Finite Math Examples Systems of Linear Equations

Solve Using a Matrix with Cramer's Rule

$4 x - y = - 4$ , $6 x - y = 0$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 4 & - 1 \\ 6 & - 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 4 \\ 0 \end{array}]$

Step 2

Find the determinant of the coefficient matrix $[\begin{array}{c} 4 & - 1 \\ 6 & - 1 \end{array}]$ .

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Step 2.1

Write $[\begin{array}{c} 4 & - 1 \\ 6 & - 1 \end{array}]$ in determinant notation.

$| \begin{array}{c} 4 & - 1 \\ 6 & - 1 \end{array} |$

Step 2.2

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$4 \cdot - 1 - 6 \cdot - 1$

Step 2.3

Simplify the determinant.

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Multiply $4$ by $- 1$ .

$- 4 - 6 \cdot - 1$

Step 2.3.1.2

Multiply $- 6$ by $- 1$ .

$- 4 + 6$

Step 2.3.2

Add $- 4$ and $6$ .

$2$

$D = 2$

Step 3

Since the determinant is not $0$ , the system can be solved using Cramer's Rule.

Step 4

Find the value of $x$ by Cramer's Rule, which states that $x = \frac{D_{x}}{D}$ .

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Step 4.1

Replace column $1$ of the coefficient matrix that corresponds to the $x$ -coefficients of the system with $[\begin{array}{c} - 4 \\ 0 \end{array}]$ .

$| \begin{array}{c} - 4 & - 1 \\ 0 & - 1 \end{array} |$

Step 4.2

Find the determinant.

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Step 4.2.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$- 4 \cdot - 1 + 0 \cdot - 1$

Step 4.2.2

Simplify the determinant.

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

Multiply $- 4$ by $- 1$ .

$4 + 0 \cdot - 1$

Step 4.2.2.1.2

Multiply $0$ by $- 1$ .

$4 + 0$

Step 4.2.2.2

Add $4$ and $0$ .

$4$

$D_{x} = 4$

Step 4.3

Use the formula to solve for $x$ .

$x = \frac{D_{x}}{D}$

Step 4.4

Substitute $2$ for $D$ and $4$ for $D_{x}$ in the formula.

$x = \frac{4}{2}$

Step 4.5

Divide $4$ by $2$ .

$x = 2$

Step 5

Find the value of $y$ by Cramer's Rule, which states that $y = \frac{D_{y}}{D}$ .

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Step 5.1

Replace column $2$ of the coefficient matrix that corresponds to the $y$ -coefficients of the system with $[\begin{array}{c} - 4 \\ 0 \end{array}]$ .

$| \begin{array}{c} 4 & - 4 \\ 6 & 0 \end{array} |$

Step 5.2

Find the determinant.

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Step 5.2.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$4 \cdot 0 - 6 \cdot - 4$

Step 5.2.2

Simplify the determinant.

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Step 5.2.2.1

Simplify each term.

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Step 5.2.2.1.1

Multiply $4$ by $0$ .

$0 - 6 \cdot - 4$

Step 5.2.2.1.2

Multiply $- 6$ by $- 4$ .

$0 + 24$

Step 5.2.2.2

Add $0$ and $24$ .

$24$

$D_{y} = 24$

Step 5.3

Use the formula to solve for $y$ .

$y = \frac{D_{y}}{D}$

Step 5.4

Substitute $2$ for $D$ and $24$ for $D_{y}$ in the formula.

$y = \frac{24}{2}$

Step 5.5

Divide $24$ by $2$ .

$y = 12$

Step 6

List the solution to the system of equations.

$x = 2$

$y = 12$

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Solve Using a Matrix with Cramer's Rule

Systems of Linear Equations

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