Cramer's Rule (Practice Problems)

Cramer's Rule is a method for solving a system of linear equations using determinants. It is applicable only when the system has the same number of equations as unknowns and the coefficient matrix has a non-zero determinant.

Solved Examples

Example 1: Solve  \left\{ \begin{array}{c} 12x-10y= 46\\ 3x+20y=-11 \\ \end{array} \right\}

The given equations in the form of AX = B 

A =\begin{bmatrix}    12 & -10 \\    3 & 20   \end{bmatrix}          , 

B = \begin{bmatrix}    46  \\    -11       \end{bmatrix}         , 

X =  \begin{bmatrix}     x \\     y    \end{bmatrix}

Then, the determinant D of matrix 

A = \begin{vmatrix}    12 & -10 \\    3 & 20  \\ \end{vmatrix}        

   = 12 × 20 - 3 × (-10) = 240 + 30 = 270

Now, find D_x and D_y 

D_x = \begin{vmatrix}    46 & -10 \\    -11& 20  \\ \end{vmatrix}        

    = [46×20 - (-10)×(-11)] = 920 - 110 = 810

D_y = \begin{vmatrix}    12 & 46 \\    3 & -11 \\ \end{vmatrix}         

     =[12×(-11) - 3×46] = -132 -138 = -270

Now, find x = D_x/D, y = D_y/D 

x = 810/270 = 3, y = -270/270 = -1

x = 3, y = -1

Example 2: Solve \left\{ \begin{array}{c} 6.6x+0.95y= 5.2\\ 4.2x+8.6y=19.3 \\ \end{array}\right\}  

The given equations in the form of AX = B 

A = \begin{bmatrix}    6.6& 0.95 \\    4.2& 8.6  \\ \end{bmatrix}               

B = \begin{bmatrix}    5.2  \\    19.3       \end{bmatrix}     

X  =  \begin{bmatrix}     x \\     y    \end{bmatrix}

Then, the determinant D of matrix 

A = \begin{vmatrix}    6.6& 0.95 \\    4.2& 8.6  \\ \end{vmatrix}        = 6.6×8.6 - 4.2×0.95 = 56.76 - 3.99 =52.77

Now, find D_x and D_y 

D_x = \begin{vmatrix}    5.2& 0.95 \\    19.3& 8.6  \\ \end{vmatrix}        = 5.2 × 8.6 - 19.3 × 0.95 = 44.72 - 18.335 = 26.385

D_y = \begin{vmatrix}    6.6& 5.2 \\    4.2& 19.3  \\ \end{vmatrix}        = 6.6×19.3 - 4.2×5.2 = 127.38 - 21.84 = 105.54

Now, find x = D_x/D  , y = D_y/D

x = 26.385/52.77 = 0.5, y = 105.54/52.77 = 2

x = 0.5,  y = 2

Example 3: Solve \left\{ \begin{array}{c} 3x^2+4y^2= 91\\ 6x^2-y^2=38 \\ \end{array} \right\}

Let x² = a, y² = b ---(1)

Then, the equation can be written as,

\left\{ \begin{array}{c} 3a+4b= 91\\ 6a-b=38 \\ \end{array} \right.         

The given equations in the form of AX = B

A = \begin{bmatrix}    3& 4 \\    6&-1  \\ \end{bmatrix}      

B = \begin{bmatrix}    91  \\    38     \end{bmatrix}     

X = \begin{bmatrix}     a \\     b    \end{bmatrix}

Then, the determinant D of matrix A = \begin{vmatrix}    3& 4 \\    6& -1   \end{vmatrix}        = 3×(-1) - 6×4 = -3-24 = -27

Now, find D_a and D_b

D_a =\begin{vmatrix}    91& 4 \\    38& -1   \end{vmatrix}        

     = 91×(-1) - 38×4 = - 91 - 152 

     = -243

D_b = \begin{vmatrix}    3& 91 \\    6& 38   \end{vmatrix}     

     = 3×38 - 6×91 = 114 - 546 

     = -432

Now, find a = D_a/D, b = D_b/D

a = -243/-27 = 9, b = -432/-27 = 16

a = 9, b = 16

We have to find x and y, so we substitute the values into equation (1).

