Cramer’s questions and answers: Gabriel Cramer was a Swiss mathematician from the 17th century, who developed a method to solve the system of linear equations. He developed a formula/rule using the mathematical concepts of determinants to find the solutions of the unknowns in linear equations. Before diving into solving Cramer’s Questions, let us Revisit Matrices and Determinants

Cramer’s Rule

1. For the system of 2 linear equations(2 unknowns)

For the following system of two linear equations with two variables, say

a1x + b1y = c1

a2x + b2y = c2

The steps for Cramer’s Rule is

Writing the above two equations in the form of AX = B, which is

\(\begin{array}{l}\begin{bmatrix} a1 & b1 \\ a2 & b2 \\ \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} =\begin{bmatrix}c1 \\ c2\end{bmatrix} \end{array} \)

where A is the Coefficient matrix

A =

\(\begin{array}{l}\begin{bmatrix} a1 & b1 \\ a2 & b2 \\ \end{bmatrix}\end{array} \)

Variable matrix = X =

X =

\(\begin{array}{l}\begin{bmatrix} x \\ y\end{bmatrix}\end{array} \)

Constant matrix = B

B =

\(\begin{array}{l}\begin{bmatrix} c1 \\ c2\end{bmatrix}\end{array} \)

And

\(\begin{array}{l}D =\begin{vmatrix}A\end{vmatrix}=\begin{vmatrix}a1 & b1 \\a2 & b2\\\end{vmatrix}= a1b2 – a2b1 \end{array} \)

\(\begin{array}{l}D_{x} =\begin{vmatrix}c1 & b1 \\c2 & b2\\\end{vmatrix}= c1b2 – c2b1 \end{array} \)

\(\begin{array}{l}D_{y} =\begin{vmatrix}a1 & c1 \\a2 & c2\\\end{vmatrix}= a1c2 – a2c1 \end{array} \)

Therefore the solutions for x and y are

x = Dx/D

y = Dy/D

Similarly, for the system of 3 variables, Cramer’s Rule can be referred at Cramer’s Rule.

Cramer’s Questions With Solutions

Here are a few Cramer’s questions with solutions that are detailed in steps. At the end of the article, you can also practise a few more problems

Question 1:

In the mid-term exam, there were 100 questions to be answered. Every correct answer was awarded 2 marks and every incorrect one with a negative mark. If the student answered all 100 questions and got 80 marks, how many did the student get correct and how many incorrect did the student get?

Solution:

Using Cramer’s rule for linear equations, this is a 2 variable problem.

Let x be the number of correct answers and y be incorrect answers. There are 100 questions.

Therefore x + y = 100

Since 2 marks are awarded to correct answers, that would be 2x and 1 negative mark for incorrect answer, that would be -y. Total marks obtained are 80.

Therefore, the other equation will be

2x – y = 80

The 2 set of linear equations are:

x + y = 100

2x – y = 80

D =

\(\begin{array}{l}\begin{vmatrix}1 & 1 \\2 & -1\\\end{vmatrix}\end{array} \)
= -1-2 = -3

D1 =

\(\begin{array}{l}\begin{vmatrix}100 & 1 \\80 & -1\\\end{vmatrix}\end{array} \)
= -100 – 80 = -180

D2 =

\(\begin{array}{l}\begin{vmatrix}1 & 100 \\2 & 80\\\end{vmatrix}\end{array} \)
= 80 – 200 = -120

x = D1/D = -180 / -3 = 60

y = D2/D = -120 / -3 = 40

Since x represents the correct answers and y represents the incorrect answers, the student got 60 questions with correct answers and 40 questions with incorrect answers.

Question 2:

Solve these two linear equations with the help of Cramer’s rule.

3x – y = 10 and x + y = 6

Solution:

The two linear equations are

3x – y = 10 and

x + y = 6

By Cramer’s rule, we know that x = D1/D and y = D2/D. Let us determine D1 and D2

D =

\(\begin{array}{l}\begin{vmatrix}3 & -1 \\1 & 1\\\end{vmatrix}\end{array} \)
= 3 + 1 = 4

D1 =

\(\begin{array}{l}\begin{vmatrix}10 & -1 \\6 & 1\\\end{vmatrix}\end{array} \)
= 10 – (-6) = 10 + 6 = 16

D2 =

\(\begin{array}{l}\begin{vmatrix}3 & 10 \\1 & 6\\\end{vmatrix}\end{array} \)
= 18 – 10 = 8

x will be D1/D and y will be D2/D. Substituting the values of D, D1 and D2

x = 16/4 = 4

y = 8/4 = 2.

Therefore the values of x and y using Cramer’s rule are 4 and 2.

Question 3:

Solve these two linear equations with the help of Cramer’s rule.

9x – 3y = 21 and 2x + 3y = 1.

Solution:

The two linear equations are

9x – 3y = 21 and

2x + 3y = 1.

