Linear Algebra Examples

$y = - 2 - 3$ , $y = 1$

Step 1

Subtract $3$ from $- 2$ .

$y = - 5, y = 1$

Step 2

Write the system of equations in matrix form.

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

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} 1 & - 5 \\ 1 - 1 & 1 + 5 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 5 \\ 0 & 6 \end{array}]$

Step 3.2

Multiply each element of $R_{2}$ by $\frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

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Step 3.2.1

Multiply each element of $R_{2}$ by $\frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 5 \\ \frac{0}{6} & \frac{6}{6} \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 5 \\ 0 & 1 \end{array}]$

Step 3.3

Perform the row operation $R_{1} = R_{1} + 5 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{1} = R_{1} + 5 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + 5 \cdot 0 & - 5 + 5 \cdot 1 \\ 0 & 1 \end{array}]$

Step 3.3.2

Simplify $R_{1}$ .

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

Step 4

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

$y = 0$

$0 = 1$

Step 5

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

No solution