Wize University Linear Algebra Textbook > Matrices

Solving Matrix Equations

Invertible Matrix TheoremPrevious Section

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Solving Matrix Equations
Matrix Multiplication
You can multiply both sides of a matrix equation by a matrix:

Watch Out!
Matrices are not commutative in general! (AB≠BAAB \ne BAAB=BA)
That means we have to be consistent when multiplying: left multiplication vs. right multiplication.
Left multiplication by a matrix C\colorOne{\bm C}C:
  Ax⃗=b⃗  ⟹  CAx⃗=Cb⃗A\vec x=\vec b 
\quad\implies\quad
\colorOne{\mathbf{C}}A\vec x
=\colorOne{\mathbf{C}}\vec bAx=b⟹CAx=Cb
Right multiplication by a matrix C\colorOne{\bm C}C :
Ax⃗=b⃗  ⟹  Ax⃗C=b⃗CA\vec x=\vec b 
\quad\implies\quad
A\vec x\colorOne{\mathbf{C}}=\vec b\colorOne{\mathbf{C}}Ax=b⟹AxC=bC
PAGE BREAK
Solving Linear Systems Using Inverse
Write the matrix equation for the linear system: Ax⃗=b⃗A \vec x = \vec bAx=b.
Wize Concept
Recall:
AAA is the coefficient matrix
x⃗\vec xx is the solution vector
b⃗\vec bb is the constant vector (the augmented column of [ A ∣ b⃗ ][\ A\ |\ \vec b\ ][ A ∣ b ])

Rearrange to solve for x⃗\vec xx, the solution vector, by cancelling AAA on the LHS:

Ax⃗=b⃗A−1A⏟=I x⃗=A−1b⃗Ix⃗=A−1b⃗\begin{aligned}
A \vec x &= \vec b\\[0.5em]
\underbrace{\colorOne{A^{-1}}A}_{=I} \ \vec x &= \colorOne{A^{-1}}\vec b\\[0.5em]
I \vec x &= A^{-1}\vec b
\end{aligned}Ax=IA−1A​​ xIx​=b=A−1b=A−1b​

x⃗=A−1b⃗\boxed{\quad
\vec x = A^{-1}\vec b
\quad}x=A−1b​
Steps
Write the system of linear equations as a matrix equation 𝐴x⃗=b⃗𝐴\vec{x} = \vec bAx=b.
Find A−1A^{-1}A−1.
The solution is x⃗=A−1b\vec x=A^{-1}bx=A−1b.

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Example: Solving Linear Systems Using Inverse
A=[121011210],A−1=[−1112−2−1−231],b⃗=[0−34]A=
\left[\begin{array}{rrr}
1&2&1\\
0&1&1\\
2&1&0
\end{array}\right]

,\quad

A^{-1}=
\left[\begin{array}{rrr}
-1&1&1\\
2&-2&-1\\
-2&3&1
\end{array}\right]

,\quad

\vec b=
\left[\begin{array}{r}
0\\
-3\\
4
\end{array}\right]A=​102​211​110​​,A−1=​−12−2​1−23​1−11​​,b=​0−34​​
Find the solution(s) to the linear system Ax⃗=b⃗A \vec x = \vec bAx=b.

The linear system we want to solve can be written:
x+2y+z=0y+z=−32x+y=4\begin{array}{rcr}
x+2y+z&=&0\\[0.5em]
y+z&=&-3\\[0.5em]
2x+y&=&4
\end{array}x+2y+zy+z2x+y​===​0−34​
In matrix form, we have:
Ax⃗=b⃗  ⟹  [121011210][xyz]=[0−34]A\vec x = \vec b 
\quad\implies\quad 

\left[\begin{array}{rrr}
1&2&1\\
0&1&1\\
2&1&0
\end{array}\right]
\left[\begin{array}{c}
x\\
y\\
z
\end{array}\right]
=
\left[\begin{array}{r}
0\\
-3\\
4
\end{array}\right]Ax=b⟹​102​211​110​​​xyz​​=​0−34​​
We have been given the inverse of AAA, so AAA must be invertible.
Wize Concept
Recall: since AAA is invertible, the linear system Ax⃗=b⃗A\vec x = \vec bAx=b must have a unique solution.

