Linear Algebra Examples

$10 = d$ , $7 = a + b + c + d$ , $- 11 = 27 a + 9 b + 3 c + d$ , $- 14 = 64 a + 16 b + 4 c + d$

Step 1

Subtract $10$ from both sides of the equation.

$0 = d - 10, 7 = a + b + c + d, - 11 = 27 a + 9 b + 3 c + d, - 14 = 64 a + 16 b + 4 c + d$

Step 2

Subtract $d$ from both sides of the equation.

$- d = - 10, 7 = a + b + c + d, - 11 = 27 a + 9 b + 3 c + d, - 14 = 64 a + 16 b + 4 c + d$

Step 3

Subtract $7$ from both sides of the equation.

$- d = - 10, 0 = a + b + c + d - 7, - 11 = 27 a + 9 b + 3 c + d, - 14 = 64 a + 16 b + 4 c + d$

Step 4

Move all terms containing variables to the left side of the equation.

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Step 4.1

Subtract $a$ from both sides of the equation.

$- d = - 10, - a = b + c + d - 7, - 11 = 27 a + 9 b + 3 c + d, - 14 = 64 a + 16 b + 4 c + d$

Step 4.2

Subtract $b$ from both sides of the equation.

$- d = - 10, - a - b = c + d - 7, - 11 = 27 a + 9 b + 3 c + d, - 14 = 64 a + 16 b + 4 c + d$

Step 4.3

Subtract $c$ from both sides of the equation.

$- d = - 10, - a - b - c = d - 7, - 11 = 27 a + 9 b + 3 c + d, - 14 = 64 a + 16 b + 4 c + d$

Step 4.4

Subtract $d$ from both sides of the equation.

$- d = - 10, - a - b - c - d = - 7, - 11 = 27 a + 9 b + 3 c + d, - 14 = 64 a + 16 b + 4 c + d$

Step 5

Add $11$ to both sides of the equation.

$- d = - 10, - a - b - c - d = - 7, 0 = 27 a + 9 b + 3 c + d + 11, - 14 = 64 a + 16 b + 4 c + d$

Step 6

Move all terms containing variables to the left side of the equation.

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Step 6.1

Subtract $27 a$ from both sides of the equation.

$- d = - 10, - a - b - c - d = - 7, - 27 a = 9 b + 3 c + d + 11, - 14 = 64 a + 16 b + 4 c + d$

Step 6.2

Subtract $9 b$ from both sides of the equation.

$- d = - 10, - a - b - c - d = - 7, - 27 a - 9 b = 3 c + d + 11, - 14 = 64 a + 16 b + 4 c + d$

Step 6.3

Subtract $3 c$ from both sides of the equation.

$- d = - 10, - a - b - c - d = - 7, - 27 a - 9 b - 3 c = d + 11, - 14 = 64 a + 16 b + 4 c + d$

Step 6.4

Subtract $d$ from both sides of the equation.

$- d = - 10, - a - b - c - d = - 7, - 27 a - 9 b - 3 c - d = 11, - 14 = 64 a + 16 b + 4 c + d$

Step 7

Add $14$ to both sides of the equation.

$- d = - 10, - a - b - c - d = - 7, - 27 a - 9 b - 3 c - d = 11, 0 = 64 a + 16 b + 4 c + d + 14$

Step 8

Move all terms containing variables to the left side of the equation.

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Step 8.1

Subtract $64 a$ from both sides of the equation.

$- d = - 10, - a - b - c - d = - 7, - 27 a - 9 b - 3 c - d = 11, - 64 a = 16 b + 4 c + d + 14$

Step 8.2

Subtract $16 b$ from both sides of the equation.

$- d = - 10, - a - b - c - d = - 7, - 27 a - 9 b - 3 c - d = 11, - 64 a - 16 b = 4 c + d + 14$

Step 8.3

Subtract $4 c$ from both sides of the equation.

$- d = - 10, - a - b - c - d = - 7, - 27 a - 9 b - 3 c - d = 11, - 64 a - 16 b - 4 c = d + 14$

Step 8.4

Subtract $d$ from both sides of the equation.

$- d = - 10, - a - b - c - d = - 7, - 27 a - 9 b - 3 c - d = 11, - 64 a - 16 b - 4 c - d = 14$

Step 9

Write the system of equations in matrix form.

