Linear Algebra Examples

$y + x = - 2$ , $3 + x = 3 - 2 y$

Step 1

Move all of the variables to the left side of each equation.

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Step 1.1

Reorder $y$ and $x$ .

$x + y = - 2$

$3 + x = 3 - 2 y$

Step 1.2

Add $2 y$ to both sides of the equation.

$x + y = - 2$

$3 + x + 2 y = 3$

Step 1.3

Move all terms not containing a variable to the right side of the equation.

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Step 1.3.1

Subtract $3$ from both sides of the equation.

$x + y = - 2$

$x + 2 y = 3 - 3$

Step 1.3.2

Subtract $3$ from $3$ .

$x + y = - 2$

$x + 2 y = 0$

$x + y = - 2$

$x + 2 y = 0$

$x + y = - 2$

$x + 2 y = 0$

Step 2

Represent the system of equations in matrix format.

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

Step 3

Find the determinant of the coefficient matrix $[\begin{array}{c} 1 & 1 \\ 1 & 2 \end{array}]$ .

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Step 3.1

Write $[\begin{array}{c} 1 & 1 \\ 1 & 2 \end{array}]$ in determinant notation.

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

Step 3.2

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

$1 \cdot 2 - 1 \cdot 1$

Step 3.3

Simplify the determinant.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Multiply $2$ by $1$ .

$2 - 1 \cdot 1$

Step 3.3.1.2

Multiply $- 1$ by $1$ .

$2 - 1$

Step 3.3.2

Subtract $1$ from $2$ .

$1$

$D = 1$

Step 4

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

Step 5

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

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Step 5.1

Replace column $1$ of the coefficient matrix that corresponds to the $x$ -coefficients of the system with $[\begin{array}{c} - 2 \\ 0 \end{array}]$ .

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

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

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

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 $- 2$ by $2$ .

$- 4 + 0 \cdot 1$

Step 5.2.2.1.2

Multiply $0$ by $1$ .

$- 4 + 0$

Step 5.2.2.2

Add $- 4$ and $0$ .

$- 4$

$D_{x} = - 4$

Step 5.3

Use the formula to solve for $x$ .

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

Step 5.4

Substitute $1$ for $D$ and $- 4$ for $D_{x}$ in the formula.

$x = \frac{- 4}{1}$

Step 5.5

Divide $- 4$ by $1$ .

$x = - 4$

Step 6

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

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Step 6.1

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

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

Step 6.2

Find the determinant.

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Step 6.2.1

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

$1 \cdot 0 - 1 \cdot - 2$

Step 6.2.2

Simplify the determinant.

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Step 6.2.2.1

Simplify each term.

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Step 6.2.2.1.1

Multiply $0$ by $1$ .

$0 - 1 \cdot - 2$

Step 6.2.2.1.2

Multiply $- 1$ by $- 2$ .

$0 + 2$

Step 6.2.2.2

Add $0$ and $2$ .

$2$

$D_{y} = 2$

Step 6.3

Use the formula to solve for $y$ .

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

Step 6.4

Substitute $1$ for $D$ and $2$ for $D_{y}$ in the formula.

$y = \frac{2}{1}$

Step 6.5

Divide $2$ by $1$ .

$y = 2$

Step 7

List the solution to the system of equations.

$x = - 4$

$y = 2$