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9.6 Determinants and Cramer’s Rule

Evaluate determinants of square matrices.
Use Cramer’s rule to solve systems of equations.

Determinants of Square Matrices

With every square matrix, we associate a number called its determinant.

Example 1

Evaluate: ∣∣∣2–√−4−3−2–√∣∣∣ $| \begin{matrix} \sqrt{2} & - 3 \\ - 4 & - \sqrt{2} \end{matrix} |$ .

Solution

=2–√(−2–√)−(−4)(−3)=−2−12=−14 $\begin{matrix} = \sqrt{2} (- \sqrt{2}) - (- 4) (- 3) \\ = - 2 - 12 \\ = - 14 \end{matrix}$

Now Try Exercise 1.

We now consider a way to evaluate determinants of square matrices of order 3 × 3 $3 \times 3$ or higher.

Evaluating Determinants Using Cofactors

Often we first find minors and cofactors of matrices in order to evaluate determinants.

Example 2

For the matrix

A = [a i j] = ⎡ ⎣ ⎢ - 8 4 - 1 0 - 6 - 3 675 ⎤ ⎦ ⎥,

$A = [a_{i j}] = [\begin{matrix} - 8 & 0 & 6 \\ 4 & - 6 & 7 \\ - 1 & - 3 & 5 \end{matrix}],$

find each of the following.

M11 $M_{11}$
M23 $M_{23}$

Solution

For M11 $M_{11}$ , we delete the first row and the first column and find the determinant of the 2×2 $2 \times 2$ matrix formed by the remaining entries.

$M 11 = = = = = ∣ ∣ ∣ - 6 - 3 75 ∣ ∣ ∣ (- 6) \cdot 5 - (- 3) \cdot 7 - 30 - (- 21) - 30 + 21 - 9$ $\begin{array}{l} M_{11} & = & | \begin{matrix} - 6 & 7 \\ - 3 & 5 \end{matrix} | \\ = & (- 6) \cdot 5 - (- 3) \cdot 7 \\ = & - 30 - (- 21) \\ = & - 30 + 21 \\ = & - 9 \end{array}$
For M23 $M_{23}$ , we delete the second row and the third column and find the determinant of the 2×2 $2 \times 2$ matrix formed by the remaining entries.

$M 23 = = = ∣ ∣ ∣ - 8 - 1 0 - 3 ∣ ∣ ∣ - 8 (- 3) - (- 1) 0 24$ $\begin{matrix} M_{23} & = & | \begin{matrix} - 8 & 0 \\ - 1 & - 3 \end{matrix} | \\ = & - 8 (- 3) - (- 1) 0 \\ = & 24 \end{matrix}$

Now Try Exercise 9.

Example 3

For the matrix given in Example 2, find each of the following.

A11 $A_{11}$
A23 $A_{23}$

Solution

In Example 2, we found that M11=−9 $M_{11} = - 9$ . Then

$A 11 = (- 1) 1 + 1 (- 9) = (1) (- 9) = - 9 .$ $A_{11} = {(- 1)}^{1 + 1} (- 9) = (1) (- 9) = - 9 .$
In Example 2, we found that M23=24 $M_{23} = 24$ . Then

$A 23 = (- 1) 2 + 3 (24) = (- 1) (24) = - 24 .$ $A_{23} = {(- 1)}^{2 + 3} (24) = (- 1) (24) = - 24 .$

Now Try Exercise 11.

Consider the matrix A given by

A = ⎡ ⎣ ⎢ a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 ⎤ ⎦ ⎥ .

$A = [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}] .$

The determinant of the matrix, denoted |A| $| A |$ , can be found by multiplying each element of the first column by its cofactor and adding:

| A | = a 11 A 11 + a 21 A 21 + a 31 A 31 .

$| A | = a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31} .$

Because

and A 11 A 21 A 31 = = = (- 1) 1 + 1 M 11 (- 1) 2 + 1 M 21 (- 1) 3 + 1 M 31 = = = M 11, - M 21, M 31,

$\begin{array}{l} A_{11} & = & {(- 1)}^{1 + 1} M_{11} & = & M_{11}, \\ A_{21} & = & {(- 1)}^{2 + 1} M_{21} & = & - M_{21}, \\ and & A_{31} & = & {(- 1)}^{3 + 1} M_{31} & = & M_{31}, \end{array}$

we can write

| A | = a 11 \cdot ∣ ∣ ∣ a 22 a 32 a 23 a 33 ∣ ∣ ∣ - a 21 \cdot ∣ ∣ ∣ a 12 a 32 a 13 a 33 ∣ ∣ ∣ + a 31 \cdot ∣ ∣ ∣ a 12 a 22 a 13 a 23 ∣ ∣ ∣ .

