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

Solving Matrix Equations

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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

Matrix Algebra P2

Find the matrix A:
[2A‾T−3I‾]−1=[3111]\left[  2\underline{A}^T -3\underline{I} \right]^{-1}
=\begin{bmatrix}
3&1\\1&1
\end{bmatrix}[2A​T−3I​]−1=[31​11​]

Practice Question: Matrix Operations

Practice Question: Matrix Operations
Find the matrix A if (AT+[1−221])T=3A−2[3−101]\left(A^T+\begin{bmatrix}
1&-2\\
2&1
\end{bmatrix}\right)^T=3A-2\begin{bmatrix}
3&-1\\
0&1
\end{bmatrix}(AT+[12​−21​])T=3A−2[30​−11​].