Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices:
XX
is the matrix representing the variables of the system, and
BB
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
To solve a system of linear equations using an inverse matrix, let
AA
be the coefficient matrix, let
XX
be the variable matrix, and let
BB
be the constant matrix. Thus, we want to solve a system
AX=BAX=B
. For example, look at the following system of equations.
a1x+b1y=c1a2x+b2y=c2\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=[a 1b1a2b2]A=\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right]
The variable matrix is
X=[xy] X=\left[\begin{array}{c}x\\ y\end{array}\right]
And the constant matrix is
B=[c1 c2]B=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right]
Then
AX=BAX=B
looks like
[a1b1a2b2] [xy ]=[c1c2]\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right]
Recall the discussion earlier in this section regarding multiplying a real number by its inverse,
(2−1) 2=(12)2=1\left({2}^{-1}\right)2=\left(\frac{1}{2}\right)2=1
. To solve a single linear equation
ax=bax=b
for
xx
, we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of
a a
. Thus,
ax=b (1a)ax=(1a)b(a−1 )ax=( a−1)b[(a−1)a]x=(a−1) b 1x=(a−1)b x=(a−1)b\begin{array}{c}\text{ }ax=b\\ \text{ }\left(\frac{1}{a}\right)ax=\left(\frac{1}{a}\right)b\\ \left({a}^{-1}\text{ }\right)ax=\left({a}^{-1}\right)b\\ \left[\left({a}^{-1}\right)a\right]x=\left({a}^{-1}\right)b\\ \text{ }1x=\left({a}^{-1}\right)b\\ \text{ }x=\left({a}^{-1}\right)b\end{array}
The only difference between a
solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable.
We will investigate this idea in detail, but it is helpful to begin with a
2×22\times 2
system and then move on to a
3×33\times 3
system.
A General Note: Solving a System of Equations Using the Inverse of a Matrix
Given a system of equations, write the coefficient matrix
AA
, the variable matrix
XX
, and the constant matrix
BB
. Then
Multiply both sides by the inverse of
AA
to obtain the solution.
(A−1)AX=(A−1)B [(A−1)A]X=(A−1)BIX=(A−1)BX=(A−1)B \begin{array}{r}\qquad \left({A}^{-1}\right)AX=\left({A}^{-1}\right)B\\ \qquad \left[\left({A}^{-1}\right)A\right]X=\left({A}^{-1}\right)B\\ \qquad IX=\left({A}^{-1}\right)B\\ \qquad X=\left({A}^{-1}\right)B\end{array}
Q & A
If the coefficient matrix does not have an inverse, does that mean the system has no solution?
No, if the coefficient matrix
is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions.
Example 7: Solving a 2 × 2 System Using the Inverse of a Matrix
Solve the given system of equations using the inverse of a matrix.
3x+8y=54x+11y=7\begin{array}{r}\qquad 3x+8y=5\\ \qquad 4x+11y=7\end{array}
Solution
Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.
A=[38 411],X=[x y],B=[57]A=\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],B=\left[\begin{array}{c}5\\ 7\end{array}\right]
Then
[38411] [xy]=[57]\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}5\\ 7\end{array}\right]
First, we need to calculate
A−1{A}^{-1}
. Using the formula to calculate the inverse of a 2 by 2 matrix, we have:
A−1=1ad−bc[d−b−ca] =13(11)−8(4)[11−8−43] =11 [11−8−43]\begin{array}{l}{A}^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]\qquad \\ \text{ }=\frac{1}{3\left(11\right)-8\left(4\right)}\left[\begin{array}{cc}11& -8\\ -4& 3\end{array}\right]\qquad \\ \text{ }=\frac{1}{1}\left[\begin{array}{cc}11& -8\\ -4& 3\end{array}\right]\qquad \end{array}
So,
A−1=[11−8 −4 3]{A}^{-1}=\left[\begin{array}{cc}11& -8\\ -4& \text{ }\text{ }3\end{array}\right]
Now we are ready to solve. Multiply both sides of the equation by
A−1{A}^{-1}
.