Ax⃗=b⃗  ⟹  x⃗=A−1b⃗  ⟹  [xyz]=[−1112−2−1−231][0−34]  ⟹  [xyz]=[12−5]\begin{array}{rrcll}
A \vec x = \vec b 

&\quad \implies \quad&

\vec x 
&=& A^{-1}\vec b\\[1em]

&\quad \implies \quad&

\left[\begin{array}{r}
x\\
y\\
z
\end{array}\right]
&=&
\left[\begin{array}{rrr}
-1&1&1\\
2&-2&-1\\
-2&3&1
\end{array}\right]
\left[\begin{array}{r}
0\\
-3\\
4
\end{array}\right]\\[2em]

&\quad \implies \quad&

\left[\begin{array}{r}
x\\
y\\
z
\end{array}\right]
&=&
\boxed{
\left[\begin{array}{r}
1\\
2\\
-5
\end{array}\right]
}

\end{array}Ax=b​⟹⟹⟹​x​xyz​​​xyz​​​===​A−1b​−12−2​1−23​1−11​​​0−34​​​12−5​​​​

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Example: Solving Matrix Equations
Find the matrix CCC such that:
(I−2C[24−1−3])−1=[3111]\left(
I-2C
\left[\begin{array}{rr}
2&4\\
-1&-3
\end{array}\right]
\right)^{-1}
=
\left[\begin{array}{rr}
3&1\\
1&1
\end{array}\right](I−2C[2−1​4−3​])−1=[31​11​]

Writing the equation symbolically:
(I−2CA)−1=B\left(I-2CA\right)^{-1}
=B(I−2CA)−1=B,  where  A=[24−1−3],B=[3111]A=
\left[\begin{array}{rr}
2&4\\
-1&-3
\end{array}\right]
,\quad
B=
\left[\begin{array}{rr}
3&1\\
1&1
\end{array}\right]A=[2−1​4−3​],B=[31​11​]
Take the inverse of both sides to cancel the inverse on the LHS:
I−2CA=B−1I-2CA
=B^{-1}I−2CA=B−1
We are trying to isolate CCC, so subtract III from both sides:
−2CA=B−1−I-2CA=B^{-1}-I−2CA=B−1−I
Multiply both sides by the scalar −12-\dfrac{1}{2}−21​:
CA=−12(B−1−I)CA = -\dfrac{1}{2}(B^{-1}-I)CA=−21​(B−1−I)
Right multiply both sides by A−1A^{-1}A−1 to cancel AAA on the LHS:
CAA−1=−12(B−1−I)A−1  ⟹  CI=−12(B−1−I)A−1  ⟹    C=−12(B−1−I)A−1\begin{aligned}
CA\colorOne{A^{-1}} &= -\dfrac{1}{2}(B^{-1}-I) \colorOne{A^{-1}}\\[1em]
\implies \quad C\colorOne{I} &= -\dfrac{1}{2}(B^{-1}-I) \colorOne{A^{-1}}\\[1em]
\implies \quad \ \  C &= -\dfrac{1}{2}(B^{-1}-I)\colorOne{A^{-1}}
\end{aligned}CAA−1⟹CI⟹  C​=−21​(B−1−I)A−1=−21​(B−1−I)A−1=−21​(B−1−I)A−1​
We can now compute CCC (use the formula for 2×22 \times 22×2 matrices to find B−1, A−1B^{-1}, \ A^{-1}B−1, A−1):
B−1=[3111]−1=12[1−1−13]=[12−12−1232]B^{-1}
=
\left[\begin{array}{rr}3&1\\1&1\end{array}\right]^{-1}
=
\dfrac{1}{2}\left[\begin{array}{rr}1&-1\\-1&3\end{array}\right]
=
\left[\begin{array}{rr}\frac{1}{2}&-\frac{1}{2}\\[0.5em]-\frac{1}{2}&\frac{3}{2}\end{array}\right]B−1=[31​11​]−1=21​[1−1​−13​]=[21​−21​​−21​23​​]
−12(B−1−I) = −12([12−12−1232]−[1001]) = −12[−12−12−1212] = [141414−14]-\dfrac{1}{2}(B^{-1}-I)
\ =\ 
-\dfrac{1}{2}\left(\left[\begin{array}{rr}\frac{1}{2}&-\frac{1}{2}\\[0.5em]-\frac{1}{2}&\frac{3}{2}\end{array}\right]-\left[\begin{array}{rr}1&0\\[0.5em]0&1\end{array}\right]\right)
\ =\ 
-\dfrac{1}{2}\left[\begin{array}{rr}-\frac{1}{2}&-\frac{1}{2}\\[0.5em]-\frac{1}{2}&\frac{1}{2}\end{array}\right]
\ =\ 
\colorOne{
\left[\begin{array}{rr}\frac{1}{4}&\frac{1}{4}\\[0.5em]\frac{1}{4}&-\frac{1}{4}\end{array}\right]
}−21​(B−1−I) = −21​([21​−21​​−21​23​​]−[10​01​]) = −21​[−21​−21​​−21​21​​] = [41​41​​41​−41​​]
A−1=[24−1−3]−1=1−2[−3−412]=[322−12−1]A^{-1}=\left[\begin{array}{rr}2&4\\-1&-3\end{array}\right]^{-1}
=
\dfrac{1}{-2}\left[\begin{array}{rr}-3&-4\\1&2\end{array}\right]
=
\colorTwo{
\left[\begin{array}{rr}\frac{3}{2}&2\\[0.5em]-\frac{1}{2}&-1\end{array}\right]
}A−1=[2−1​4−3​]−1=−21​[−31​−42​]=[23​−21​​2−1​]
C = −12(B−1−I)A−1 = [141414−14][322−12−1] = [14141234]C
\ =\  \colorOne{-\dfrac{1}{2}(B^{-1}-I)} \colorTwo{A^{-1}}
\ =\ 
\colorOne{
\left[\begin{array}{rr}\frac{1}{4}&\frac{1}{4}\\[0.5em]\frac{1}{4}&-\frac{1}{4}\end{array}\right]
}
\colorTwo{
\left[\begin{array}{rr}\frac{3}{2}&2\\[0.5em]-\frac{1}{2}&-1\end{array}\right]
}
\ =\ 
\left[\begin{array}{rr}\frac{1}{4}&\frac{1}{4}\\[0.5em]\frac{1}{2}&\frac{3}{4}\end{array}\right]C = −21​(B−1−I)A−1 = [41​41​​41​−41​​][23​−21​​2−1​] = [41​21​​41​43​​]
∴ C=[14141234]\therefore \ 
\boxed{C=\left[\begin{array}{rr}\frac{1}{4}&\frac{1}{4}\\[0.5em]\frac{1}{2}&\frac{3}{4}\end{array}\right]}∴ C=[41​21​​41​43​​]​