Now x² = a = 9, x = √9 = 3

y² = b = 16, y = √16 = +4 or -4           

Example 4: Solve \left\{ \begin{array}{c} 3x-4y+8z= 34\\ 4x+y-2z=1\\ -6x-13y+20z=61\\ \end{array} \right. 

The given equations in the form of AX = B

A = \begin{bmatrix}    3& -4& 8 \\    4&1&-2  \\-6&-13&20 \end{bmatrix}      

B =\begin{bmatrix}    34  \\    1 \\    61     \end{bmatrix}     

X =\begin{bmatrix}     x \\     y \\     z    \end{bmatrix}     

Then, the determinant D of matrix A = \begin{vmatrix}    3& -4& 8 \\    4&1&-2  \\-6&-13&20 \end{vmatrix}         = 3(20 - 26) - (-4)(80 - 12) + 8(-52-(-6)) = 3×(-6) + 4×68 - 46×8 = -18 + 272 - 368

= -114

Now, find D_x , D_y and D_z

D_x = \begin{vmatrix}    34& -4& 8 \\    1&1&-2  \\61&-13&20 \end{vmatrix}         

    = 34(20 - 26) - (-4)(20 + 122) + 8(-13 - 61) 

    = 34 × (-6) + 4 × 142 + 8 × (-74) = -204 + 568 - 592 = -228 

D_y = \begin{vmatrix}    3& 34& 8 \\    4&1&-2  \\-6&61&20 \end{vmatrix}         

    = 3(20 + 2 × 61) - 34(80 - 12) + 8(61 × 4 + 6) 

    = 3 × 142 - 34 × 68 + 8 × 250 = 426 - 2312 + 2000 = 114

D_z = \begin{vmatrix}    3& -4& 34 \\    4&1&1  \\-6&-13&61 \end{vmatrix}          

    = 3(61+13) - (-4)(61×4 + 6) + 34(-52+6) 

    = 3 × 74 + 4 × 250 + 34 × (-46) = 222 + 1000 -1564  = -342

Now, find x = D_x/D, y = D_y/D, z = D_z/D

x = -228/-114 = 2, y = 114/-114 = -1,  z = -342/-114 = 3   

x = 2,  y = -1,  z = 3

Example 5: Solve \left\{ \begin{array}{c} 3x-8y+10z= 8\\ -x+10y+9z=-15\\ 2x-6y+z=11\\ \end{array} \right. 

The given equations in the form of AX = B 

A = \begin{bmatrix}    3& -8& 10 \\    -1&10&9  \\2&-6&1 \end{bmatrix}      

B = \begin{bmatrix}    8  \\    -15 \\    11     \end{bmatrix}     

X = \begin{bmatrix}     x \\     y \\     z    \end{bmatrix}     

Then, the determinant D of matrix A = \begin{vmatrix}    3& -8& 10 \\    -1&10&9  \\2&-6&1 \end{vmatrix}         = 3(10+54) + 8(-1-18) +10(6-20)                                                                     

 = 3 × 64 - 8 × 19 + 10 × (-14) = 192 -152 - 140 = -100

Now, find D_x , D_y and D_z 

D_x = \begin{vmatrix}    8& -8& 10 \\    -15&10&9  \\11&-6&1 \end{vmatrix}         

     = 8(10+54) + 8(-15-99) + 10(90 -110) 

     = 8 × 64 + 8 × (-114) + 10 × (-20) = 512 - 912 - 200 = -600 

D_y = \begin{vmatrix}    3& 8& 10 \\    -1&-15&9  \\2&11&1 \end{vmatrix}          

    = 3(-15-99) - 8(-1-18) + 10(-11+30) 

    = 3 × (-114) + 8 × 19 + 10 × 19 = -342 + 152 +190 = 0

D_z = \begin{vmatrix}    3& -8& 8 \\    -1&10&-15  \\2&-6&11 \end{vmatrix}          

    = 3(110-90) + 8(-11+30) + 8(6-20) 

    = 3 × 20 + 8 × 19 + 8 × (-14) = 60 + 152 - 112 = 100 

Now, find x = D_x/D, y = D_y/D, z = D_z/D 

x = -600/-100 = 6, y = 0/-100 = 0, z = 100/-100 = -1

x = 6, y = 0, z = -1

Example 6: Solve \left\{ \begin{array}{c} 2x+4y-6z= 19\\ 3x+6y-9z=30\\ 4x-7y+z=15\\ \end{array} \right. 