$| A | = a_{11} \cdot | \begin{matrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{matrix} | - a_{21} \cdot | \begin{matrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{matrix} | + a_{31} \cdot | \begin{matrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{matrix} | .$

It can be shown that we can determine |A| $| A |$ by choosing any row or column, multiplying each element in that row or column by its cofactor, and adding. This is called expanding across a row or down a column. We just expanded down the first column. We now define the determinant of a square matrix of any order.

Example 4

Evaluate |A| $| A |$ by expanding across the third row.

A = ⎡ ⎣ ⎢ - 8 4 - 1 0 - 6 - 3 675 ⎤ ⎦ ⎥

$A = [\begin{array}{l} - 8 & 0 & 6 \\ 4 & - 6 & 7 \\ - 1 & - 3 & 5 \end{array}]$

Solution

We have

| A | = = = = = (- 1) A 31 + (- 3) A 32 + 5 A 33 (- 1) (- 1) 3 + 1 \cdot ∣ ∣ ∣ 0 - 6 67 ∣ ∣ ∣ + (- 3) (- 1) 3 + 2 \cdot ∣ ∣ ∣ - 8 4 67 ∣ ∣ ∣ + 5 (- 1) 3 + 3 \cdot ∣ ∣ ∣ - 8 4 0 - 6 ∣ ∣ ∣ (- 1) \cdot 1 \cdot [0 \cdot 7 - (- 6) 6] + (- 3) (- 1) [- 8 \cdot 7 - 4 \cdot 6] + 5 \cdot 1 \cdot [- 8 (- 6) - 4 \cdot 0] - [36] + 3 [- 80] + 5 [48] - 36 - 240 + 240 = - 36 .

$\begin{array}{l} | A | & = & (- 1) A_{31} + (- 3) A_{32} + 5 A_{33} \\ = & (- 1) {(- 1)}^{3 + 1} \cdot | \begin{matrix} 0 & 6 \\ - 6 & 7 \end{matrix} | + (- 3) {(- 1)}^{3 + 2} \cdot | \begin{matrix} - 8 & 6 \\ 4 & 7 \end{matrix} | \\ + 5 {(- 1)}^{3 + 3} \cdot | \begin{matrix} - 8 & 0 \\ 4 & - 6 \end{matrix} | \\ = & (- 1) \cdot 1 \cdot [0 \cdot 7 - (- 6) 6] + (- 3) (- 1) [- 8 \cdot 7 - 4 \cdot 6] \\ + 5 \cdot 1 \cdot [- 8 (- 6) - 4 \cdot 0] \\ = & - [36] + 3 [- 80] + 5 [48] \\ = & - 36 - 240 + 240 = - 36 . \end{array}$

The value of this determinant is −36 $- 36$ no matter which row or column we expand on.

Now Try Exercise 13.

Cramer’s Rule

Determinants can be used to solve systems of linear equations. Consider a system of two linear equations:

a 1 x a 2 x + + b 1 y b 2 y = = c 1, c 2 .

$\begin{array}{l} a_{1} x & + & b_{1} y & = & c_{1}, \\ a_{2} x & + & b_{2} y & = & c_{2} . \end{array}$

Solving this system using the elimination method, we obtain

x = c 1 b 2 - c 2 b 1 a 1 b 2 - a 2 b 1 and y = a 1 c 2 - a 2 c 1 a 1 b 2 - a 2 b 1 .

$x = \frac{c_{1} b_{2} - c_{2} b_{1}}{a_{1} b_{2} - a_{2} b_{1}} and y = \frac{a_{1} c_{2} - a_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}} .$

The numerators and the denominators of these expressions can be written as determinants:

x = ∣ ∣ ∣ c 1 c 2 b 1 b 2 ∣ ∣ ∣ ∣ ∣ ∣ a 1 a 2 b 1 b 2 ∣ ∣ ∣ and y = ∣ ∣ ∣ a 1 a 2 c 1 c 2 ∣ ∣ ∣ ∣ ∣ ∣ a 1 a 2 b 1 b 2 ∣ ∣ ∣ .

