Wize High School Algebra II Textbook (Common Core) > Matrices (Extension Topic)

Matrix Inverse Algorithm

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Matrix Inverse Algorithm
Matrix Inverse (2×2\colorOne{2\times2}2×2)
The inverse of a  2×22\times22×2 matrix can be found using the following formula:
[abcd]−1=1ad−bc[d−b−ca]\left[\begin{array}{rr}
\colorOne{a} & b\\
c & \colorTwo{d}
\end{array}\right]^{-1}
=
\dfrac{1}
{\colorOne{a}\colorTwo{d} - bc}

\left[
\begin{array}{rr}
\colorTwo{d} & \colorFour{-}b\\
\colorFour{-}c & \colorOne{a}
\end{array}
\right][ac​bd​]−1=ad−bc1​[d−c​−ba​]
Note: a 2×22\times22×2 matrix is invertible if and only if  ad−bc≠0{\colorOne{a}\colorTwo{d} - bc \ne 0}ad−bc=0.
Example
Find the inverse of A=[21−43]A=
\left[
\begin{array}{rr}
\colorOne{2}&1\\
-4&\colorTwo{3}
\end{array}
\right]A=[2−4​13​].

A−1=1(2)(3)−(1)(−4)[3−1−(−4)2]=110[3−142]=[310−110210110]\begin{aligned}

A^{-1}

&=
\frac{1}
{(\colorOne{2})(\colorTwo{3})-(1)(-4)}
\left[
\begin{array}{cc}
\colorTwo{3} & \colorFour{-}1\\
\colorFour{-}(-4) & \colorOne{2}
\end{array}
\right]\\[2em]

&=
\dfrac{1}{10}
\left[\begin{array}{rr}
3&-1\\
4&2
\end{array}
\right]\\[2em]

&=
\left[
\begin{array}{rr}
\dfrac{3}{10}&-\dfrac{1}{10}\\[1.5em]
\dfrac{2}{10}&\dfrac{1}{10}
\end{array}
\right]

\end{aligned}A−1​=(2)(3)−(1)(−4)1​[3−(−4)​−12​]=101​[34​−12​]=​103​102​​−101​101​​​​

PAGE BREAK
Matrix Inverse (Any Size)
Use Gauss-Jordan elimination to row reduce AAA into III. Applying the same EROs to III will give us A−1A^{-1}A−1!
Steps
Augment the matrix An×nA_{n \times n}An×n​ with InI_nIn​: [AI]\left[\begin{array}{c|c}
A&I
\end{array}\right][A​I​]

Example (3×33\times 33×3):  [a11a12a131  0  0  a21a22a230  1  0  a31a32a330  0  1  ]\left[\begin{array}{rrr|rrr}
a_{11} & a_{12} & a_{13} & 1\ \ & 0\ \ & 0\ \ \\
a_{21} & a_{22} & a_{23} & 0\ \ & 1\ \ & 0\ \ \\
a_{31} & a_{32} & a_{33} & 0\ \ & 0\ \ & 1\ \ 
\end{array}\right]
​a11​a21​a31​​a12​a22​a32​​a13​a23​a33​​1  0  0  ​0  1  0  ​0  0  1  ​​
Use Gauss-Jordan elimination to row reduce AAA to III, creating the matrix: [IX]\left[\begin{array}{c|c}
I&X
\end{array}\right][I​X​]

Example (3×33\times 33×3):  [1  0  0  x11x12x130  1  0  x21x22x230  0  1  x31x32x33]\left[\begin{array}{ccc|ccc}
 1\ \ & 0\ \ & 0\ \ & x_{11} & x_{12} & x_{13} \\
 0\ \ & 1\ \ & 0\ \ & x_{21} & x_{22} & x_{23} \\
 0\ \ & 0\ \ & 1\ \ & x_{31} & x_{32} & x_{33}
\end{array}\right]​1  0  0  ​0  1  0  ​0  0  1  ​x11​x21​x31​​x12​x22​x32​​x13​x23​x33​​​
A−1=XA^{-1}=XA−1=X
Wize Tip
If AAA cannot be row reduced to III, then A−1A^{-1}A−1 does not exist, i.e.  AAA is not invertible.