The given equations in the form of AX = B  

A = \begin{bmatrix}    2& 4& -6 \\    3&6&-9  \\4&-7&1 \end{bmatrix}      

B = \begin{bmatrix}    19  \\    30 \\    15    \end{bmatrix}     

X =  \begin{bmatrix}     x \\     y \\     z    \end{bmatrix}     

Then, the determinant D of matrix A =  = 2(6 - 63) - 4(3 + 36) - 6(-21 - 24) = 2 × (-57) - 4 × 39 - 6 × (-45) = -114 - 156 + 270  = 0

Since |D| = 0, 

which means the given system of equations does not have a unique solution, which is invalid in Cramer's Rule as it is defined only for the system of equations that have a unique solution. This means that the given system of equations either has an infinite solution or no solution.

Example 7: Solve: \left\{ \begin{array}{c} x+y+z= 6\\ 5x-6y+8z=17\\ 2x+3y-z=5\\ \end{array} \right. 

The given equations in the form of AX = B  

A = \begin{bmatrix}    1& 1& 1 \\    5&-6&8  \\2&3&-1 \end{bmatrix}      

B = \begin{bmatrix}    6  \\    17 \\    5    \end{bmatrix}     

X = \begin{bmatrix}     x \\     y \\     z    \end{bmatrix}

Then, the determinant D of matrix A = \begin{vmatrix}    1& 1& 1 \\    5&-6&8  \\2&3&-1 \end{vmatrix}         = 1(6 - 24) - 1(-5 - 16) + 1(15 + 12)

= -18 + 21 + 27 = 30

Now, find D_x , D_y and D_z  

D_x = \begin{vmatrix}    6& 1& 1 \\    17&-6&8  \\5&3&-1 \end{vmatrix}         

     = 6(6-24) -1(-17-40) +1(51+30) 

     = 6(-18) + 57 + 81 = -108 + 138 = 30

D_y = \begin{vmatrix}    1& 6& 1 \\    5&17&8  \\2&5&-1 \end{vmatrix}         

     = 1(-17 - 40) - 6(-5 - 16) + 1(25 - 34) 

     = -57 + 126 - 9 = 60

D_z = \begin{vmatrix}    1& 1& 6 \\    5&-6&17  \\2&3&5 \end{vmatrix}         

     = 1(-30 - 51) - 1(25 - 34) + 6(15 + 12) 

     = -81 + 9 + 162 = 90  

Now, find x = D_x/D, y = D_y/D, z = D_z/D 

x = 30/30 = 1, y = 60/30 = 2, z = 90/30 = 3

x = 1,  y = 2, z = 3

Practice Problems

Question 1: Solve the following system using Cramer's Rule:

• 2x+3y = 5
• 4x+y = 11

Question 2: Find the values of 𝑥 y, and z using Cramer's Rule:

• x−2y+3z = 7
• 2x+y−z = 4
• -x+y+z = 2

Question 3: Determine if the following system has a unique solution using Cramer's Rule:

• 3x+5y = 9
• 6x+10y = 18

Question 4: Apply Cramer's Rule to solve:

• 5x+2y−z = 4
• -3x+4y+2z = −1
• 7x−y+5z = 10

Question 5: Use Cramer's Rule to solve for x and 𝑦:

• 4x+y = 3
• x−2y = 1

Answers:

1. x = 14/5, y = -1/5
2. x = 32/13, y = 23/13, z= 35/13
3. The given system of equations does not have a unique solution, which is invalid in Cramer's Rule as it is defined only for the system of equations that have a unique solution. This means that the given system of equations either has an infinite solution or no solution.
4. x = 177/193, y = 14/193, z = 141/193
5. x = 7/9, y = -1/9