$x = \frac{| \begin{matrix} c_{1} & b_{1} \\ c_{2} & b_{2} \end{matrix} |}{| \begin{matrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{matrix} |} and y = \frac{| \begin{matrix} a_{1} & c_{1} \\ a_{2} & c_{2} \end{matrix} |}{| \begin{matrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{matrix} |} .$

If we let

D = ∣ ∣ ∣ a 1 a 2 b 1 b 2 ∣ ∣ ∣, D x = ∣ ∣ ∣ c 1 c 2 b 1 b 2 ∣ ∣ ∣, and D y = ∣ ∣ ∣ a 1 a 2 c 1 c 2 ∣ ∣ ∣,

$D = | \begin{matrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{matrix} |, D_{x} = | \begin{matrix} c_{1} & b_{1} \\ c_{2} & b_{2} \end{matrix} |, and D_{y} = | \begin{matrix} a_{1} & c_{1} \\ a_{2} & c_{2} \end{matrix} |,$

we have

x = D x D and y = D y D .

$x = \frac{D_{x}}{D} and y = \frac{D_{y}}{D} .$

This procedure for solving systems of equations is known as Cramer’s rule.

Note that the denominator D contains the coefficients of x and y, in the same position as in the original equations. For x, the numerator is obtained by replacing the x-coefficients in D (the a’s) with the c’s. For y, the numerator is obtained by replacing the y-coefficients in D (the b’s) with the c’s.

Example 5

Solve using Cramer’s rule:

$\begin{array}{l} 2 x & + & 5 y & = & 7, \\ 5 x & - & 2 y & = & - 3. \end{array}$

Solution

We have

$\begin{matrix} x = & \frac{| \begin{matrix} 7 & 5 \\ - 3 & - 2 \end{matrix} |}{| \begin{matrix} 2 & 5 \\ 5 & - 2 \end{matrix} |} = \frac{7 (- 2) - (- 3) 5}{2 (- 2) - 5 \cdot 5} = \frac{1}{- 29} = - \frac{1}{29}, \\ y = & \frac{| \begin{matrix} 2 & 7 \\ 5 & - 3 \end{matrix} |}{| \begin{matrix} 2 & 5 \\ 5 & - 2 \end{matrix} |} = \frac{2 (- 3) - 5 \cdot 7}{- 29} = \frac{- 41}{- 29} = \frac{41}{29} . \end{matrix}$

The solution is $(- \frac{1}{29}, \frac{41}{29})$ .

Now Try Exercise 29.

Cramer’s rule works only when a system of equations has a unique solution. This occurs when $D \neq 0$ . If $D = 0$ and $D_{x}$ and $D_{y}$ are also 0, then the equations are dependent. If $D = 0$ and $D_{x}$ and/or $D_{y}$ is not 0, then the system is inconsistent.

Cramer’s rule can be extended to a system of n linear equations in n variables. We consider a $3 \times 3$ system.

Note that the determinant $D_{x}$ is obtained from D by replacing the x-coefficients with $d_{1}$ , $d_{2}$ , and $d_{3}$ . $D_{y}$ and $D_{z}$ are obtained in a similar manner. As with a system of two equations, Cramer’s rule cannot be used if $D = 0$ . If $D = 0$ and $D_{x}$ , $D_{y}$ , and $D_{z}$ are 0, then the equations are dependent. If $D = 0$ and one of $D_{x}$ , $D_{y}$ , or $D_{z}$ is not 0, then the system is inconsistent.

