Linear Algebra Examples

$4 + y = 3$ , $y = - 3 - x + 2$

Step 1

Subtract $4$ from both sides of the equation.

$y = 3 - 4, y = - 3 - x + 2$

Step 2

Subtract $4$ from $3$ .

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

Step 3

Add $- 3$ and $2$ .

$y = - 1, y = - x - 1$

Step 4

Add $x$ to both sides of the equation.

$y = - 1, y + x = - 1$

Step 5

Write the system of equations in matrix form.

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

Step 6

Find the reduced row echelon form.

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Step 6.1

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

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

Step 6.2

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

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Step 6.2.1

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

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

Step 6.2.2

Simplify $R_{1}$ .

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

Step 7

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

$x = 0$

$y = - 1$

Step 8

The solution is the set of ordered pairs that makes the system true.

$(0, - 1)$

Step 9

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

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