NAME _____________________________________________ DATE ____________________________ PERIOD _____________
6-3 Study Guide and Intervention
Solving Linear Systems Using Inverses and Cramer’s Rule
Use Inverse Matrices A square system has the same number of equations as variables. If a square matrix has an
inverse, the system has one unique solution.
Example : Use an inverse matrix to solve each system of equations, if possible.
a. 3x – 7y = –16 b. –2x + y + z = 0
–x + 2y = 8 x + 2z = 9
x – 2y – 9z = –31
Write the system in matrix form. Write the system in matrix form.
A ⋅ X = B A ⋅ X = B
𝑥 −2 1 1 𝑥 0
3 −7 −16
[ ] ⋅ [𝑦 ] = [ ] [ 1 0 2 ] ⋅ [ 𝑦 ] = [ 9]
−1 2 8
1 −2 −9 𝑧 −31
Use the formula for an inverse of a 2 × 2 matrix to find Use a graphing calculator to find 𝐴−1 .
the inverse 𝐴−1 .
1 𝑑 −𝑏]
𝐴−1 = 𝑎𝑑 − 𝑐𝑏 [
−𝑐 𝑎
1 2 7
= (2)(3) − (−7)(−1) [ ]
1 3
Multiply 𝐴−1 by B to solve the system. Multiply 𝐴−1 by B to solve the system.
X= 𝐴−1 ⋅ B X= 𝐴−1 ⋅ B
−2 −7 −16 −24 4 7 2 0 1
=[ ]⋅[ ]=[ ]
−1 −3 8 −8 = [ 11 17 5] ⋅ [ 9] = [−2]
So, the solution of the system is (–24, –8). −2 −3 −1 −3 4
So, the solution of the system is (1, –2, 4).
Exercises
Use an inverse matrix to solve each system of equations, if possible.
1. –2x + 5y = 24 2. x – y + 2z = 5
3x – y = –10 x – z = –4
3x + 2y + z = 0
3. 3x + y = 7 4. x + y – z = –5
–2x – 5y = 43 2x – 3y + 2z = 20
y + 4z = 18
Chapter 6 16 Glencoe Precalculus
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
6-3 Study Guide and Intervention (continued)
Solving Linear Systems Using Inverses and Cramer’s Rule
Use Cramer’s Rule Another method, known as Cramer’s Rule, can be used to solve a square system of equations.
Let A be the coefficient matrix of a system of n linear equations in n variables given by AX = B. If det(A) ≠ 0, then the
unique solution of the system is given by
|𝐴1 | |𝐴2| |𝐴3 | |𝐴𝑛|
𝑥1 = |𝐴|
, 𝑥2 = |𝐴|
, 𝑥3 = |𝐴|
…, 𝑥𝑛 = |𝐴|
,
where Ai is the matrix obtained by replacing the ith column of A with the column of constants B. If det(A) = 0,
then AX = B has either no solution or infinitely many solutions.
Example: Use Cramer’s Rule to find the solution of the system of linear equations, if a unique solution exists.
–2𝒙𝟏 + 𝒙𝟐 = –7
5𝒙𝟏 – 2𝒙𝟐 = 17
−2 1
The coefficient matrix is A = [ ]. Calculate the determinant of A.
5 −2
|𝐴| = |−2 1
| = (–2) (–2) – 5(1) or –1
5 −2
Because the determinant of A does not equal zero, you can apply Cramer’s Rule.
−7 1
|𝐴1| | | −7(−2) − 17(1) −3
17 −2
𝑥1 = |𝐴|
= |−1|
= = or 3
−1 −1
−2 −7
|𝐴2 | | | −2(17) − 5(−7) 1
5 17
𝑥2 = |𝐴|
= −1
= −1
= −1 or –1
Therefore, the solution is 𝑥1 = 3 and 𝑥2 = –1 or (3, –1).
Exercises
Use Cramer’s Rule to find the solution of each system of linear equations, if a unique solution exists.
1. x – 2y = –5 2. 3x – 3y = –18
–2x – 5y = –8 –x + 4y = 9
3. 3x + y = 21 4. –2x – 4y = 2
–x + 2y = 14 x + 3y = –3
Chapter 6 17 Glencoe Precalculus