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Example: Matrix Inverse Algorithm
Find the inverse of A=[1325]A=\left[\begin{array}{rr}1 & 3\\2 & 5\end{array}\right]A=[12​35​] using the formula for a 2×22\times22×2 matrix, and using the Gauss-Jordan elimination algorithm.

Using the Formula
A−1=1(1)(5)−(3)(2)[5−3−21]=(−1)[5−3−21]=[−532−1]A^{-1}
=\dfrac{1}{(1)(5)-(3)(2)}
\left[\begin{array}{rr}
5 & -3\\
-2 & 1
\end{array}\right] 

= 
(-1)
\left[\begin{array}{rr}
5 & -3\\
-2 & 1
\end{array}\right]

=\left[\begin{array}{rr}
-5 & 3\\
2 & -1
\end{array}\right]A−1=(1)(5)−(3)(2)1​[5−2​−31​]=(−1)[5−2​−31​]=[−52​3−1​]
Using the Algorithm
[AI]=[13102501]R2−2R1⟶[13100−1−21](−1)R2⟶[13100   12−1]R1−3R2⟶[10−530   12−1]=[IA−1]\begin{aligned}

\left[
\begin{array}{r|r}
A&I
\end{array}
\right]

=&
\left[
\begin{array}{rr|rr}
1&3&1&0\\
2&5&0&1
\end{array}
\right]

\begin{array}{l}
\\
R_2-2R_1\\
\end{array}\\[2em]

\longrightarrow&
\left[
\begin{array}{rr|rr}
1&3&1&0\\
0&-1&-2&1
\end{array}
\right]

\begin{array}{l}
\\
(-1)R_2\\
\end{array}\\[2em]

\longrightarrow&
\left[
\begin{array}{rr|rr}
1&3&1&0\\
0&\ \ \ 1&2&-1
\end{array}
\right]

\begin{array}{l}
R_1-3R_2\\
\\
\end{array}\\[2em]

\longrightarrow&
\left[
\begin{array}{rr|rr}
1&0&-5&3\\
0&\ \ \ 1&2&-1
\end{array}
\right]
=
\left[
\begin{array}{r|r}
I&A^{-1}
\end{array}
\right]

\end{aligned}[A​I​]=⟶⟶⟶​[12​35​10​01​]R2​−2R1​​[10​3−1​1−2​01​](−1)R2​​[10​3   1​12​0−1​]R1​−3R2​​[10​0   1​−52​3−1​]=[I​A−1​]​
∴    A−1=[−532−1]\therefore\;\;A^{-1}=\left[\begin{array}{rr}-5 & 3\\2 & -1\end{array}\right]∴A−1=[−52​3−1​]
Note: You can always check your answer.
AA−1=[1325][−532−1]=[1001]=I✓AA^{-1}
=
\left[
\begin{array}{rr}
1 & 3\\
2 & 5
\end{array}
\right]

\left[
\begin{array}{rr}
-5 & 3\\
2 & -1
\end{array}
\right]
=
\left[
\begin{array}{rr}
1 & 0\\
0 & 1
\end{array}
\right]
=
I \quad \colorThree{\checkmark}AA−1=[12​35​][−52​3−1​]=[10​01​]=I✓
A−1A=[−532−1][1325]=[1001]=I✓A^{-1}A=
\left[\begin{array}{rr}
-5 & 3\\
2 & -1
\end{array}
\right]
\left[
\begin{array}{rr}
1 & 3\\
2 & 5
\end{array}
\right]
=
\left[
\begin{array}{rr}
1 & 0\\
0 & 1
\end{array}
\right]
=
I \quad \colorThree{\checkmark}A−1A=[−52​3−1​][12​35​]=[10​01​]=I✓

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Example: Matrix Inverse Algorithm
Find the inverse of the matrix A=[112011214] A=\begin{bmatrix}
1 & 1 & 2 \\
0 & 1 & 1 \\
2 & 1 & 4 \\
\end{bmatrix}A=​102​111​214​​, if it exists.