(A−1)AX=(A−1)B[11−8−43 ] [38411 ] [xy]=[11−8−43 ] [57][1001] [xy]=[ 11(5)+(−8)7−4(5)+3(7) ][xy]=[−11]\begin{array}{l}\left({A}^{-1}\right)AX=\left({A}^{-1}\right)B\qquad \\ \left[\begin{array}{rr}\qquad 11& \qquad -8\\ \qquad -4& \qquad 3\end{array}\right]\text{ }\left[\begin{array}{cc}3& 8\\ 4& 11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{rr}\qquad 11& \qquad -8\\ \qquad -4& \qquad 3\end{array}\right]\text{ }\left[\begin{array}{c}5\\ 7\end{array}\right]\qquad \\ \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{r}\qquad 11\left(5\right)+\left(-8\right)7\\ \qquad -4\left(5\right)+3\left(7\right)\end{array}\right]\qquad \\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{r}\qquad -1\\ \qquad 1\end{array}\right]\qquad \end{array}
The solution is
(−1,1)\left(-1,1\right)
.
Q & A
Can we solve for
XX
by finding the productBA−1?B{A}^{-1}?
No, recall that matrix multiplication is not commutative, so
A−1 B≠BA−1{A}^{-1}B\ne B{A}^{-1}
. Consider our steps for solving the matrix equation.(A−1)AX=(A−1)B[(A −1)A]X=(A−1)BIX=(A−1) BX=(A−1)B\begin{array}{r}\qquad \left({A}^{-1}\right)AX=\left({A}^{-1}\right)B\\ \qquad \left[\left({A}^{-1}\right)A\right]X=\left({A}^{-1}\right)B\\ \qquad IX=\left({A}^{-1}\right)B\\ \qquad X=\left({A}^{-1}\right)B\end{array}
Notice in the first step we multiplied both sides of the equation by
A−1{A}^{-1}
, but theA−1{A}^{-1}
was to the left ofAA
on the left side and to the left ofBB
on the right side. Because matrix multiplication is not commutative, order matters.Example 8: Solving a 3 × 3 System Using the Inverse of a Matrix
Solve the following system using the inverse of a matrix.
5x+15y+56z=35−4x−11y−41z=−26 −x−3y−11z=−7\begin{array}{r}\qquad 5x+15y+56z=35\\ \qquad -4x - 11y - 41z=-26\\ \qquad -x - 3y - 11z=-7\end{array}
Solution
Write the equation
AX=BAX=B
.
[51556−4−11−41 −1−3−11] [xyz]=[35 −26−7]\left[\begin{array}{ccc}5& 15& 56\\ -4& -11& -41\\ -1& -3& -11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{r}\qquad 35\\ \qquad -26\\ \qquad -7\end{array}\right]
First, we will find the inverse of
AA
by augmenting with the identity.
[51556 −4−11−41−1 −3−11∣100010001]\left[\begin{array}{rrr}\qquad 5& \qquad 15& \qquad 56\\ \qquad -4& \qquad -11& \qquad -41\\ \qquad -1& \qquad -3& \qquad -11\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]
Multiply row 1 by
15\frac{1}{5}
.
[13565−4−11−41 −1−3−11∣1500010001]\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ -4& -11& -41\\ -1& -3& -11\end{array}|\begin{array}{ccc}\frac{1}{5}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]
Multiply row 1 by 4 and add to row 2.
[1356 501195−1−3 −11∣1500 4510001 ]\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ 0& 1& \frac{19}{5}\\ -1& -3& -11\end{array}|\begin{array}{ccc}\frac{1}{5}& 0& 0\\ \frac{4}{5}& 1& 0\\ 0& 0& 1\end{array}\right]
Add row 1 to row 3.
[1356501 1950015∣ 15004510 1501]\left[\begin{array}{ccc}1& 3& \frac{56}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& \frac{1}{5}\end{array}|\begin{array}{ccc}\frac{1}{5}& 0& 0\\ \frac{4}{5}& 1& 0\\ \frac{1}{5}& 0& 1\end{array}\right]
Multiply row 2 by −3 and add to row 1.
[10−15011950 015∣−115 −30451015 01]\left[\begin{array}{ccc}1& 0& -\frac{1}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& \frac{1}{5}\end{array}|\begin{array}{ccc}-\frac{11}{5}& -3& 0\\ \frac{4}{5}& 1& 0\\ \frac{1}{5}& 0& 1\end{array}\right]
Multiply row 3 by 5.
