Fundamentals of Matrix Algebra, 2011a

2.5 Solving Matrix Equaons AX = B
We can easily check to see if our answer is correct by mulplying AX. .
. Example 52 Solve the matrix equaon AX = B where
⎡
A = ⎣ 1 0 2 ⎤
⎡
0 −1 −2 ⎦ and B = ⎣ −1 2 ⎤
2 −6 ⎦ .
2 −1 0
2 −4
S
To solve, let’s again form the augmented matrix
[ A B
]
,
put it into reduced row echelon form, and interpret the result.
⎡
⎣ 1 0 2 −1 2 ⎤
0 −1 −2 2 −6 ⎦
2 −1 0 2 −4
−→ rref
⎡
⎣ 1 0 0 1 0 ⎤
0 1 0 0 4 ⎦
0 0 1 −1 1
We see from this that
.
⎡
X = ⎣ 1 0 ⎤
0 4 ⎦ .
−1 1
Why does this work? To see the answer, let’s define five matrices.
A =
[ ] [ ]
1 2 1
, ⃗u = , ⃗v =
3 4 1
[ ] [ ]
−1 5
, ⃗w =
1 6
and X =
[ 1 −1
] 5
1 1 6
Noce that ⃗u, ⃗v and ⃗w are the first, second and third columns of X, respecvely.
Now consider this list of matrix products: A⃗u, A⃗v, A⃗w and AX.
[ ] [ ] 1 2 1
A⃗u =
3 4 1
[ ] 3
=
7
[ ] [ ]
1 2 −1
A⃗v =
3 4 1
[ ] 1
=
1
[ ] [ ] 1 2 5
A⃗w =
3 4 6
[ ] 17
=
39
[ ] [ ]
1 2 1 −1 5
AX =
3 4 1 1 6
[ ] 3 1 17
=
7 1 39
So again note that the columns of X are ⃗u,⃗v and ⃗w; that is, we can write
X = [ ⃗u ⃗v ⃗w ] . 99