Linear Algebra Examples

$a + 3 b = 1$ , $3 a - 2 b = 14$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 3 & 1 \\ 3 & - 2 & 14 \end{array}]$

Step 2

Find the reduced row echelon form.

Tap for more steps...

Step 2.1

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

Tap for more steps...

Step 2.1.1

Perform the row operation $R_{2} = R_{2} - 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 3 & 1 \\ 3 - 3 \cdot 1 & - 2 - 3 \cdot 3 & 14 - 3 \cdot 1 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 3 & 1 \\ 0 & - 11 & 11 \end{array}]$

Step 2.2

Multiply each element of $R_{2}$ by $- \frac{1}{11}$ to make the entry at $2, 2$ a $1$ .

Tap for more steps...

Step 2.2.1

Multiply each element of $R_{2}$ by $- \frac{1}{11}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 3 & 1 \\ - \frac{1}{11} \cdot 0 & - \frac{1}{11} \cdot - 11 & - \frac{1}{11} \cdot 11 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

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

Step 2.3

Perform the row operation $R_{1} = R_{1} - 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

Tap for more steps...

Step 2.3.1

Perform the row operation $R_{1} = R_{1} - 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Step 2.3.2

Simplify $R_{1}$ .

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

Step 3

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

$a = 4$

$b = - 1$

Step 4

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

$(4, - 1)$