[10−15 01195001∣−115−30 4510105]\left[\begin{array}{ccc}1& 0& -\frac{1}{5}\\ 0& 1& \frac{19}{5}\\ 0& 0& 1\end{array}|\begin{array}{ccc}-\frac{11}{5}& -3& 0\\ \frac{4}{5}& 1& 0\\ 1& 0& 5\end{array}\right]
Multiply row 3 by
15 \frac{1}{5}
and add to row 1.
[10001195001∣−2−314510105]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& \frac{19}{5}\\ 0& 0& 1\end{array}|\begin{array}{ccc}-2& -3& 1\\ \frac{4}{5}& 1& 0\\ 1& 0& 5\end{array}\right]
Multiply row 3 by
−195-\frac{19}{5}
and add to row 2.
[100010001 ∣−2−31− 31−19105]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}-2& -3& 1\\ -3& 1& -19\\ 1& 0& 5\end{array}\right]
So,
A−1=[−2−31−31 −19105]{A}^{-1}=\left[\begin{array}{ccc}-2& -3& 1\\ -3& 1& -19\\ 1& 0& 5\end{array}\right]
Multiply both sides of the equation by
A−1{A}^{-1}
. We want
A−1AX=A−1B:{A}^{-1}AX={A}^{-1}B:
[ −2−31−3 1−1910 5] [5 1556−4−11−41−1−3−11] [xyz]=[−2−31 −31−191 05] [35−26−7]\left[\begin{array}{rrr}\qquad -2& \qquad -3& \qquad 1\\ \qquad -3& \qquad 1& \qquad -19\\ \qquad 1& \qquad 0& \qquad 5\end{array}\right]\text{ }\left[\begin{array}{rrr}\qquad 5& \qquad 15& \qquad 56\\ \qquad -4& \qquad -11& \qquad -41\\ \qquad -1& \qquad -3& \qquad -11\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{rrr}\qquad -2& \qquad -3& \qquad 1\\ \qquad -3& \qquad 1& \qquad -19\\ \qquad 1& \qquad 0& \qquad 5\end{array}\right]\text{ }\left[\begin{array}{r}\qquad 35\\ \qquad -26\\ \qquad -7\end{array}\right]
Thus,
A−1B=[−70+78−7 −105−26+13335+0−35]=[ 120]{A}^{-1}B=\left[\begin{array}{r}\qquad -70+78 - 7\\ \qquad -105 - 26+133\\ \qquad 35+0 - 35\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 0\end{array}\right]
The solution is
(1,2,0)\left(1,2,0\right)
.
Try It 4
Solve the system using the inverse of the coefficient matrix.
2x−17y+11z=0 −x +11y−7z=8 3y−2z=−2\begin{array}{l}\text{ }2x - 17y+11z=0\qquad \\ \text{ }-x+11y - 7z=8\qquad \\ \text{ }3y - 2z=-2\qquad \end{array}
Solution
How To: Given a system of equations, solve with matrix inverses using a calculator.
- Save the coefficient matrix and the constant matrix as matrix variables
[A]\left[A\right]
and[B]\left[B\right]
. - Enter the multiplication into the calculator, calling up each matrix variable as needed.
- If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message.
Example 9: Using a Calculator to Solve a System of Equations with Matrix Inverses
Solve the system of equations with matrix inverses using a calculator
2x+3y+z=323x+3y+z=−272x+ 4y+z=−2\begin{array}{l}2x+3y+z=32\qquad \\ 3x+3y+z=-27\qquad \\ 2x+4y+z=-2\qquad \end{array}
Solution
On the matrix page of the calculator, enter the coefficient matrix as the matrix variable
[A]\left[A\right]
, and enter the constant matrix as the matrix variable
[B]\left[B\right]
.
[A]=[231331241], [B]=[32−27 −2]\left[A\right]=\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right],\text{ }\left[B\right]=\left[\begin{array}{c}32\\ -27\\ -2\end{array}\right]
On the home screen of the calculator, type in the multiplication to solve for
XX
, calling up each matrix variable as needed.
[A]−1× [B]{\left[A\right]}^{-1}\times \left[B\right]
Evaluate the expression.
[− 59−34252]\left[\begin{array}{c}-59\\ -34\\ 252\end{array}\right]