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Cramer’s Rule

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  • Difficulty Level : Medium
  • Last Updated : 13 May, 2022
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Matrices are one of the important topics in mathematics. Matrices and determinants are used to find the solutions to different systems of equations. Cramer’s Rule is used to find the unknowns in the given system of linear equations. Let’s how to apply Cramer’s Rule and its explanation. It requires some prior knowledge of matrices, determinants, and the system of linear equations.

Cramer’s Rule 

Cramer’s rule is a rule which is used to find the unknowns from the given set of linear equations. This rule is valid only if the given system of equations has a unique solution. It doesn’t work with the system of equations with infinitely many solutions or no solution. This rule is used to find solutions for any number of variables with the same number of equations. This rule uses determinants to find the solution of the given equations or the value of unknowns.

Cramer’s Rule Formula

Cramer’s Rule Formula to solve system AX = B ( where, A= coefficient matrix, B =column matrix of constants (RHS), X = column matrix of unknowns) or to find the values of the variables involves the following steps,

Steps of Cramer’s Rule Formula 

  1. Write the given system of equations in AX = B form.
  2. Find the value of determinant (D) of matrix A. (Note: If the determinant is zero, then a system of equations does not have a unique solution, which is invalid in Cramer’s Rule).
  3. Now, find the value Dx which is the determinant of matrix A in which constants of the given linear equations replace the coefficient of x. 
  4. Now, find the value Dy which is the determinant of matrix A in which the coefficient of y is replaced by constants of the given linear equations. 
  5. Now, find the value Dz which is the determinant of matrix A in which the coefficient of z is replaced by constants of the given linear equations. (find this determinant only if 3 variables present in the given equation).
  6. Similarly, find determinants for all the unknowns if more than three unknowns are present.
  7. Find the values of x = Dx/D , y = Dy/D , z = Dz/D.

Sample Questions 

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

Solution: 

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 Dx and D

Dx \begin{vmatrix}    46 & -10 \\    -11& 20  \\ \end{vmatrix}  = [46×20 – (-10)×(-11)] = 920 – 110 = 810

Dy\begin{vmatrix}    12 & 46 \\    3 & -11 \\ \end{vmatrix}   =[12×(-11) – 3×46] = -132 -138 = -270

Now, find x = Dx/D, y = Dy/D 

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

x = 3, y = -1

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

Solution:  

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 Dx and Dy 

Dx\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

Dy\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 = Dx/D  , y = Dy/D

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

x = 0.5,  y = 2

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

Solution: 

Let x2 = a, y2 = b

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 Da and Db

Da =\begin{vmatrix}    91& 4 \\    38& -1   \end{vmatrix}  = 91×(-1) – 38×4 = – 91 – 152 = -243

Db\begin{vmatrix}    3& 91 \\    6& 38   \end{vmatrix} = 3×38 – 6×91 = 114 – 546 = -432

Now, find a = Da/D, b = Db/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           

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

Solution: 

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 Dx , Dy and Dz

Dx \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 

Dy\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

Dz\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 = Dx/D, y = Dy/D, z = Dz/D

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

x = 2,  y = -1,  z = 3

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

Solution: 

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 Dx , Dy and D

Dx\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 

Dy\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

Dz\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 = Dx/D, y = Dy/D, z = Dz/D 

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

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

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

Solution: 

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.

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

Solution: 

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 Dx , Dy and Dz  

Dx\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

Dy\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

Dz\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 = Dx/D, y = Dy/D, z = Dz/D 

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

x = 1,  y = 2, z = 3


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