Linear Algebra Examples

${(3)}^{2} a + 3 b + c = 4$ , ${(5)}^{2} a + 5 b + c = 4$ , ${(2)}^{2} a + 2 b + c = 1$

Step 1

Move variables to the left and constant terms to the right.

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Step 1.1

Raise $3$ to the power of $2$ .

$9 a + 3 b + c = 4$

${(5)}^{2} a + 5 b + c = 4$

${(2)}^{2} a + 2 b + c = 1$

Step 1.2

Raise $5$ to the power of $2$ .

$9 a + 3 b + c = 4$

$25 a + 5 b + c = 4$

${(2)}^{2} a + 2 b + c = 1$

Step 1.3

Raise $2$ to the power of $2$ .

$9 a + 3 b + c = 4$

$25 a + 5 b + c = 4$

$4 a + 2 b + c = 1$

$9 a + 3 b + c = 4$

$25 a + 5 b + c = 4$

$4 a + 2 b + c = 1$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 9 & 3 & 1 & 4 \\ 25 & 5 & 1 & 4 \\ 4 & 2 & 1 & 1 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{9}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{9}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{9}{9} & \frac{3}{9} & \frac{1}{9} & \frac{4}{9} \\ 25 & 5 & 1 & 4 \\ 4 & 2 & 1 & 1 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{1}{9} & \frac{4}{9} \\ 25 & 5 & 1 & 4 \\ 4 & 2 & 1 & 1 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - 25 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} - 25 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{1}{9} & \frac{4}{9} \\ 25 - 25 \cdot 1 & 5 - 25 (\frac{1}{3}) & 1 - 25 (\frac{1}{9}) & 4 - 25 (\frac{4}{9}) \\ 4 & 2 & 1 & 1 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{1}{9} & \frac{4}{9} \\ 0 & - \frac{10}{3} & - \frac{16}{9} & - \frac{64}{9} \\ 4 & 2 & 1 & 1 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{1}{9} & \frac{4}{9} \\ 0 & - \frac{10}{3} & - \frac{16}{9} & - \frac{64}{9} \\ 4 - 4 \cdot 1 & 2 - 4 (\frac{1}{3}) & 1 - 4 (\frac{1}{9}) & 1 - 4 (\frac{4}{9}) \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{1}{9} & \frac{4}{9} \\ 0 & - \frac{10}{3} & - \frac{16}{9} & - \frac{64}{9} \\ 0 & \frac{2}{3} & \frac{5}{9} & - \frac{7}{9} \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $- \frac{3}{10}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $- \frac{3}{10}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{1}{9} & \frac{4}{9} \\ - \frac{3}{10} \cdot 0 & - \frac{3}{10} (- \frac{10}{3}) & - \frac{3}{10} (- \frac{16}{9}) & - \frac{3}{10} (- \frac{64}{9}) \\ 0 & \frac{2}{3} & \frac{5}{9} & - \frac{7}{9} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{1}{9} & \frac{4}{9} \\ 0 & 1 & \frac{8}{15} & \frac{32}{15} \\ 0 & \frac{2}{3} & \frac{5}{9} & - \frac{7}{9} \end{array}]$

Step 3.5

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

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Step 3.5.1

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

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{1}{9} & \frac{4}{9} \\ 0 & 1 & \frac{8}{15} & \frac{32}{15} \\ 0 - \frac{2}{3} \cdot 0 & \frac{2}{3} - \frac{2}{3} \cdot 1 & \frac{5}{9} - \frac{2}{3} \cdot \frac{8}{15} & - \frac{7}{9} - \frac{2}{3} \cdot \frac{32}{15} \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{1}{9} & \frac{4}{9} \\ 0 & 1 & \frac{8}{15} & \frac{32}{15} \\ 0 & 0 & \frac{1}{5} & - \frac{11}{5} \end{array}]$

Step 3.6

Multiply each element of $R_{3}$ by $5$ to make the entry at $3, 3$ a $1$ .

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Step 3.6.1

Multiply each element of $R_{3}$ by $5$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{1}{9} & \frac{4}{9} \\ 0 & 1 & \frac{8}{15} & \frac{32}{15} \\ 5 \cdot 0 & 5 \cdot 0 & 5 (\frac{1}{5}) & 5 (- \frac{11}{5}) \end{array}]$

Step 3.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{1}{9} & \frac{4}{9} \\ 0 & 1 & \frac{8}{15} & \frac{32}{15} \\ 0 & 0 & 1 & - 11 \end{array}]$

Step 3.7

Perform the row operation $R_{2} = R_{2} - \frac{8}{15} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.7.1

Perform the row operation $R_{2} = R_{2} - \frac{8}{15} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{1}{9} & \frac{4}{9} \\ 0 - \frac{8}{15} \cdot 0 & 1 - \frac{8}{15} \cdot 0 & \frac{8}{15} - \frac{8}{15} \cdot 1 & \frac{32}{15} - \frac{8}{15} \cdot - 11 \\ 0 & 0 & 1 & - 11 \end{array}]$

Step 3.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & \frac{1}{9} & \frac{4}{9} \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & - 11 \end{array}]$

Step 3.8

Perform the row operation $R_{1} = R_{1} - \frac{1}{9} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 3.8.1

Perform the row operation $R_{1} = R_{1} - \frac{1}{9} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{1}{9} \cdot 0 & \frac{1}{3} - \frac{1}{9} \cdot 0 & \frac{1}{9} - \frac{1}{9} \cdot 1 & \frac{4}{9} - \frac{1}{9} \cdot - 11 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & - 11 \end{array}]$

Step 3.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{3} & 0 & \frac{5}{3} \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & - 11 \end{array}]$

Step 3.9

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

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Step 3.9.1

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

$[\begin{array}{c} 1 - \frac{1}{3} \cdot 0 & \frac{1}{3} - \frac{1}{3} \cdot 1 & 0 - \frac{1}{3} \cdot 0 & \frac{5}{3} - \frac{1}{3} \cdot 8 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & - 11 \end{array}]$

Step 3.9.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - 1 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & - 11 \end{array}]$

Step 4

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

$a = - 1$

$b = 8$

$c = - 11$

Step 5

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

$(- 1, 8, - 11)$