Linear Algebra Examples

$\frac{1}{5} a + \frac{1}{2} Y = \frac{15}{5}$ , $\frac{1}{5} a + 3 y = \frac{88}{5}$

Step 1

Simplify.

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Step 1.1

Simplify each term.

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Step 1.1.1

Combine $\frac{1}{5}$ and $a$ .

$\frac{a}{5} + \frac{1}{2} Y = \frac{15}{5}, \frac{1}{5} a + 3 y = \frac{88}{5}$

Step 1.1.2

Combine $\frac{1}{2}$ and $Y$ .

$\frac{a}{5} + \frac{Y}{2} = \frac{15}{5}, \frac{1}{5} a + 3 y = \frac{88}{5}$

Step 1.2

Divide $15$ by $5$ .

$\frac{a}{5} + \frac{Y}{2} = 3, \frac{1}{5} a + 3 y = \frac{88}{5}$

Step 2

Combine $\frac{1}{5}$ and $a$ .

$\frac{a}{5} + \frac{Y}{2} = 3, \frac{a}{5} + 3 y = \frac{88}{5}$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} \frac{1}{5} & 0 & \frac{1}{2} & 3 \\ \frac{1}{5} & 3 & 0 & \frac{88}{5} \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

Multiply each element of $R_{1}$ by $5$ to make the entry at $1, 1$ a $1$ .

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Step 4.1.1

Multiply each element of $R_{1}$ by $5$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} 5 (\frac{1}{5}) & 5 \cdot 0 & 5 (\frac{1}{2}) & 5 \cdot 3 \\ \frac{1}{5} & 3 & 0 & \frac{88}{5} \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{5}{2} & 15 \\ \frac{1}{5} & 3 & 0 & \frac{88}{5} \end{array}]$

Step 4.2

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

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Step 4.2.1

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

$[\begin{array}{c} 1 & 0 & \frac{5}{2} & 15 \\ \frac{1}{5} - \frac{1}{5} \cdot 1 & 3 - \frac{1}{5} \cdot 0 & 0 - \frac{1}{5} \cdot \frac{5}{2} & \frac{88}{5} - \frac{1}{5} \cdot 15 \end{array}]$

Step 4.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & \frac{5}{2} & 15 \\ 0 & 3 & - \frac{1}{2} & \frac{73}{5} \end{array}]$

Step 4.3

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

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Step 4.3.1

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

$[\begin{array}{c} 1 & 0 & \frac{5}{2} & 15 \\ \frac{0}{3} & \frac{3}{3} & \frac{- \frac{1}{2}}{3} & \frac{\frac{73}{5}}{3} \end{array}]$

Step 4.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & \frac{5}{2} & 15 \\ 0 & 1 & - \frac{1}{6} & \frac{73}{15} \end{array}]$

Step 5

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

$a + \frac{5 Y}{2} = 15$

$y - \frac{Y}{6} = \frac{73}{15}$

Step 6

Subtract $\frac{5 Y}{2}$ from both sides of the equation.

$a = 15 - \frac{5 Y}{2}$

$y - \frac{Y}{6} = \frac{73}{15}$

Step 7

Add $\frac{Y}{6}$ to both sides of the equation.

$y = \frac{73}{15} + \frac{Y}{6}$

$a = 15 - \frac{5 Y}{2}$

Step 8

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

$(15 - \frac{5 Y}{2}, \frac{73}{15} + \frac{Y}{6}, Y)$

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.

$X = [\begin{array}{c} a \\ y \\ Y \end{array}] = [\begin{array}{c} 15 - \frac{5 Y}{2} \\ \frac{73}{15} + \frac{Y}{6} \\ Y \end{array}]$