College Algebra Lecture Notes
Section 6.3
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Section 6.3: Solving Linear Systems Using Matrix Equations
Big Idea: .
Big Skill: .
A. Multiplication and Identity Matrices
The identity matrix I is a square matrix that has 1’s on the diagonal and 0’s for all other entries.
Different order I are indicated with a subscript:
1 0 0 0 
1 0 0 
0 1 0 0 
1 0 




I2  
 , I 3  0 1 0  , I 4  0 0 1 0 
0
1


0 0 1 


0 0 0 1 
If A is an m  n matrix, then ImA = A and AIn = A.
If we want AI = IA = A, then A must be a square matrix.
Practice:
 1 2 3  1 0 0 
1. Compute  4 5 6  0 1 0 
7 8 9  0 0 1 
B. The Inverse of a Matrix
Given an n  n matrix A, if there exists another n  n matrix called A-1 such that AA-1 = A-1A = In,
then A-1 is the (multiplicative) inverse of matrix A.
The inverse of a 2  2 matrix can be found by solving for the entries that result in AA-1 = In.
OR
a b 
1  d b 
The inverse of a 2  2 matrix A  
is A1 
, provided that ad  bc  0 .

ad  bc  c a 
c d 
College Algebra Lecture Notes
Section 6.3
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Practice:
1 2 
2. Find the inverse of A  
 using both techniques, then show it is an inverse by
3 4 
computing AA-1.
College Algebra Lecture Notes
Section 6.3
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The inverse of a larger square matrix can be found by creating an augmented matrix with the
original matrix on the left hand side and the identity matrix on the right hand side, and then
doing row operations until the left hand side is the identity matrix. The right hand side will then
be the inverse (see page 614).
Practice:
1 2 
3. Find the inverse of A  
 using this new technique.
3 4 
College Algebra Lecture Notes
Section 6.3
2 1 0
4. Find the inverse of A   1 3 2  .


 3 1 2 
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College Algebra Lecture Notes
Section 6.3
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C. Solving Systems Using Matrix Equations
A system of linear equations can be written as a square matrix of variable coefficients times a
column matrix of variables equals a column matrix of constants.
Practice:
 x  4 y  z  10

5. Write as a matrix multiplication equation: 2 x  5 y  3z  7 .
8x 
y  2 z  11

Once we write the system like this, we can write the solution in a way that looks like a regular
linear equation, but we have to keep in mind that the quantities are matrices.
AX  B
A1  AX   A1 B
 A A X  A
1
1
B
IX  A1 B
X  A1 B
So, the solution to a system of equations is the inverse matrix left-multiplied times the matrix of
constants.
College Algebra Lecture Notes
Section 6.3
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Practice:
 x  4 y  z  10

6. Solve 2 x  5 y  3z  7 using a matrix equation solved by hand and with a
8x 
y  2 z  11

graphing calculator.
College Algebra Lecture Notes
Section 6.3
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D. Determinants and Singular Matrices
The determinant of a matrix is a single number that can be computed to tell whether the matrix
has an inverse or not. Just like the fact that the number zero does not have a multiplicative
inverse (i.e., 1/0 is undefined), a matrix with a zero determinant also has no inverse.
a b 
The determinant of a 2  2 matrix A  
 is:
c d 
det  A   A

a
b
c
d
 ad  bc
College Algebra Lecture Notes
Section 6.3
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