Fundamentals of Matrix Algebra, 2011a

2.4 Vector Soluons to Linear Systems
Since A⃗x is equal to ⃗b, we have
⎡
⎣ x ⎤
1 + x 2 + x 3
x 1 − x 2 + 2x 3
⎦ =
2x 1 + x 3
⎡
⎣ 2
Knowing that two vectors are equal only when their corresponding entries are equal,
we know
x 1 + x 2 + x 3 = 2
x 1 − x 2 + 2x 3 = −3
2x 1 + x 3 = 1.
This should look familiar; it is a system of linear equaons! Given the matrix-vector
equaon A⃗x = ⃗b, we can recognize A as the coefficient matrix from a linear system
and ⃗b as the vector of the constants from the linear system. To solve a matrix–vector
equaon (and the corresponding linear system), we simply augment the matrix A with
the vector ⃗b, put this matrix into reduced row echelon form, and interpret the results.
We convert the above linear system into an augmented matrix and find the reduced
row echelon form:
⎡
⎣ 1 1 1 2 ⎤ ⎡
⎤
1 −1 2 −3
2 0 1 1
⎦
−→ rref
This tells us that x 1 = 1, x 2 = 2 and x 3 = −1, so
⎡ ⎤
⃗x =
⎣ 1 2 ⎦ .
−1
−3
1
⎤
⎦ .
⎣ 1 0 0 1
0 1 0 2
0 0 1 −1
We should check our work; mulply out A⃗x and verify that we indeed get ⃗b:
⎡
1 1
⎤ ⎡
1 1
⎤
⎡
2
⎤
⎣ 1 −1 2 ⎦ ⎣ 2 ⎦ does equal ⎣ −3 ⎦ .
2 0 1 −1
1
We should pracce.
. Example 43 .Solve the equaon A⃗x = ⃗b for ⃗x where
⎡
A = ⎣ 1 2 3 ⎤ ⎡
−1 2 1 ⎦ and ⎣ 5 ⎤
−1 ⎦ .
1 1 0
2
⎦ .
S The soluon is rather straighorward, even though we did a lot of
work before to find the answer. Form the augmented matrix [ A ⃗b ] and interpret its
reduced row echelon form.
⎡
⎤ ⎡
⎤
1 2 3 5
⎣ −1 2 1 −1 ⎦
−→ 1 0 0 2
rref ⎣ 0 1 0 0 ⎦
1 1 0 2
0 0 1 1
81