Example 6

Solve using Cramer’s rule:

$\begin{array}{r} x & - & 3 y & + & 7 z & = & 13, \\ x & + & y & + & z & = & 1, \\ x & - & 2 y & + & 3 z & = & 4. \end{array}$

Solution

We have

$\begin{array}{l} D = & | \begin{matrix} 1 & - 3 & 7 \\ 1 & 1 & 1 \\ 1 & - 2 & 3 \end{matrix} | = - 10, & D_{x} = & | \begin{matrix} 13 & - 3 & 7 \\ 1 & 1 & 1 \\ 4 & - 2 & 3 \end{matrix} | = 20, \\ D_{y} = & | \begin{matrix} 1 & 13 & 7 \\ 1 & 1 & 1 \\ 1 & 4 & 3 \end{matrix} | = - 6, & D_{z} = & | \begin{matrix} 1 & - 3 & 13 \\ 1 & 1 & 1 \\ 1 & - 2 & 4 \end{matrix} | = - 24. \end{array}$

Then

$\begin{matrix} x = \frac{D_{x}}{D} = \frac{20}{- 10} = - 2, & y = \frac{D_{y}}{D} = \frac{- 6}{- 10} = \frac{3}{5}, & z = \frac{D_{z}}{D} = \frac{- 24}{- 10} = \frac{12}{5} . \end{matrix}$

The solution is $(- 2, \frac{3}{5}, \frac{12}{5})$ .

In practice, it is not necessary to evaluate $D_{z}$ . When we have found values for x and y, we can substitute them into one of the equations to find z.

Now Try Exercise 37.

9.6 Exercise Set

Evaluate the determinant.

1. $| \begin{matrix} 5 & 3 \\ - 2 & - 4 \end{matrix} |$
2. $| \begin{matrix} - 8 & 6 \\ - 1 & 2 \end{matrix} |$
3. $| \begin{matrix} 4 & - 7 \\ - 2 & 3 \end{matrix} |$
4. $| \begin{matrix} - 9 & - 6 \\ 5 & 4 \end{matrix} |$
5. $| \begin{matrix} - 2 & - \sqrt{5} \\ - \sqrt{5} & 3 \end{matrix} |$
9. $| \begin{matrix} \sqrt{5} & - 3 \\ 4 & 2 \end{matrix} |$
7. $| \begin{matrix} x & 4 \\ x & x^{2} \end{matrix} |$
8. $| \begin{matrix} y^{2} & - 2 \\ y & 3 \end{matrix} |$

Use the following matrix for Exercises 9–16:

$A = [\begin{array}{r} 7 & - 4 & - 6 \\ 2 & 0 & - 3 \\ 1 & 2 & - 5 \end{array}] .$

9.Find $M_{11}$ , $M_{32}$ , and $M_{22}$ .
10. Find $M_{13}$ , $M_{31}$ , and $M_{23}$ .
11.Find $A_{11}$ , $A_{32}$ , and $A_{22}$ .
12. Find $A_{13}$ , $A_{31}$ , and $A_{23}$ .
13. Evaluate $| A |$ by expanding across the second row.
14. Evaluate $| A |$ by expanding down the second column.
15.Evaluate $| A |$ by expanding down the third column.
16. Evaluate $| A |$ by expanding across the first row.

Use the following matrix for Exercises 17–22:

$A = [\begin{array}{r} 1 & 0 & 0 & - 2 \\ 4 & 1 & 0 & 0 \\ 5 & 6 & 7 & 8 \\ - 2 & - 3 & - 1 & 0 \end{array}] .$

17.Find $M_{12}$ and $M_{44}$ .
18. Find $M_{41}$ and $M_{33}$ .
19.Find $A_{22}$ and $A_{34}$ .
20. Find $A_{24}$ and $A_{43}$ .
21.Evaluate $| A |$ by expanding across the first row.
22. Evaluate $| A |$ by expanding down the third column.

Evaluate the determinant.

23. $| \begin{matrix} 3 & 1 & 2 \\ - 2 & 3 & 1 \\ 3 & 4 & - 6 \end{matrix} |$
24. $| \begin{matrix} 3 & - 2 & 1 \\ 2 & 4 & 3 \\ - 1 & 5 & 1 \end{matrix} |$
25. $| \begin{matrix} x & 0 & - 1 \\ 2 & x & x^{2} \\ - 3 & x & 1 \end{matrix} |$
26. $| \begin{matrix} x & 1 & - 1 \\ x^{2} & x & x \\ 0 & x & 1 \end{matrix} |$

Solve using Cramer’s rule.