Solve the following system of linear equations using the inverse of the coefficient matrix.
3x+4y=1−2x−y=0\begin{array}{rrr}
3x+4y&=&1\\
-2x-y&=&0
\end{array}3x+4y−2x−y​==​10​

⟨−15, 25⟩\left\lang -\dfrac{1}{5},\ \dfrac{2}{5}\right\rang⟨−51​, 52​⟩

⟨15, −25⟩\left\lang \dfrac{1}{5},\ -\dfrac{2}{5}\right\rang⟨51​, −52​⟩

⟨−15, −15⟩\left\lang-\dfrac{1}{5},\ -\dfrac{1}{5}\right\rang⟨−51​, −51​⟩

⟨25, 25⟩\left\lang\dfrac{2}{5},\ \dfrac{2}{5}\right\rang⟨52​, 52​⟩

None of the above

I don't know

Let [3−104]\begin{bmatrix}
3&-1\\
0&4
\end{bmatrix}[30​−14​] be the inverse of the matrix [abcd]\begin{bmatrix}
a&b\\c&d
\end{bmatrix}[ac​bd​].

Find the solution to the following system of linear equations:
ax+by=−2cx+dy=2\begin{array}{rrr}
ax+by&=&-2\\
cx+dy&=&2
\end{array}ax+bycx+dy​==​−22​

There are no solutions

There is a unique solution ⟨1,−1⟩ \lang 1,-1 \rang⟨1,−1⟩

There is a unique solution ⟨−8,8⟩ \lang -8,8 \rang⟨−8,8⟩

There is a unique solution ⟨2,−5⟩ \lang 2, -5 \rang⟨2,−5⟩

There are infinitely many solutions

I don't know

Solve for AAA in the following equation:
A[−122−5]=I2−[2−302]A
\left[
\begin{array}{rr}
-1&2\\
2&-5
\end{array}
\right]
=
I_2-
\left[
\begin{array}{rr}
2&-3\\
0&2
\end{array}
\right]A[−12​2−5​]=I2​−[20​−32​]
[Fill in the entries for matrix AAA]

