Write as a Vector Equality x+y=3 , 2x-y=3

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

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & 3 \\ 2 & - 1 & 3 \end{array}]$

Step 2

Find the reduced row echelon form.

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

$[\begin{array}{c} 1 & 1 & 3 \\ 2 - 2 \cdot 1 & - 1 - 2 \cdot 1 & 3 - 2 \cdot 3 \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 3 \\ 0 & - 3 & - 3 \end{array}]$

Multiply each element of $R_{2}$ by $- \frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & 3 \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 3 \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 3 \\ 0 & 1 & 1 \end{array}]$

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 3 - 1 \\ 0 & 1 & 1 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 2 \\ 0 & 1 & 1 \end{array}]$

Step 3

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

$x = 2$

$y = 1$

Step 4

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

$(2, 1)$

Step 5

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 2 \\ 1 \end{array}]$

Write as a Vector Equality x+y=3 , 2x-y=3

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