By Cramer’s rule, we know that x = D1/D and y = D2/D. Let us determine D, D1 and D2

D =

\(\begin{array}{l}\begin{vmatrix}9 & 3 \\2 & 3\\\end{vmatrix}\end{array} \)
= 27- (-6) = 27 + 6 = 33

D1 =

\(\begin{array}{l}\begin{vmatrix}21 & -3 \\1 & 3\\\end{vmatrix}\end{array} \)
= 63 – (-3) = 63 + 3 = 66

D2 =

\(\begin{array}{l}\begin{vmatrix}9 & 21 \\2 & 1\\\end{vmatrix}\end{array} \)
= 9 – 42 = -33

x will be D1/D and y will be D2/D. Substituting the values of D, D1 and D2

x = 66/33 = 2

y = -33/33 = -1.

Therefore the values of x and y using Cramer’s rule are 2 and -1 respectively.

Question 4:

Using Carmer’s rule, Solve these two linear equations.

5x + 4y = 1 and 3x – 6y = 2.

Solution:

The given linear equations are

5x + 4y = 1 and

3x – 6y = 2.

We know from Cramer’s rule that x = D1/D and y = D2/D where D, D1 and D2 are determinants obtained by the coefficients of the given linear equations, where

D =

\(\begin{array}{l}\begin{vmatrix}5 & 4 \\3 & 6\\\end{vmatrix}\end{array} \)
= -30 – 6 = -42

D1 =

\(\begin{array}{l}\begin{vmatrix}1 & 4 \\2 & -6\\\end{vmatrix}\end{array} \)
= -6 -8 = -14

D2 =

\(\begin{array}{l}\begin{vmatrix}5 & 1 \\3 & 2\\\end{vmatrix}\end{array} \)
= 10 -3 = 7

By Cramer’s rule, x will be D1/D and y will be D2/D. Substituting the values of D, D1 and D2

x = -14/- 42 = 1/3

y = -42/ 7 = – 1/6

Question 5:

With the help of Cramer’s rule, Solve the following equations

6x – 5y = 8 and -6x + y = 0

Solution:

The two equations are 6x – 5y = 8 and -6x + y = 0.

Let us find the determinants D, D1 and D2

D =

\(\begin{array}{l}\begin{vmatrix}6 & -5 \\-6 & 1\\\end{vmatrix}\end{array} \)
= -6 – 30 = -24

D1 =

\(\begin{array}{l}\begin{vmatrix}8 & -5 \\0 & 1\\\end{vmatrix}\end{array} \)
= 8 – 0 = 8

D2 =

\(\begin{array}{l}\begin{vmatrix}6 & 8 \\-6 & 0\\\end{vmatrix}\end{array} \)
= 0 -(- 48) = 48

We know from Cramer’s rule that x = D1/D and y = D2/D

Therefore x = 8/-24 = -1/3.

y = 48/-24 = -2.

Question 6:

A couple went to a restaurant for dinner. Husband ordered 2 small pizzas where 1 had veggies and the other pizza had double extra cheese. Along with pizza, he ordered a fruit bowl. This was a combo which cost him 200 Rs. Wife ordered from a combo which had 1 veggie pizza and 3 fruit bowls. She decided to take two fruit bowls as a parcel. This combo cost 150 Rs. Can you determine the cost of pizza and fruit bowl. Can this problem be solved using Cramer’s rule?

Solution:

Let us first try to put the statements in equations. There are only 2 variables, they are pizza and fruit bowl

Let us represent pizza with x and fruit bowl as y

Husband’s combo offer is 2 pizzas and 1 fruit bowl with cost as Rs 200

Wife’s combo offer is 1 pizza and 3 fruit bowls with cost as Rs 150

The equations are

2x + y = 200

x + 3y = 150

We know from Cramer’s rule that x = D1/D and y = D2/D where D, D1 , D2 are determinants. The determinants are obtained such that

D =

\(\begin{array}{l}\begin{vmatrix}2 & 1 \\1 & 3\\\end{vmatrix}\end{array} \)
= 6 – 1 = 5

D1 =

\(\begin{array}{l}\begin{vmatrix}200 & 1 \\150 & 3\\\end{vmatrix}\end{array} \)
= 600 – 150 = 450

D2 =

\(\begin{array}{l}\begin{vmatrix}2 & 200 \\1 & 150\\\end{vmatrix}\end{array} \)
= 300 – 200 = 100

x = D1/D = 450/5 = 90

y = D2/D = 100/5 = 20

But x is pizza and y is fruit bowl. So the cost of one pizza is 90 Rs and one fruit bowl is 20 Rs.

Yes. This problem can be solved by Cramer’s rule.

Question 7:

Solve using Cramer’s rule

2x + 3y = 20

7x + 2y = 53

Solution:

From Cramer’s rule, x will be D1/D and y will be D2/D, where D, D1, D2 are the determinants obtained by the coefficients of the given equations.

