Wize University Linear Algebra Textbook > Matrices

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​​.

I have answered this question

Extra Practice

Practice: Method of Inverse

Practice Question: Method of Inverse
Given the matrix A=[2130101−12]A=\left[\begin{array}{rrr}
2&1&3\\
0&1&0\\
1&-1&2
\end{array}\right]A=​201​11−1​302​​,

Practice: Method of Inverse

Practice Question: Method of Inverse
Given the matrix A=[2130101−12]A=\left[\begin{array}{rrr}
2&1&3\\
0&1&0\\
1&-1&2
\end{array}\right]A=​201​11−1​302​​,

Practice: Matrix Inverse

Practice Question: Matrix Inverse
For what value(s) of k will the matrix[1210150k1]\begin{bmatrix}
1&2&1\\
0&1&5\\
0&k&1
\end{bmatrix}​100​21k​151​​ be invertible?

Practice: Matrix Inverse

Practice Question: Matrix Inverse
For what value(s) of k will the matrix[1210150k1]\begin{bmatrix}
1&2&1\\
0&1&5\\
0&k&1
\end{bmatrix}​100​21k​151​​ be invertible?

Practice: Matrix Inverse

Practice Question: Matrix Inverse
For what value(s) of k will the matrix[1210150k1]\begin{bmatrix}
1&2&1\\
0&1&5\\
0&k&1
\end{bmatrix}​100​21k​151​​ be invertible?

Practice: Matrix Inverse

Practice Question: Matrix Inverse
For what value(s) of k will the matrix[1210150k1]\begin{bmatrix}
1&2&1\\
0&1&5\\
0&k&1
\end{bmatrix}​100​21k​151​​ be invertible?

Practice: Matrix Inverse

Practice Question: Matrix Inverse
For what value(s) of k will the matrix[1210150k1]\begin{bmatrix}
1&2&1\\
0&1&5\\
0&k&1
\end{bmatrix}​100​21k​151​​ be invertible?

Practice: Matrix Inverse

Practice Question: Matrix Inverse
For what value(s) of k will the matrix[1210150k1]\begin{bmatrix}
1&2&1\\
0&1&5\\
0&k&1
\end{bmatrix}​100​21k​151​​ be invertible?

133 - FML 3 - 18.1W e.g. 10

 Find the value(s) of c such that A is NOT invertible given A=  [c−c1−1−1−11−cc]\bcb{\boldsymbol{ A =  \begin{bmatrix} c & -c & 1 \\ -1 & -1 & -1 \\ 1 & -c & c  \end{bmatrix} }}A=  ​c−11​−c−1−c​1−1c​​

Practice: Matrix Inverse

Find all values of k so that the matrix [−2003102−1k]\begin{bmatrix}
-2&0&0\\
3&1&0\\
2&-1&k
\end{bmatrix}​−232​01−1​00k​​ has an inverse.

Find the inverse of A=[101011110]A=\begin{bmatrix}
1&0&1\\
0&1&1\\1&1&0
\end{bmatrix}A=​101​011​110​​ .

For each of the following matrices, find the inverse or state why the inverse does not exist.
A=[1−1−2100012]A=\begin{bmatrix}
1&-1&-2\\
1&0&0\\
0&1&2
\end{bmatrix}A=​110​−101​−202​​
B=[3191506]B=\begin{bmatrix}
3&1&9\\
15&0&6
\end{bmatrix}B=[315​10​96​]

Practice: Matrix Inverse

Find all values of k for which the matrix [2−6−1k2−22]\begin{bmatrix}
2&-6\\
-1&k^2-22
\end{bmatrix}[2−1​−6k2−22​] has no inverse.

Practice: Matrix Inverse

Find the inverse of the following matrices:
a.) A=[4534]A=\begin{bmatrix}
4&5\\
3&4
\end{bmatrix}A=[43​54​]
b.) B=[−1−2024113−2]B=\begin{bmatrix}
-1&-2&0\\
2&4&1\\
1&3&-2
\end{bmatrix}B=​−121​−243​01−2​​

Matrix Inverse

Practice: Matrix Inverse
If [1ax2by0cz]\left[\begin{array}{c}
1&a&x\\
2&b&y\\
0&c&z
\end{array}\right]
​120​abc​xyz​​ is the inverse of the matrix[1p0−3q42r1]\left[\begin{array}{c}
1&p&0\\
-3&q&4\\
2&r&1
\end{array}\right]​1−32​pqr​041​​, find the value of rrr.

Let A=[214103421]A=\begin{bmatrix}2&1&4\\1&0&3\\4&2&1\end{bmatrix}A=​214​102​431​​. Find A−1A^{-1}A−1 if possible.

