2. Learning Intention and Success
Criteria
Learning Intention: Students will understand how
the inverse of a matrix can be used to solve a matrix
equation using matrix multiplication.
Success Criteria: You can determine the unknown
matrix in a matrix equation, using matrix inverses
3. Prior Knowledge
Recall: For any non-singular square matrix A, we can
find the inverse, 𝐴−1
.
𝐴−1
has the properties that
𝐴 × 𝐴−1 = 𝐼
𝐴−1
× 𝐴 = 𝐼
Also, recall that for any matrix, B, and compatible
identity matrix, I.
𝐵 × 𝐼 = 𝐵
𝐼 × 𝐵 = 𝐵
4. Using the Inverse
Like we can solve the equation 2𝑥 = 8, by doing the
inverse of multiplying by 2(dividing by 2), we can
solve the matrix equations 𝐴𝑋 = 𝐵 by using the
inverse of A.
For matrices, A, B and X we can solve the matrix
equation
𝐴𝑋 = 𝐵
5. Using the Inverse
For matrices, A, B and X we can solve the matrix
equation
Multiply both
sides ON THE
FRONT, by 𝐴−1 to
“cancel out” the ASimplify to the
identity. In practice,
we usually skip this
step
Actually evaluate
𝐴−1 × 𝐵 to solve
for X
6. Example 1
Solve the matrix equation
2 −1
3 4
𝑋 =
7
−6
Note that since we have 2 × 2 𝑚 × 𝑛 = 2 × 1 , that X
must be a (2 × 1).
Step 1: Find the inverse of
2 −1
3 4
By hand or on the CAS you should get that
2 −1
3 4
−1
=
1
11
4 1
−3 2
7. Example 1 continued…
Step 2: Front (pre) multiply on both sides by the
inverse
Step 3: Simplify left side (we often skip this step)
2 −1
3 4
𝑋 =
7
−6
1
11
4 1
−3 2
2 −1
3 4
𝑋 =
1
11
4 1
−3 2
7
−6
1 0
0 1
𝑋 =
1
11
4 1
−3 2
7
−6
8. Example 1 continued…
Step 4: Simplify fully. This is usually done on the CAS.
1 0
0 1
𝑋 =
1
11
4 1
−3 2
7
−6
𝐼𝑋 =
1
11
22
−33
𝑋 =
2
−3
9. Example 2
Solve the matrix equation
−1 4
2 −8
𝑋 =
6
10
Step 1: Find the inverse
−1 4
2 −8
−1
=
1
−1 × −8 − 4 × 2
−8 −4
−2 −1
−1 4
2 −8
−1
=
1
0
−8 −4
−2 −1
The determinant is zero, so the inverse does not exist.
10. Example 2 continued
Answer:
−1 4
2 −8
has determinant zero, so is a singular
matrix. There is no unique solution to this equation.
This could mean that there is an infinite number of
solutions or that there are no possible solutions.
In Further, we don’t really need to know which it is
If you want to learn more about which it is see the next
slide.
11. Example 2 continued extension
−1 4
2 −8
𝑋 =
6
10
Let’s call
−1 4
2 −8
= 𝐴 and
6
10
= 𝐵
Row 2 of A is double and negative row 1 of A, since
−2 −1 4 = [2 −8]
If there is infinite solutions, then that should also be
true for the rows of B.
Since 10 ≠ 6 × −2 = −12, there are no solutions
You can also just try any value and see if it works, if it
works, then it would be infinite.