Chapter Ten — Applications of Matrix Algebra
The Humongous Book of Algebra Problems
233
Apply the row operation to make n
22
= 1
Apply the row operation to make n
32
= 0.
Apply the row operation to make n
33
= 1.
Apply the row operation to make n
23
= 0.
Chapter Ten — Applications of Matrix Algebra
The Humongous Book of Algebra Problems
234
Apply the row operation to make n
12
= 0 and, therefore, express
N in reduced-row echelon form.
The inverse matrix N
–1
appears right of the dotted line.
Chapter Ten — Applications of Matrix Algebra
The Humongous Book of Algebra Problems
235
Note: Problems 10.39–10.41 refer to the system of equations defined below.
10.39 Express the system as a matrix equation of the form .
Construct matrix A using the coefficients of the system and use the constants of
the system to generate matrix B.
Substitute A and B into the equation .
Note: Problems 10.39–10.41 refer to matrix A, matrix B, and the system of equations defined
in Problem 10.39.
10.40 Calculate A
–1
, the inverse of matrix A.
Construct the augmented matrix and express matrix A in reduced-row
echelon form.
Begin with the row operation to make a
11
= 1.
Apply the row operation to make a
21
= 0.
Apply the row operation to make a
12
= 0.
This matrix
equation is just another
way to write the
system of equations.
You could
use the row
operation
instead, but that’s a
lot more work than just
swapping the rows. Make
sure to swap the rows
in the augmented part
of the matrix across
the dotted line as
well.
Chapter Ten — Applications of Matrix Algebra
The Humongous Book of Algebra Problems
236
The inverse matrix, A
–1
, appears right of the dotted line in the augmented
matrix.
Note: Problems 10.39–10.41 refer to matrix A, matrix B, and the system of equations defined
in Problem 10.39.
10.41 Solve the matrix equation generated in Problem 10.39.
According to Problem 10.39, the system of equations can be expressed as the
following matrix equation.
To solve this equation, isolate the column matrix left of the equal sign.
To eliminate matrix , multiply both sides of the equation by the
inverse matrix calculated in Problem 10.40: .
Notice that
matrix A and
matrix A
–1
are the
same! Some matrices
are their own
inverses.
Multiplying a
matrix by
an identity
matrix (like I
2
) is
basically the same
as multiplying a real
number by 1. You’d write
1(3) as 3 and just drop
the 1 because there’s
no need for it. Same
thing goes with the
identity matrix
here.
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