I don't know

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​​,

133 - FML 3 - 18.1W e.g. 16

 Let   ⁣M‾=[k330(k−1)300(k−2)]\bcb{\boldsymbol{ \ul{M} = \begin{bmatrix} k & 3 & 3 \\ 0 & (k-1) & 3 \\ 0 & 0 & (k-2) \end{bmatrix}}}M​=​k00​3(k−1)0​33(k−2)​​, for what value(s) of k\bcb{\boldsymbol{ k}}k does the system   ⁣M‾x⃗=b⃗\bcb{\boldsymbol{ \ul{M} \vec{x} = \vec{b} }}M​x=b have a unique solution?
(a)  k=0,  1, 2(b)  k≠0, 1 ,2(c)  k≠0(d)  None of the above\text{(a)}\;\bcb{\boldsymbol{  k = 0, \, \, 1,\, 2}}\qquad\qquad\qquad
\text{(b)}\;\bcb{\boldsymbol{  k \ne 0,\, 1\, ,2}}\qquad\qquad\qquad
\text{(c)}\;\bcb{\boldsymbol{  k \ne 0}}\qquad\qquad\qquad
\text{(d)}\; \text{\bf{None of the above}}(a)k=0,1,2(b)k=0,1,2(c)k=0(d)None of the above
 ~ 

Practice: Solving Matrix Equations
Solve for AAA in the following equation:
A[−122−5]=I2−[2−302]A
\left[
\begin{array}{rr}
-1&2\\
2&-5
\end{array}
\right]
=
I_2-
\left[
\begin{array}{rr}
2&-3\\
0&2
\end{array}
\right]A[−12​2−5​]=I2​−[20​−32​]

Find the inverse of[10221−13−215]and use it to solve the system\text{Find the inverse of}\begin{bmatrix}
1&0&2\\
2&1&-1\\
3&-2&15
\end{bmatrix}\text{and use it to solve the system}Find the inverse of​123​01−2​2−115​​and use it to solve the system
𝑥+2𝑧=3𝑥+2𝑧=3x+2z=3
2𝑥+𝑦−𝑧=42𝑥+𝑦−𝑧=42x+y−z=4

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​​

Given the matrix 𝐴=[2130101−12].\text{Given the matrix 𝐴}=\begin{bmatrix}
2&1&3\\
0&1&0\\
1&-1&2\\
\end{bmatrix}.Given the matrix A=​201​11−1​302​​.
a) find the inverse of 𝐴. 
b) solve the system of linear equations          2𝑥+𝑦+3𝑧=32𝑥+𝑦+3𝑧=32x+y+3z=3                                                                  

Find the general solution the following system of equations using inverse of the coefficient matrix
2x+y+4z=−3x+3z=−14x+2y+z=2\begin{array}{l}2x+y+4z=-3\\x+3z=-1\\4x+2y+z=2\end{array}2x+y+4z=−3x+3z=−14x+2y+z=2​

Use the method of inverse to solve the system of linear equations x+2z=32x+y−z=43x−2y+15z=0\begin{array}{c}
x+2z&=&3\\
2x+y-z&=&4\\
3x-2y+15z&=&0
\end{array}x+2z2x+y−z3x−2y+15z​===​340​

Use the fact that the matrix
                                                 A=[41−420−1−3−14]A=\begin{bmatrix}
4&1&-4\\
2&0&-1\\
-3&-1&4
\end{bmatrix}A=​42−3​10−1​−4−14​​  
has inverse matrix

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 Operations

Find the values of aaa and bbb such that [a2−11][23]=[10b]T\left[\begin{array}{c}
a&2\\
-1&1
\end{array}\right]
\left[\begin{array}{c}
2\\
3
\end{array}\right]
=
\left[\begin{array}{c}
10&b
\end{array}\right]^T[a−1​21​][23​]=[10​b​]T