Find the inverse of [021255121]\begin{bmatrix}
0 &2& 1\\
2 &5 &5\\
1 &2 &1
\end{bmatrix}​021​252​151​​

The following matrix A depends on a parameter p. Find all the values of the parameter p for which the
 corresponding matrix is not invertible.
A=[123410002000300040010𝑝𝑝+1𝑝+20−𝑝−2𝑝+1−3𝑝+2]A=\begin{bmatrix}
1&2&3&4\\
1000&2000&3000&4001\\
0&𝑝&𝑝 + 1&𝑝 + 2\\
0&−𝑝&−2𝑝 + 1&−3𝑝 + 2

\end{bmatrix}A=​1100000​22000p−p​33000p+1−2p+1​44001p+2−3p+2​​

For what value or values of a (if any) is the matrix below invertible?
[1210350𝑎1]\begin{bmatrix}
1& 2& 1\\
0 &3& 5\\
0 &𝑎 &1
\end{bmatrix}​100​23a​151​​

The following matrix A depends on a parameter p. Find all the values of the parameter 
 p for which the corresponding matrix is not invertible.
A=[123410002000300040010𝑝𝑝+1𝑝+20−𝑝−2𝑝+1−3𝑝+2]A=\begin{bmatrix}
1&2&3&4\\
1000&2000&3000&4001\\
0&𝑝&𝑝 + 1&𝑝 + 2\\
0&−𝑝&−2𝑝 + 1&−3𝑝 + 2

\end{bmatrix}A=​1100000​22000p−p​33000p+1−2p+1​44001p+2−3p+2​​

Practice Question 3: Matrix Inverse

Practice Question: Matrix Inverse
For what value(s) of k will the matrix[1210150k1]\begin{bmatrix}
1&2&1\\
0&1&5\\
0&k&1
\end{bmatrix}​100​21k​151​​ be invertible?

Find all the values of k for which the matrix A has an inverse
A=[210102102k]A=\begin{bmatrix}
2&1&0\\
1&0&2\\
1&0&2k
\end{bmatrix}A=​211​100​022k​​

Practice: Method of Inverse

Practice Question: Method of Inverse
Given the matrix A=[2130101−12]A=\left[\begin{array}{rrr}
2&1&3\\
0&1&0\\
1&-1&2
\end{array}\right]A=​201​11−1​302​​,

Find all the values of 𝑘 such that the matrix 𝐴=[33−610−k2]has an inverse.\text{Find all the values of 𝑘 such that the matrix 𝐴}=\begin{bmatrix}
3&3\\
-6&10-k^2
\end{bmatrix}\text{has an inverse.}Find all the values of k such that the matrix A=[3−6​310−k2​]has an inverse.

Matrix Inverse

If [1ax2by0cz]\left[\begin{array}{c}
1&a&x\\
2&b&y\\
0&c&z
\end{array}\right]
​120​abc​xyz​​ is the inverse of the matrix[1p0−3q42r1]\left[\begin{array}{c}
1&p&0\\
-3&q&4\\
2&r&1
\end{array}\right]​1−32​pqr​041​​, find the value of rrr.

Practice: Adjoint Matrix

Let B=[1−102111000−11−21−2−1]B=\left[\begin{array}{rrrr}
1&-1&0&2\\
1&1&1&0\\
0&0&-1&1\\
-2&1&-2&-1
\end{array}\right]B=​110−2​−1101​01−1−2​201−1​​, and note that det(B)=−1\text{det}(B)=-1det(B)=−1
Use the determinant and the adjoint of BBB to find the entry in position (4,3)(4,3)(4,3) of the matrix B−1B^{-1}B−1

Find the inverse of the matrix [10221−13−215]\begin{bmatrix}
1&0&2\\
2&1&-1\\
3&-2&15
\end{bmatrix}​123​01−2​2−115​​.

Matrix Inversion Algorithm

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.

Find the inverse of the matrix A=[10221−13−215]A=\left[\begin{array}{rrr}
1&0&2\\
2&1&-1\\
3&-2&15
\end{array}\right]A=​123​01−2​2−115​​

Matrix operations

Suppose that  A=[1−103−20041]A=\begin{bmatrix}1 & -1 & 0\\3 & -2 & 0\\0 & 4 & 1\end{bmatrix}A=​130​−1−24​001​​and BBBis a 3×23 \times 23×2matrix. If the product AB=[34011−2]AB = \begin{bmatrix}3 & 4\\0 & 1\\1 & -2\end{bmatrix}AB=​301​41−2​​, find the (1,2)-entry of BBB(i.e. B12B_{12}B12​).

