Solved Examples on Cramer’s Rule

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

Solution: 

The given equations in the form of AX = B 

A =[Tex]\begin{bmatrix}    12 & -10 \\    3 & 20   \end{bmatrix}         [/Tex] , 

B = [Tex]\begin{bmatrix}    46  \\    -11       \end{bmatrix}        [/Tex] , 

X = [Tex] \begin{bmatrix}     x \\     y    \end{bmatrix}[/Tex]

Then, the determinant D of matrix 

A = [Tex]\begin{vmatrix}    12 & -10 \\    3 & 20  \\ \end{vmatrix}       [/Tex] 

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

Now, find D_x and D_y 

D_x = [Tex]\begin{vmatrix}    46 & -10 \\    -11& 20  \\ \end{vmatrix}       [/Tex] 

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

D_y = [Tex]\begin{vmatrix}    12 & 46 \\    3 & -11 \\ \end{vmatrix}       [/Tex]  

     =[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 [Tex]\left\{ \begin{array}{c} 6.6x+0.95y= 5.2\\ 4.2x+8.6y=19.3 \\ \end{array} \right. [/Tex]

Solution:  

The given equations in the form of AX = B 

A = [Tex]\begin{bmatrix}    6.6& 0.95 \\    4.2& 8.6  \\ \end{bmatrix}       [/Tex]        

B = [Tex]\begin{bmatrix}    5.2  \\    19.3       \end{bmatrix}     [/Tex]

X  =  [Tex]\begin{bmatrix}     x \\     y    \end{bmatrix}[/Tex]

Then, the determinant D of matrix 

A = [Tex]\begin{vmatrix}    6.6& 0.95 \\    4.2& 8.6  \\ \end{vmatrix}       [/Tex] = 6.6×8.6 – 4.2×0.95 = 56.76 – 3.99 =52.77

Now, find D_x and D_y 

D_x = [Tex]\begin{vmatrix}    5.2& 0.95 \\    19.3& 8.6  \\ \end{vmatrix}       [/Tex] = 5.2 × 8.6 – 19.3 × 0.95 = 44.72 – 18.335 = 26.385

D_y = [Tex]\begin{vmatrix}    6.6& 5.2 \\    4.2& 19.3  \\ \end{vmatrix}       [/Tex] = 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 [Tex]\left\{ \begin{array}{c} 3x^2+4y^2= 91\\ 6x^2-y^2=38 \\ \end{array} \right. [/Tex]

Solution: 

Let x² = a, y² = b

Then, the equation can be written as,

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

The given equations in the form of AX = B

A = [Tex]\begin{bmatrix}    3& 4 \\    6&-1  \\ \end{bmatrix}      [/Tex]

B = [Tex]\begin{bmatrix}    91  \\    38     \end{bmatrix}     [/Tex]

X = [Tex]\begin{bmatrix}     a \\     b    \end{bmatrix}[/Tex]

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

Now, find D_a and D_b

D_a =[Tex]\begin{vmatrix}    91& 4 \\    38& -1   \end{vmatrix}       [/Tex] 

     = 91×(-1) – 38×4 = – 91 – 152 

     = -243

D_b = [Tex]\begin{vmatrix}    3& 91 \\    6& 38   \end{vmatrix}     [/Tex]

     = 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

Now x² = a = 9, x = √9 = 3

y² = b = 16, y = √16 = 4           

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

Solution: 

The given equations in the form of AX = B

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

B =[Tex]\begin{bmatrix}    34  \\    1 \\    61     \end{bmatrix}     [/Tex]

X =[Tex]\begin{bmatrix}     x \\     y \\     z    \end{bmatrix}     [/Tex]

Then, the determinant D of matrix A = [Tex]\begin{vmatrix}    3& -4& 8 \\    4&1&-2  \\-6&-13&20 \end{vmatrix}        [/Tex] = 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 = [Tex]\begin{vmatrix}    34& -4& 8 \\    1&1&-2  \\61&-13&20 \end{vmatrix}        [/Tex] 

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

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

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

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

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

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

    = 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 [Tex]\left\{ \begin{array}{c} 3x-8y+10z= 8\\ -x+10y+9z=-15\\ 2x-6y+z=11\\ \end{array} \right. [/Tex]

Solution: 

The given equations in the form of AX = B 

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

B = [Tex]\begin{bmatrix}    8  \\    -15 \\    11     \end{bmatrix}     [/Tex]

X = [Tex]\begin{bmatrix}     x \\     y \\     z    \end{bmatrix}     [/Tex]

Then, the determinant D of matrix A = [Tex]\begin{vmatrix}    3& -8& 10 \\    -1&10&9  \\2&-6&1 \end{vmatrix}        [/Tex] = 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 = [Tex]\begin{vmatrix}    8& -8& 10 \\    -15&10&9  \\11&-6&1 \end{vmatrix}        [/Tex] 

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

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

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

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

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

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

    = 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 [Tex]\left\{ \begin{array}{c} 2x+4y-6z= 19\\ 3x+6y-9z=30\\ 4x-7y+z=15\\ \end{array} \right. [/Tex]

Solution: 

The given equations in the form of AX = B  

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

B = [Tex]\begin{bmatrix}    19  \\    30 \\    15    \end{bmatrix}     [/Tex]

X =  [Tex]\begin{bmatrix}     x \\     y \\     z    \end{bmatrix}     [/Tex]

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: [Tex]\left\{ \begin{array}{c} x+y+z= 6\\ 5x-6y+8z=17\\ 2x+3y-z=5\\ \end{array} \right. [/Tex]

Solution: 

The given equations in the form of AX = B  

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

B = [Tex]\begin{bmatrix}    6  \\    17 \\    5    \end{bmatrix}     [/Tex]

X = [Tex]\begin{bmatrix}     x \\     y \\     z    \end{bmatrix}[/Tex]

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

= -18 + 21 + 27 = 30

Now, find D_x , D_y and D_z  

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

     = 6(6-24) -1(-17-40) +1(51+30) 

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

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

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

     = -57 + 126 – 9 = 60

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

     = 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

Cramer’s Rule is used to find the unknowns in the given system of linear equations. Cramer’s Rule is the most commonly used formula for finding the solution for the given system of linear equations in matrix form. Cramer’s Rule uses the concept of the determinant to find its solution.

Let’s know How to Apply Cramer’s Rule and its explanation. It requires some prior knowledge of matrices, determinants, and the system of linear equations.