1. Switch row 1 and row 2.
$\left[\begin{array}{cc}1& 3\\ 3& 6\text{\hspace{0.17em}}\text{​}\end{array}\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{​}\text{​}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$
2. Multiply row 1 by −3 and add it to row 2.
$\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{cc}1& 0\\ -3& 1\end{array}\right]$
3. There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.

## Finding the multiplicative inverse of 3×3 matrices

Unfortunately, we do not have a formula similar to the one for a $\text{\hspace{0.17em}}2\text{}×\text{}2\text{\hspace{0.17em}}$ matrix to find the inverse of a $\text{\hspace{0.17em}}3\text{}×\text{}3\text{\hspace{0.17em}}$ matrix. Instead, we will augment the original matrix with the identity matrix and use row operations    to obtain the inverse.

Given a $\text{\hspace{0.17em}}3\text{}×\text{}3\text{\hspace{0.17em}}$ matrix

$A=\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right]$

augment $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ with the identity matrix

To begin, we write the augmented matrix    with the identity on the right and $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ on the left. Performing elementary row operations    so that the identity matrix    appears on the left, we will obtain the inverse matrix on the right. We will find the inverse of this matrix in the next example.

Given a $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ matrix, find the inverse

1. Write the original matrix augmented with the identity matrix on the right.
2. Use elementary row operations so that the identity appears on the left.
3. What is obtained on the right is the inverse of the original matrix.
4. Use matrix multiplication to show that $\text{\hspace{0.17em}}A{A}^{-1}=I\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{A}^{-1}A=I.$

## Finding the inverse of a 3 × 3 matrix

Given the $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ matrix $\text{\hspace{0.17em}}A,\text{\hspace{0.17em}}$ find the inverse.

$A=\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right]$

Augment $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ with the identity matrix, and then begin row operations until the identity matrix replaces $\text{\hspace{0.17em}}A.\text{\hspace{0.17em}}$ The matrix on the right will be the inverse of $\text{\hspace{0.17em}}A.\text{\hspace{0.17em}}$

$-{R}_{2}+{R}_{1}={R}_{1}\to \left[\begin{array}{ccc}1& 0& 0\\ 2& 3& 1\\ 2& 4& 1\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right]$
$-{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc}1& 0& 0\\ 2& 3& 1\\ 0& 1& 0\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill 1& \hfill 0& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\end{array}\right]$
$-2{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 3& 1\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 3& \hfill -2& \hfill 0\end{array}\right]$
$-3{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 6& \hfill -2& \hfill -3\end{array}\right]$

Thus,

${A}^{-1}=B=\left[\begin{array}{ccc}-1& \text{\hspace{0.17em}}1& \text{\hspace{0.17em}}0\\ -1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}6& -2& -3\end{array}\text{\hspace{0.17em}}\right]$

Find the inverse of the $\text{\hspace{0.17em}}3×3\text{\hspace{0.17em}}$ matrix.

$A=\left[\begin{array}{ccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2& -17& 11\\ -1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}11& -7\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3& -2\end{array}\right]$

${A}^{-1}=\left[\begin{array}{ccc}1& 1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}2\\ 2& 4& -3\\ 3& 6& -5\end{array}\right]$

## Solving a system of linear equations using the inverse of a matrix

Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: $\text{\hspace{0.17em}}X\text{\hspace{0.17em}}$ is the matrix representing the variables of the system, and $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ is the matrix representing the constants. Using matrix multiplication , we may define a system of equations with the same number of equations as variables as

$AX=B$

To solve a system of linear equations using an inverse matrix , let $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ be the coefficient matrix    , let $\text{\hspace{0.17em}}X\text{\hspace{0.17em}}$ be the variable matrix, and let $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ be the constant matrix. Thus, we want to solve a system $\text{\hspace{0.17em}}AX=B.\text{\hspace{0.17em}}$ For example, look at the following system of equations.

$\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\ {a}_{2}x+{b}_{2}y={c}_{2}\end{array}$

From this system, the coefficient matrix is

$A=\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right]$

The variable matrix is

$X=\left[\begin{array}{c}x\\ y\end{array}\right]$

And the constant matrix is

$B=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right]$

Then $\text{\hspace{0.17em}}AX=B\text{\hspace{0.17em}}$ looks like

Recall the discussion earlier in this section regarding multiplying a real number by its inverse, $\text{\hspace{0.17em}}\left({2}^{-1}\right)\text{\hspace{0.17em}}2=\left(\frac{1}{2}\right)\text{\hspace{0.17em}}2=1.\text{\hspace{0.17em}}$ To solve a single linear equation $\text{\hspace{0.17em}}ax=b\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$ Thus,

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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Sherica
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Sherica
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Tamia
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
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Samantha
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