Invertible matrix

For what value(s) of aaawill the matrix [2−43a]\begin{bmatrix}2 & -4\\3 & a\end{bmatrix}[23​−4a​] be invertible?

Matrix Inverse

Practice: Matrix Inverse
If [1ax2by0cz]\left[\begin{array}{c}
1&a&x\\
2&b&y\\
0&c&z
\end{array}\right]
​120​abc​xyz​​ is the inverse of the matrix[1p0−3q42r1]\left[\begin{array}{c}
1&p&0\\
-3&q&4\\
2&r&1
\end{array}\right]​1−32​pqr​041​​, find the value of rrr.

Practice Question: Matrix Inverse

Practice Question: Matrix Inverse
If [1ax2by0cz]\left[\begin{array}{}
1&a&x\\
2&b&y\\
0&c&z
\end{array}\right]
​120​abc​xyz​​ is the inverse of the matrix[1p0−3q42r1]\left[\begin{array}{}
1&p&0\\
-3&q&4\\
2&r&1
\end{array}\right]​1−32​pqr​041​​, find the value of rrr.

Suppose that A=[1−1031−4]A=\begin{bmatrix}1 &-1&0\\3&1&-4\end{bmatrix}A=[13​−11​0−4​] and RRRis the reduced row echelon form of AAA. Find the invertible matrix UUUsuch that R=UAR=UAR=UA.

For what value(s) of aaawill the matrix [2−43a]\begin{bmatrix}2 & -4\\3 & a\end{bmatrix}[23​−4a​]be invertible?

Practice: Matrix Inverse

Find all values of k for which the matrix [2−6−1k2−22]\begin{bmatrix}
2&-6\\
-1&k^2-22
\end{bmatrix}[2−1​−6k2−22​] has no inverse.

Practice: Matrix Inverse

Practice Question: Matrix Inverse
Find all the values of kkk such that the matrix[210102102k]\begin{bmatrix}
2&1&0\\
1&0&2\\
1&0&2k
\end{bmatrix}​211​100​022k​​ has an inverse.

Practice: Matrix Inverse

Practice Question: Matrix Inverse
For what value(s) of k will the matrix[1210150k1]\begin{bmatrix}
1&2&1\\
0&1&5\\
0&k&1
\end{bmatrix}​100​21k​151​​ be invertible?

Practice: Matrix Inverse

Practice Question: Matrix Inverse
For what value(s) of kkk will the matrix[1210150k1]\left[\begin{array}{rrr}
1&2&1\\
0&1&5\\
0&k&1
\end{array}\right]​100​21k​151​​ be invertible?

Practice Question 1: Matrix Inverse

Practice Question: Matrix Inverse
Find all the values of kkk such that the matrix[210102102k]\begin{bmatrix}
2&1&0\\
1&0&2\\
1&0&2k
\end{bmatrix}​211​100​022k​​ has an inverse.

Practice: Matrix Inverse

Find all values of k so that the matrix [−2003102−1k]\begin{bmatrix}
-2&0&0\\
3&1&0\\
2&-1&k
\end{bmatrix}​−232​01−1​00k​​ has an inverse.

Practice: Matrix Inverse (~2016 Test2 #15)

Practice: Matrix Inverse
Find all the values of kkk such that the matrix[210102102k]\begin{bmatrix}
2&1&0\\
1&0&2\\
1&0&2k
\end{bmatrix}​211​100​022k​​ has an inverse.

Wize University Linear Algebra Textbook > Matrices

Matrix Inverse Algorithm

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Extra Practice

Practice: Method of Inverse

Practice: Method of Inverse

Practice: Matrix Inverse

Practice: Matrix Inverse

Practice: Matrix Inverse

Practice: Matrix Inverse

Practice: Matrix Inverse

Practice: Matrix Inverse

133 - FML 3 - 18.1W e.g. 10

Practice: Matrix Inverse

Practice: Matrix Inverse

Practice: Matrix Inverse

Matrix Inverse

Practice Question 3: Matrix Inverse

Practice: Method of Inverse

Matrix Inverse

Practice: Adjoint Matrix

Matrix Inversion Algorithm

Matrix operations

Invertible matrix

Matrix Inverse

Practice Question: Matrix Inverse

Practice: Matrix Inverse

Practice: Matrix Inverse

Practice: Matrix Inverse

Practice: Matrix Inverse

Practice Question 1: Matrix Inverse

Practice: Matrix Inverse

Practice: Matrix Inverse (~2016 Test2 #15)