Linear Algebra Examples

$x + y = - 4$ , $y = - 3 - x - 3$

Step 1

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

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Step 1.1

Add $x$ to both sides of the equation.

$x + y = - 4$

$y + x = - 3 - 3$

Step 1.2

Subtract $3$ from $- 3$ .

$x + y = - 4$

$y + x = - 6$

Step 1.3

Reorder $y$ and $x$ .

$x + y = - 4$

$x + y = - 6$

$x + y = - 4$

$x + y = - 6$

Step 2

Represent the system of equations in matrix format.

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

$1 \cdot 1 - 1 \cdot 1$

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 $1$ by $1$ .

$1 - 1 \cdot 1$

Step 3.3.1.2

Multiply $- 1$ by $1$ .

$1 - 1$

Step 3.3.2

Subtract $1$ from $1$ .

$0$

$D = 0$

Step 4

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