Linear Algebra Examples

$3 (1) + 6 (1) + 6 = 0$ , $8 (y) = 4 (2) + 24$

Step 1

Simplify.

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Step 1.1

Simplify each term.

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Step 1.1.1

Multiply $3$ by $1$ .

$3 + 6 (1) + 6 = 0, 8 (y) = 4 (2) + 24$

Step 1.1.2

Multiply $6$ by $1$ .

$3 + 6 + 6 = 0, 8 (y) = 4 (2) + 24$

Step 1.2

Add $3$ and $6$ .

$9 + 6 = 0, 8 (y) = 4 (2) + 24$

Step 1.3

Add $9$ and $6$ .

$15 = 0, 8 (y) = 4 (2) + 24$

Step 2

Subtract $15$ from both sides of the equation.

$0 = - 15, 8 (y) = 4 (2) + 24$

Step 3

Simplify.

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Step 3.1

Multiply $4$ by $2$ .

$0 = - 15, 8 y = 8 + 24$

Step 3.2

Add $8$ and $24$ .

$0 = - 15, 8 y = 32$

Step 4

Write the system of equations in matrix form.

$[\begin{array}{c} 0 & - 15 \\ 8 & 32 \end{array}]$

Step 5

Find the reduced row echelon form.

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Step 5.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 8 & 32 \\ 0 & - 15 \end{array}]$

Step 5.2

Multiply each element of $R_{1}$ by $\frac{1}{8}$ to make the entry at $1, 1$ a $1$ .

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Step 5.2.1

Multiply each element of $R_{1}$ by $\frac{1}{8}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{8}{8} & \frac{32}{8} \\ 0 & - 15 \end{array}]$

Step 5.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 4 \\ 0 & - 15 \end{array}]$

Step 5.3

Multiply each element of $R_{2}$ by $- \frac{1}{15}$ to make the entry at $2, 2$ a $1$ .

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Step 5.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{15}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 4 \\ - \frac{1}{15} \cdot 0 & - \frac{1}{15} \cdot - 15 \end{array}]$

Step 5.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 4 \\ 0 & 1 \end{array}]$

Step 5.4

Perform the row operation $R_{1} = R_{1} - 4 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 5.4.1

Perform the row operation $R_{1} = R_{1} - 4 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 4 \cdot 0 & 4 - 4 \cdot 1 \\ 0 & 1 \end{array}]$

Step 5.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Step 6

Use the result matrix to declare the final solutions to the system of equations.

$y = 0$

$0 = 1$

Step 7

Since $0 \neq 1$ , there are no solutions.

No solution