http://www.ck12.org Chapter 8. Systems and Matrices
x+ 2 y+ 3 z= 96
x+ 0 y+z= 36
0 x+ 2 y−z=− 12By writing this system as a matrix equation you get:
1 2 3
1 0 1
0 2 − 1
·
x
y
z
=
96
36
− 12
A·
x
y
z
=
96
36
− 12
If this were a normal linear equation where you had a constant times the variable equals a constant, you would
multiply both sides by the multiplicative inverse of the coefficient. Do the same in this case.
A−^1 ·A·
x
y
z
=A−^1 ·
96
36
− 12
x
y
z
=A−^1 ·
96
36
− 12
All that is left is for you to perform the matrix multiplication to get the solution. See Example B.
Example A
Show the steps for finding the inverse matrixAfrom the guidance section.
Solution:
1 2 3
1 0 1
0 2 − 1
∣∣
∣∣
∣∣
1 0 0
0 1 0
0 0 1
→
→ −I →
→
1 2 3
0 − 2 − 2
0 2 − 1
∣∣
∣∣
∣∣
1 0 0
−1 1 0
0 0 1
→
→
→ +II →
1 2 3
0 − 2 − 2
0 0 − 3
∣∣
∣∣
∣∣
1 0 0
−1 1 0
−1 1 1
→
→ ÷(− 2 ) →
→ ÷(− 3 ) →
1 2 3
0 1 1
0 0 1
∣∣
∣∣
∣∣
1 0 0
(^12) − (^120)
(^13) − (^13) − (^13)
→ − 3 III →
→ −III →
→
1 2 0
0 1 0
0 0 1
∣∣
∣∣
∣∣
01 1 1
61 −^1613
3 −^13 −^13
→ − 2 II →
→
→
1 0 0
0 1 0
0 0 1
∣∣
∣∣
∣∣
− 113 43 13
61 −^1613
3 −^13 −^13
The matrix on the right is the inverse matrixA−^1.
A−^1 =
−^134313
(^16) − (^1613)
(^13) − (^13) − (^13)
Example B
Solve the following system of equations using inverse matrices.