D =

\(\begin{array}{l}\begin{vmatrix}2 & 3 \\7 & 2\\\end{vmatrix}\end{array} \)
= 4 – 21 = -17

D1 =

\(\begin{array}{l}\begin{vmatrix}20 & 3 \\53 & 2\\\end{vmatrix}\end{array} \)
= 40 – 159 = -119

D2 =

\(\begin{array}{l}\begin{vmatrix}2 & 20 \\7 & 53\\\end{vmatrix}\end{array} \)
= 106 – 140 = -34

x = D1/D = – 119/ -17 = 7

y = D2/D = -34/-17 = 2

Question 8:

Solve for x + y = 4 and -x + y = 2 using Cramer’s rule.

Solution:

x will be D1/D and y will be D2/D, where D, D1, D2 are the determinants obtained by the coefficients of the given equations from Cramer’s rule.

Let us find the values of all the determinants D, D1, D2

D =

\(\begin{array}{l}\begin{vmatrix}1 & 1 \\-1 & 1\\\end{vmatrix}\end{array} \)
= 1 – (-1)= 1 + 1 = 2

D1 =

\(\begin{array}{l}\begin{vmatrix}4 & 1 \\2 & 1\\\end{vmatrix}\end{array} \)
= 4 – 2 = 2

D2 =

\(\begin{array}{l}\begin{vmatrix}1 & 4 \\-1 & 2\\\end{vmatrix}\end{array} \)
= 2 – (-4) = 2 + 4 = 6

x = D1 /D = 2/2 = 1

y = D2 /D = 6/2 = 3

For the equations x + y = 4 and -x + y = 2, the value of x = 1 and y = 3, which is obtained by Cramer’s rule.

Question 9:

Solve with the help of Cramer’s rule the following 2 linear equations.

x + y = 3; 2x – y = 6;

Solution:

Given 2 equations are

x + y = 3 and

2x – y = 6

By Cramer’s rule, we know that x will be D1/D and y will be D2/D, where D, D1, D2 are the determinants. The values of D, D1, D2 are

D =

\(\begin{array}{l}\begin{vmatrix}1 & 1 \\2 & -1\\\end{vmatrix}\end{array} \)
= -1 – 2 = -3

D1 =

\(\begin{array}{l}\begin{vmatrix}3 & 1 \\6 & 1\\\end{vmatrix}\end{array} \)
= 3 – 6 = -3

D2 =

\(\begin{array}{l}\begin{vmatrix}1 & 3 \\2 & 6\\\end{vmatrix}\end{array} \)
= 6 – 6 = 0

x = D1/D = -3/-3 = 1

y = D2/D = 0/-3 = 0

So the values of x and y for the equations x + y = 3 and 2x – y = 6 are 1 and 0 respectively.

Question 10:

Solve the linear equations with 3 variables using Cramer’s rule. The equations are

x + y = 2

x – y + z = 4

x + y – z = 6

Solution:

Given 3 variable linear equations are,

x + y = 2

x – y + z = 4

x + y – z = 6

By Cramer’s rule for 3 variables, x = D1/D, y = D2/D, z = D3/D, where D, D1, D2 and D3 are determinants. Let us find the values of all the determinants to solve the given equations.

D =

\(\begin{array}{l}\begin{vmatrix}1 & 1 & 0 \\1 & -1 & 1 \\1 & 1 & -1 \\\end{vmatrix}\end{array} \)

D = 1(1-1) -1 (-1-1) + 0

D = 0 + 2 + 0

D = 2

D1 =

\(\begin{array}{l}\begin{vmatrix}2 & 1 & 0 \\4 & -1 & 1 \\6 & 1 & -1 \\\end{vmatrix}\end{array} \)

D1 = 2(1-1) -1 (-4-6) + 0

D1 = 0 -1(-10) + 0

D1 = 10

D2 =

\(\begin{array}{l}\begin{vmatrix}1 & 2 & 0 \\1 & 4 & 1 \\1 & 6 & -1 \\\end{vmatrix}\end{array} \)

D2 = 1(-4-6) -2 (-1-1) + 0

D2 = 1 (-10) + 4 + 0

D2 = -6

D3 =

\(\begin{array}{l}\begin{vmatrix}1 & 1 & 2 \\1 & -1 & 4 \\1 & 1 & 6 \\\end{vmatrix}\end{array} \)

D3 = 1 (-6-4) -1 (6-4) + 2(1+1)

D3 = -10) -2 + 4

D3 = -8

x = D1/D = 10/2 = 5

y = D2/D = -6/2 = -3

z = D3/D = -8/2 = -4

For the equations, x + y = 2, x – y + z = 4, and x + y – z = 6 , the values of x, y and z are 5, -3 and -4 respectively. This is solved by Cramer’s rule.

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Practice Questions on Cramer’s

  1. Solve the system of equations using Cramer’s rule.

    2. x + y = 10

    x – 2y = 15

  2. Solve the system of equations using Cramer’s rule.

    3. 2x – 3y = 9

    4x – 7y = 18

  3. Solve the system of equations using Cramer’s rule.

    4. 3x + 6y = -1

    x + y = 12

  4. Solve the system of equations for the linear equations with 3 variables using Cramer’s rule.

    5. x – z = 11

    7x – 4y – z = -15

    4x + 6y + 5z = -6