The document provides examples for solving matrix equations using inverses. It introduces the concept of pre-multiplying both sides of a matrix equation AX=B by the inverse of A to solve for X. Example 1 demonstrates this process, finding the inverse of the coefficient matrix and multiplying both sides to isolate X. Example 2 shows that the inverse does not exist when the coefficient matrix is singular, meaning there may be no solution or infinite solutions depending on whether the rows of B satisfy the relationship of the rows of the singular A.

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.