Practice: Matrix Operations

Solve for a, b, ca,\ b,\ ca, b, c in the following matrix equation:
[1a−12−2130−1][−12b]−[19−1]T=[−8c−2]\begin{bmatrix}
1 &a&-1\\
2&-2&1\\
3&0&-1
\end{bmatrix}

\begin{bmatrix}
-1\\2\\b
\end{bmatrix}-
\begin{bmatrix}
1&9&-1
\end{bmatrix}^T
=\begin{bmatrix}
-8\\c\\-2
\end{bmatrix}​123​a−20​−11−1​​​−12b​​−[1​9​−1​]T=​−8c−2​​

Practice: Matrix Inverse

Practice: Solving Matrix Equations
Solve for AAA in the following equation:
A[−122−5]=I2−[2−302]A
\left[
\begin{array}{rr}
-1&2\\
2&-5
\end{array}
\right]
=
I_2-
\left[
\begin{array}{rr}
2&-3\\
0&2
\end{array}
\right]A[−12​2−5​]=I2​−[20​−32​]

Practice Question: Using Matrix inverse

Practice Question: Using Matrix Inverse
Let A=[120−1]A=\left[\begin{array}{}
1&2\\0&-1
\end{array}\right]A=[10​2−1​] and suppose that 3AB−[1−120]=I23AB-

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

=I_23AB−[12​−10​]=I2​. 
Find the matrix BBB.

Solving Matrix Equations

Let [3−104]\begin{bmatrix}
3&-1\\
0&4
\end{bmatrix}[30​−14​] be the inverse of the matrix [abcd]\begin{bmatrix}
a&b\\c&d
\end{bmatrix}[ac​bd​].
Find the solution to the following system of linear equations:
ax+by=−2cx+dy=2\begin{array}{rrr}
ax+by&=&-2\\
cx+dy&=&2
\end{array}ax+bycx+dy​==​−22​

Solving Matrix Equations

Solve the following system of linear equations using the inverse of the coefficient matrix.
3x+4y=1−2x−y=0\begin{array}{rrr}
3x+4y&=&1\\
-2x-y&=&0
\end{array}3x+4y−2x−y​==​10​

Practice: Matrix Inverse

Solve for AAA in the following equation:
A[−122−5]=I2−[2−302]A\begin{bmatrix}
-1&2\\
2&-5
\end{bmatrix}
=
I_2-\begin{bmatrix}
2&-3\\
0&2
\end{bmatrix}A[−12​2−5​]=I2​−[20​−32​]

Inverse of a Matrix

e.g. In each case find the matrix A:
 ⁣(a)[2[11−23]−5A‾−1]T=(4A‾T)−1 ⁣;(b)[2A‾T−3I‾]−1=[3111]\!\begin{array}{rr}
(a) \left[2\begin{bmatrix}
1&1\\-2&3
\end{bmatrix}-5\underline{A}^{-1}\right]^T=\left( 4\underline{A}^T \right)^{-1}\!;&
\hspace{1cm}
(b) \left[  2\underline{A}^T -3\underline{I} \right]^{-1}
=\begin{bmatrix}
3&1\\1&1
\end{bmatrix}\end{array}(a)[2[1−2​13​]−5A​−1]T=(4A​T)−1;​(b)[2A​T−3I​]−1=[31​11​]​

Practice: Solving Matrix Equations

Find the matrix CCC such that:
(I−2C[24−1−3])−1=[3111]\left(I-2C\left[\begin{array}{rr}2&4\\-1&-3\end{array}\right]\right)^{-1}
=\left[\begin{array}{rr}3&1\\1&1\end{array}\right](I−2C[2−1​4−3​])−1=[31​11​]

Use the fact that the matrix
A=[41−420−1−3−14]A=\begin{bmatrix}
4&1&-4\\
2&0&-1\\
-3&-1&4
\end{bmatrix}A=​42−3​10−1​−4−14​​
has inverse matrix

Cramer's Rule

Find yyy given the system of linear equations:
[0−1142021−2][xyz]=[214]\left[
\begin{array}{rrr}
0&-1&1\\
4&2&0\\
2&1&-2
\end{array}
\right]

\begin{bmatrix}
x\\y\\z\\
\end{bmatrix}
=
\begin{bmatrix}
2\\1\\4\\
\end{bmatrix}​042​−121​10−2​​​xyz​​=​214​​

Wize University Linear Algebra Textbook > Matrices

Solving Matrix Equations

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133 - FML 3 - 18.1W e.g. 16

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