This article elaborates on how to solve systems of linear equations by applying a theorem called the Cramer’s rule. This rule is an alternative to the elimination and substitution techniques. At the first stage of applying the rule, we have to rewrite the given system of linear equations as a matrix equation. More specifically, suppose that we have a system of linear equations in k unknowns as follows.

It can be expressed as the matrix equation below.

It has the form of AX = B, where

Note: A is called coefficient matrix.

 

Theorem (Cramer’s Rule)

Let AX = B be a k×k system of linear equations with nonsingular coefficient matrix A = (aij). Let Ai be the matrix obtained from A by replacing the i th column of A by the column vector B. Let D = |A| and Di = |Ai| for i = 1, 2, 3, …, k. Then the system has the unique solution x_i = \frac{D_i}{D}. (Note: D is the determinant of A and Di is the determinant of Ai for i = 1, 2, …, k.)

 

Example 1

Find the solution set for the system of linear equations below.

\left \{ \begin{matrix}3x - 5y = 1 \\ -2x + 7y = 3 \end{matrix}

Answer

Let x = x1 and y = x2. Then the system of linear equations can be expressed as AX = B where

To find A1 , replace the first column of A with B. This gives A_1 = \begin{pmatrix} 1 & -5 \\ 3 & 7 \end{pmatrix}. Similarly, to find A2 , replace the second column of A with B. This gives A_2 = \begin{pmatrix} 3 & 1 \\ -2 & 3 \end{pmatrix}. Now, compute D, D1 , and D2 as follows.

D = |A| = \left| \begin{matrix} 3 & -5 \\ -2 & 7 \end{matrix} \right| = 3 \cdot 7 - (-5) \cdot (-2) = 11

D_1 = |A_1| = \left| \begin{matrix} 1 & -5 \\ 3 & 7 \end{matrix} \right| = 1 \cdot 7 - (-5) \cdot 3 = 22

D_2 = |A_2| = \left| \begin{matrix} 3 & 1 \\ -2 & 3 \end{matrix} \right| = 3 \cdot 3 - 1 \cdot (-2) = 11

Consequently, x_1 = \frac{D_1}{D} = \frac{22}{11} = 2 and x_2 = \frac{D_2}{D} = \frac{11}{11} = 1. Thus, the solution set is {(2,1)}.

 

Example 2

Find the solution set for the system of linear equations below.

\left \{ \begin{matrix}3x_1 + 2x_2 = 5 \\ -7x_1 + 3x_2 = -27 \end{matrix}

Answer

The corresponding matrix equation for the system is AX = B where A = \begin{pmatrix} 3 & 2 \\ -7 & 3 \end{pmatrix}, X = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}, and B = \begin{pmatrix} 5 \\ -27 \end{pmatrix}. By following the similar procedure as in Example 1, we then have:

D = |A| = \left| \begin{matrix} 3 & 2 \\ -7 & 3 \end{matrix} \right| = 3 \cdot 3 - 2 \cdot (-7) = 23

D_1 = |A_1| = \left| \begin{matrix} 5 & 2 \\ -27 & 3 \end{matrix} \right| = 5 \cdot 3 - 2 \cdot (-27) = 69

D_2 = |A_2| = \left| \begin{matrix} 3 & 5 \\ -7 & -27 \end{matrix} \right| = 3 \cdot (-27) - 5 \cdot (-7) = -46

Consequently, x_1 = \frac{D_1}{D} = \frac{69}{23} = 3 and x_2 = \frac{D_2}{D} = \frac{-46}{23} = -2. Thus, the solution set is {(3,-2)}.

 

Example 3

Find the solution set for the system of linear equations below.

\left \{ \begin{matrix}3x - 5y + 2z = 3 \\ 2x + 3y - 4z = 3 \\ 4x - 5y - 3z = 0 \end{matrix}

Answer

Let x = x1, y = x2, and z = x3. Then the system of linear equations can be expressed as AX = B where A = \begin{pmatrix} 3 & -5 & 2 \\ 2 & 3 & -4 \\ 4 & -5 & -3 \end{pmatrix}, X = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, and B = \begin{pmatrix} 3 \\ 3 \\ 0 \end{pmatrix}.

By following the similar procedure as in Example 1, we then have:

Now, compute D, D1 , D2 and D3 as follows.

Accordingly, x_1 = \frac{D_1}{D} = \frac{-162}{-81} = 2, x_2 = \frac{D_2}{D} = \frac{-81}{-81} = 1, and x_3 = \frac{D_3}{D} = \frac{-81}{-81} = 1. So, the solution set is {(2,1,1)}.

 

In fact, there are some other ways to solve a system of linear equations, such as by using the inverse of the coefficient matrix and Gauss-Jordan elimination method. These alternatives will be discussed in some other articles on this website.

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