[AI]=[   1   1   2   1   0   0011010214001]R3−2R1⟶[   1   1   2   1   0   00110100−10−201]R1−R2R3+R2⟶[   1   0   1   1   −1   0011010001−211]R1−R3R2−R3⟶[   1   0   0   3   −2   −101020−1001−211]=[IA−1]\begin{aligned}

\left[
\begin{array}{r|r}
A&I
\end{array}
\right]

=&
\left[\begin{array}{rrr | rrr}
\ \ \ 1 &\ \ \  1 &\ \ \  2 &\ \ \  1 &\ \ \  0 &\ \ \  0 \\
0 & 1 & 1 & 0 & 1 &0\\ 
2 & 1 & 4 & 0 & 0 &1 \\
\end{array}\right]

\begin{array}{l}
\\
\\
R_3 - 2R_1\\
\end{array}\\[2.5em]

\longrightarrow&
\left[\begin{array}{rrr | rrr}
\ \ \ 1 &\ \ \  1 &\ \ \  2 &\ \ \ 1 &\ \ \  0 &\ \ \  0 \\
0 & 1 & 1 & 0 & 1 &0\\ 
0 & -1 & 0 & -2 & 0 &1 \\
\end{array}\right]

\begin{array}{l}
R_1 - R_2\\
\\
R_3 + R_2\\
\end{array}\\[2.5em]

\longrightarrow&
\left[\begin{array}{rrr | rrr}
\ \ \ 1 &\ \ \  0 &\ \ \  1 &\ \ \ 1 &\ \ \  -1 &\ \ \  0 \\
0 & 1 & 1 & 0 & 1 &0\\ 
0 & 0 & 1 & -2 & 1 &1 \\
\end{array}\right]

\begin{array}{l}
R_1 - R_3\\
R_2 - R_3\\
\\
\end{array}\\[2.5em]

\longrightarrow&
\left[\begin{array}{rrr | rrr}
\ \ \ 1 &\ \ \  0 &\ \ \  0 &\ \ \ 3 &\ \ \  -2 &\ \ \  -1 \\
0 & 1 & 0 & 2 & 0 &-1\\ 
0 & 0 & 1 & -2 & 1 &1 \\
\end{array}\right]
=
\left[
\begin{array}{r|r}
I&A^{-1}
\end{array}
\right]

\end{aligned}[A​I​]=⟶⟶⟶​​   102​   111​   214​   100​   010​   001​​R3​−2R1​​​   100​   11−1​   210​   10−2​   010​   001​​R1​−R2​R3​+R2​​​   100​   010​   111​   10−2​   −111​   001​​R1​−R3​R2​−R3​​​   100​   010​   001​   32−2​   −201​   −1−11​​=[I​A−1​]​
Since we were able to reduce AAA into III, the inverse exists and A−1=[3−2−120−1−211]A^{-1}
=
\left[\begin{array}{rrr}
3&-2&-1\\
2&0&-1\\
-2&1&1
\end{array}\right]A−1=​32−2​−201​−1−11​​.
Exercise: check if AA−1=IAA^{-1}=IAA−1=I and A−1A=IA^{-1}A=IA−1A=I.

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Find the inverse of the matrix M=[−1−42310−2−2−71]M=\left[\begin{array}{rrr}
-1 & -4&2\\[0.5em]
3 & 10&-2\\[0.5em]
-2&-7&1
\end{array}\right]M=​−13−2​−410−7​2−21​​.

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Extra Practice

Practice: Matrix Inverse Algorithm

Find the inverse of the matrix M=[−1−42310−2−2−71]M=\left[\begin{array}{rrr}
-1 & -4&2\\[0.5em]
3 & 10&-2\\[0.5em]
-2&-7&1
\end{array}\right]M=​−13−2​−410−7​2−21​​