Linear Algebra Examples

$5 x - y = - 2$ ,

$3 x - 4 y = 0$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 5 & - 1 \\ 3 & - 4 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 2 \\ 0 \end{array}]$

Step 2

Find the determinant of the coefficient matrix

$[\begin{array}{c} 5 & - 1 \\ 3 & - 4 \end{array}]$ .

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Step 2.1

Write

$[\begin{array}{c} 5 & - 1 \\ 3 & - 4 \end{array}]$ in determinant notation.

$| \begin{array}{c} 5 & - 1 \\ 3 & - 4 \end{array} |$

Step 2.2

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$5 \cdot - 4 - 3 \cdot - 1$

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

$5$ by

$- 4$ .

$- 20 - 3 \cdot - 1$

Step 2.3.1.2

Multiply

$- 3$ by

$- 1$ .

$- 20 + 3$

Step 2.3.2

Add

$- 20$ and

$3$ .

$- 17$

$D = - 17$

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

$[\begin{array}{c} - 2 \\ 0 \end{array}]$ .

$| \begin{array}{c} - 2 & - 1 \\ 0 & - 4 \end{array} |$

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

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$- 2 \cdot - 4 + 0 \cdot - 1$

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

$- 2$ by

$- 4$ .

$8 + 0 \cdot - 1$

Step 4.2.2.1.2

Multiply

$0$ by

$- 1$ .

$8 + 0$

Step 4.2.2.2

Add

$8$ and

$0$ .

$8$

$D_{x} = 8$

Step 4.3

Use the formula to solve for

$x$ .

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

Step 4.4

Substitute

$- 17$ for

$D$ and

$8$ for

$D_{x}$ in the formula.

$x = \frac{8}{- 17}$

Step 4.5

Move the negative in front of the fraction.

$x = - \frac{8}{17}$

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

$[\begin{array}{c} - 2 \\ 0 \end{array}]$ .

$| \begin{array}{c} 5 & - 2 \\ 3 & 0 \end{array} |$

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

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$5 \cdot 0 - 3 \cdot - 2$

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

$5$ by

$0$ .

$0 - 3 \cdot - 2$

Step 5.2.2.1.2

Multiply

$- 3$ by

$- 2$ .

$0 + 6$

Step 5.2.2.2

Add

$0$ and

$6$ .

$6$

$D_{y} = 6$

Step 5.3

Use the formula to solve for

$y$ .

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

Step 5.4

Substitute

$- 17$ for

$D$ and

$6$ for

$D_{y}$ in the formula.

$y = \frac{6}{- 17}$

Step 5.5

Move the negative in front of the fraction.

$y = - \frac{6}{17}$

Step 6

List the solution to the system of equations.

$x = - \frac{8}{17}$

$y = - \frac{6}{17}$

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