﻿ Cramer's Rule Worksheet | Problems & Solutions
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# Cramer's Rule Worksheet

Cramer's Rule Worksheet
• Page 1
1.
Evaluate the determinant.
$|\begin{array}{cc}2& 2\\ 2& 3\end{array}|$
 a. 3 b. 2 c. 1 d. 4

#### Solution:

|2223 |

= 2(3) - 2(2) = 2.
[|abcd | = ad - bc.]

Correct answer : (2)
2.
Evaluate the determinant.
$|\begin{array}{cc}4& \mathrm{- 3}\\ 6& 6\end{array}|$
 a. 42 b. 44 c. - 44 d. - 42

#### Solution:

|4- 366 |

= 4(6) - 6(- 3) = 42.
[|abcd | = ad - bc.]

Correct answer : (1)
3.
Evaluate.
$|\begin{array}{cc}\mathrm{- 9}& 8\\ \mathrm{- 3}& 2\end{array}|$
 a. 5 b. 6 c. 8 d. 7

#### Solution:

|- 98- 32 |

= - 9(2) - 8(- 3)
[|abcd | = ad - bc.]

= - 18 + 24 = 6

Correct answer : (2)
4.
Evaluate the determinant: $|\begin{array}{cc}9& 0\\ 2& 0\end{array}|$
 a. 9 b. 12 c. 10

#### Solution:

|9020 |

= 9(0) - 2(0) = 0
[|abcd | = ad - bc.]

Correct answer : (2)
5.
Solve the system using Cramer's rule
x - y + 6 = - z
y + 1 = -z
x + 1 = z
 a. (- 2, 2, - 2) b. (- 3, 1, - 2) c. (- 4, - 1, - 3) d. (1, 3, - 4)

#### Solution:

(1-1101110-1)(xyz) = (-6-1-1)
[Write the matrix equation.]

D = |a1b1a2b2 | = |1-1101110-1 |
[Find D.]

D = 1 |110-1 | + 1 |011-1 | + 1 |0110 |

D = 1 [- 1 - 0] + 1[0 - 1] + 1[0 - 1] = - 1 - 1 - 1 = - 3

Dx = |c1b1c2b2 | = |-6-11-111-10-1 |

Dx = - 6 |110-1 | + 1 |-11-1-1 | + 1 |-11-10 |

Dx = - 6 [- 1 - 0] + 1[1 + 1] + 1[0 + 1] = 6 + 2 + 1 = 9

Dy = |a1c1a2c2 | = |1-610-111-1-1 |
[Find Dy.]

Dy = 1 |-11-1-1 | + 6 |011-1 | + 1 |0-11-1 |

Dy = 1 [1 + 1] + 6[0 -1] + 1[0 + 1] = 2 - 6 + 1 = - 3

Dz = |1-1-601-110-1 |
[Find Dz.]

Dz = 1 |1-10-1 | + 1 |0-11-1 | - 6 |0110 |

Dz = 1 [- 1 - 0] + 1[0 + 1] - 6[0 - 1] = - 1 + 1 + 6 = 6

x = DxD = 9 / -3 = - 3 ; y = DyD = -3 / -3 = 1;z = DzD = 6 / -3 = -2

The solution to the system is (- 3, 1, - 2).

Correct answer : (2)
6.
The sum of two numbers is 14 while their difference is 32. Find the numbers using Cramer's rule.
 a. 25 and - 11 b. 24 and - 10 c. 26 and - 12 d. 23 and - 9

#### Solution:

Let the numbers be x and y.

x + y = 14

x - y = 32

 A X = C (111-1) (xy) = (1432)

[The matrix equation is AX = C.]

D = |a1b1a2b2 | = |111-1 | = 1(- 1) - 1(1) = - 2
[Find D.]

Dx = |c1b1c2b2 | = |14132- 1 | = 14(- 1) - 32(1) = - 46
[Substitute first column of A with the column of C.]

Dy = |a1c1a2c2 | = |114132 | = 1(32) - 1(14) = 18
[Substitute second column of A with the column of C.]

x = DxD = - 46- 2 = 23 and y = DyD = 18- 2 = - 9

The two numbers are 23 and - 9.

Correct answer : (4)
7.
William sells 3 pens and 4 pencils for $38, 5 pens and 6 pencils for$60. Find the selling price of 1 pen and 1 pencil. (Use Cramer's rule.)
 a. (5, 5) b. (8, 6) c. (6, 6) d. (6, 5)

#### Solution:

Let the selling price of 1 pen = $x Let the selling price of 1 pencil =$y

3x + 4y = 38

5x + 6y = 60

(3456)(xy) = (3860)
[Write the matrix equation AX = C.]

D = |3456 | = 3(6) - 5(4) = - 2
[Find D.]

Dx = |384606 | = 38(6) - 60(4) = - 12
[Find Dx.]

Dy = |338560 | = 3(60) - 5(38) = - 10
[Find Dy.]

x = DxD = - 12- 2 = 6 and y = DyD = - 10- 2 = 5.
[Find x, y.]

So, the selling price of 1 pen is $6 and the selling price of 1 pencil =$5.

The solution is (6, 5).

Correct answer : (4)
8.
Evaluate.
$|\begin{array}{cc}6& 2\\ 4& 0\end{array}|$
 a. 8 b. - 8 c. 9 d. - 9

#### Solution:

|6240 |

= 6(0) - 2(4) = - 8
[|abcd | = ad - bc.]

Correct answer : (2)
9.
Evaluate the determinant: $|\begin{array}{cc}\frac{1}{4}& \frac{\mathrm{- 5}}{4}\\ \frac{1}{6}& \frac{7}{6}\end{array}|$
 a. $\frac{1}{3}$ b. 1 c. $\frac{1}{2}$

#### Solution:

|14- 541676 |

= (1 / 4 × 7 / 6) - (1 / 6 × - 5 / 4)
[|abcd | = ad - bc.]

= 7 / 24 + 5 / 24 = 12 / 24 = 1 / 2

Correct answer : (3)
10.
If $|\begin{array}{cc}a& 2\\ 7& 1\end{array}|$ = - 18, then find the value of $a$.
 a. 6 b. - 6 c. - 4 d. 4

#### Solution:

|a271 | = - 18

a(1) - 2(7) = - 18
[|abcd | = ad - bc.]

a - 14 = - 18

a = - 18 + 14

a = - 4.

Correct answer : (3)

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