Linear Algebra Examples

$5 b - 1 b = - 8$ , $- 1 a + 5 b = 16$

Step 1

Simplify.

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Step 1.1

Rewrite $- 1 b$ as $- b$ .

$5 b - b = - 8, - 1 a + 5 b = 16$

Step 1.2

Subtract $b$ from $5 b$ .

$4 b = - 8, - 1 a + 5 b = 16$

Step 2

Rewrite $- 1 a$ as $- a$ .

$4 b = - 8, - a + 5 b = 16$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} 0 & 4 & - 8 \\ - 1 & 5 & 16 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

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

$[\begin{array}{c} - 1 & 5 & 16 \\ 0 & 4 & - 8 \end{array}]$

Step 4.2

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

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Step 4.2.1

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

$[\begin{array}{c} - - 1 & - 1 \cdot 5 & - 1 \cdot 16 \\ 0 & 4 & - 8 \end{array}]$

Step 4.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 5 & - 16 \\ 0 & 4 & - 8 \end{array}]$

Step 4.3

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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Step 4.3.1

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 5 & - 16 \\ \frac{0}{4} & \frac{4}{4} & \frac{- 8}{4} \end{array}]$

Step 4.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 5 & - 16 \\ 0 & 1 & - 2 \end{array}]$

Step 4.4

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 4.4.1

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

$[\begin{array}{c} 1 + 5 \cdot 0 & - 5 + 5 \cdot 1 & - 16 + 5 \cdot - 2 \\ 0 & 1 & - 2 \end{array}]$

Step 4.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 26 \\ 0 & 1 & - 2 \end{array}]$

Step 5

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

$a = - 26$

$b = - 2$

Step 6

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

$(- 26, - 2)$

Step 7

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.

$X = [\begin{array}{c} a \\ b \end{array}] = [\begin{array}{c} - 26 \\ - 2 \end{array}]$