27. $\begin{array}{l} - 2 x & + & 4 y & = & 3, \\ 3 x & - & 7 y & = & 1 \end{array}$
28. $\begin{array}{l} 5 x & - & 4 y & = & - 3, \\ 7 x & + & 2 y & = & 6 \end{array}$
29. $\begin{array}{l} 2 x & - & y & = & 5, \\ x & - & 2 y & = & 1 \end{array}$
30. $\begin{array}{l} 3 x & + & 4 y & = & - 2, \\ 5 x & - & 7 y & = & 1 \end{array}$
31. $\begin{array}{l} 2 x & + & 9 y & = & - 2, \\ 4 x & - & 3 y & = & 3 \end{array}$
32. $\begin{array}{l} 2 x & + & 3 y & = & - 1, \\ 3 x & + & 6 y & = & - 0.5 \end{array}$
33. $\begin{array}{l} 2 x & + & 5 y & = & 7, \\ 3 x & - & 2 y & = & 1 \end{array}$
34. $\begin{array}{l} 3 x & + & 2 y & = & 7, \\ 2 x & + & 3 y & = & - 2 \end{array}$
35. $\begin{array}{l} 3 x + 2 y - z = 4 \\ 3 x - 2 y + z = 5 \\ 4 x - 5 y - z = - 1 \end{array}$
36. $\begin{array}{l} 3 x & - & y & + & 2 z & = & 1, \\ x & - & y & + & 2 z & = & 3, \\ - 2 x & + & 3 y & + & z & = & 1 \end{array}$
37. $\begin{array}{l} 3 x & + & 5 y & - & z & = & - 2, \\ x & - & 4 y & + & 2 z & = & 13, \\ 2 x & + & 4 y & + & 3 z & = & 1 \end{array}$
38. $\begin{array}{l} 3 x & + & 2 y & + & 2 z & = & 1, \\ 5 x & - & y & - & 6 z & = & 3, \\ 2 x & + & 3 y & + & 3 z & = & 4 \end{array}$
39. $\begin{array}{l} x & - & 3 y & - & 7 z & = & 6, \\ 2 x & + & 3 y & + & z & = & 9, \\ 4 x & + & y & = & 7 \end{array}$
40. $\begin{array}{l} x & - & 2 y & - & 3 z & = & 4, \\ 3 x & - & 2 z & = & 8, \\ 2 x & + & y & + & 4 z & = & 13 \end{array}$
41. $\begin{array}{l} 6 y & + & 6 z & = & - 1, \\ 8 x & + & 6 z & = & - 1, \\ 4 x & + & 9 y & = & 8 \end{array}$
42. $\begin{array}{l} 3 x & + & 5 y & = & 2, \\ 2 x & - & 3 z & = & 7, \\ 4 y & + & 2 z & = & - 1 \end{array}$

Skill Maintenance

Determine whether the function is one-to-one, and if it is, find a formula for $f^{- 1} (x)$ . [5.1]

43. $f (x) = 3 x + 2$
44. $f (x) = x^{2} - 4$
45. $f (x) = | x | + 3$
46. $f (x) = \sqrt[3]{x} + 1$

Simplify. Write answers in the form $a + b i$ , where a and b are real numbers. [3.1]

47. $(3 - 4 i) - (- 2 - i)$
48. $(5 + 2 i) + (1 - 4 i)$
49. $(1 - 2 i) (6 + 2 i)$
50. $\frac{3 + i}{4 - 3 i}$

Synthesis

Solve.

51. $| \begin{matrix} y & 2 \\ 3 & y \end{matrix} | = y$
52. $| \begin{matrix} x & - 3 \\ - 1 & x \end{matrix} | \geq 0$
53. $| \begin{matrix} 2 & x & 1 \\ 1 & 2 & - 1 \\ 3 & 4 & - 2 \end{matrix} | = - 6$
54. $| \begin{matrix} m + 2 & - 3 \\ m + 5 & - 4 \end{matrix} | = 3 m - 5$

Rewrite the expression using a determinant. Answers may vary.

55. $a^{2} + b^{2}$
56. $\frac{1}{2} h (a + b)$
57. $2 π r^{2} + 2 π r h$
58. $x^{2} y^{2} - Q^{2}$

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Table of Contents for 9.6 Determinants and Cramer&#8217;s RuleDeterminants and Cramer’s Rule

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9.6 Determinants and Cramer’s RuleDeterminants and Cramer’s Rule