Fundamentals of Matrix Algebra, 2011a

Chapter 2
Matrix Arithmec
We know how to solve this; put the appropriate matrix into reduced row echelon form
and interpret the result.
[ ] [ ]
1 1 0 −→ 1 0 1
rref
2 1 1
0 1 −1
We read from this that
⃗x =
[ 1
−1
]
.
Wrien in a more general form, we found our soluon by forming the augmented
matrix [
A ⃗b ]
and interpreng its reduced row echelon form:
[
A ⃗b ] −→ rref
[ I ⃗x
]
Noce that when the reduced row echelon form of A is the identy matrix I we have
exactly one soluon. This, again, is the best case scenario.
We apply the same general technique to solving the matrix equaon AX = B for X.
We’ll assume that A is a square matrix (B need not be) and we’ll form the augmented
matrix [ A B
]
.
Pung this matrix into reduced row echelon form will give us X, much like we found ⃗x
before. [
A B
] −→ rref
[
I X
]
As long as the reduced row echelon form of A is the identy matrix, this technique
works great. Aer a few examples, we’ll discuss why this technique works, and we’ll
also talk just a lile bit about what happens when the reduced row echelon form of A
is not the identy matrix.
First, some examples.
. Example 51 .Solve the matrix equaon AX = B where
[ ]
[ ]
1 −1
−8 −13 1
A =
and B =
.
5 3
32 −17 21
S To solve AX = B for X, we form the proper augmented matrix, put
it into reduced row echelon form, and interpret the result.
[ ] [ ]
1 −1 −8 −13 1 −→ 1 0 1 −7 3
rref
5 3 32 −17 21
0 1 9 6 2
We read from the reduced row echelon form of the matrix that
[ ]
1 −7 3
X =
.
9 6 2
98