$[\begin{array}{c} 0 & 0 & 0 & - 1 & - 10 \\ - 1 & - 1 & - 1 & - 1 & - 7 \\ - 27 & - 9 & - 3 & - 1 & 11 \\ - 64 & - 16 & - 4 & - 1 & 14 \end{array}]$

Step 10

Find the reduced row echelon form.

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Step 10.1

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

$[\begin{array}{c} - 1 & - 1 & - 1 & - 1 & - 7 \\ 0 & 0 & 0 & - 1 & - 10 \\ - 27 & - 9 & - 3 & - 1 & 11 \\ - 64 & - 16 & - 4 & - 1 & 14 \end{array}]$

Step 10.2

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

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Step 10.2.1

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

$[\begin{array}{c} - - 1 & - - 1 & - - 1 & - - 1 & - - 7 \\ 0 & 0 & 0 & - 1 & - 10 \\ - 27 & - 9 & - 3 & - 1 & 11 \\ - 64 & - 16 & - 4 & - 1 & 14 \end{array}]$

Step 10.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 0 & 0 & - 1 & - 10 \\ - 27 & - 9 & - 3 & - 1 & 11 \\ - 64 & - 16 & - 4 & - 1 & 14 \end{array}]$

Step 10.3

Perform the row operation $R_{3} = R_{3} + 27 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 10.3.1

Perform the row operation $R_{3} = R_{3} + 27 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 0 & 0 & - 1 & - 10 \\ - 27 + 27 \cdot 1 & - 9 + 27 \cdot 1 & - 3 + 27 \cdot 1 & - 1 + 27 \cdot 1 & 11 + 27 \cdot 7 \\ - 64 & - 16 & - 4 & - 1 & 14 \end{array}]$

Step 10.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 0 & 0 & - 1 & - 10 \\ 0 & 18 & 24 & 26 & 200 \\ - 64 & - 16 & - 4 & - 1 & 14 \end{array}]$

Step 10.4

Perform the row operation $R_{4} = R_{4} + 64 R_{1}$ to make the entry at $4, 1$ a $0$ .

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Step 10.4.1

Perform the row operation $R_{4} = R_{4} + 64 R_{1}$ to make the entry at $4, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 0 & 0 & - 1 & - 10 \\ 0 & 18 & 24 & 26 & 200 \\ - 64 + 64 \cdot 1 & - 16 + 64 \cdot 1 & - 4 + 64 \cdot 1 & - 1 + 64 \cdot 1 & 14 + 64 \cdot 7 \end{array}]$

Step 10.4.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 0 & 0 & - 1 & - 10 \\ 0 & 18 & 24 & 26 & 200 \\ 0 & 48 & 60 & 63 & 462 \end{array}]$

Step 10.5

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 18 & 24 & 26 & 200 \\ 0 & 0 & 0 & - 1 & - 10 \\ 0 & 48 & 60 & 63 & 462 \end{array}]$

Step 10.6

Multiply each element of $R_{2}$ by $\frac{1}{18}$ to make the entry at $2, 2$ a $1$ .

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Step 10.6.1

Multiply each element of $R_{2}$ by $\frac{1}{18}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ \frac{0}{18} & \frac{18}{18} & \frac{24}{18} & \frac{26}{18} & \frac{200}{18} \\ 0 & 0 & 0 & - 1 & - 10 \\ 0 & 48 & 60 & 63 & 462 \end{array}]$

Step 10.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 1 & \frac{4}{3} & \frac{13}{9} & \frac{100}{9} \\ 0 & 0 & 0 & - 1 & - 10 \\ 0 & 48 & 60 & 63 & 462 \end{array}]$

Step 10.7

Perform the row operation $R_{4} = R_{4} - 48 R_{2}$ to make the entry at $4, 2$ a $0$ .

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Step 10.7.1

Perform the row operation $R_{4} = R_{4} - 48 R_{2}$ to make the entry at $4, 2$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 1 & \frac{4}{3} & \frac{13}{9} & \frac{100}{9} \\ 0 & 0 & 0 & - 1 & - 10 \\ 0 - 48 \cdot 0 & 48 - 48 \cdot 1 & 60 - 48 (\frac{4}{3}) & 63 - 48 (\frac{13}{9}) & 462 - 48 (\frac{100}{9}) \end{array}]$

Step 10.7.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 1 & \frac{4}{3} & \frac{13}{9} & \frac{100}{9} \\ 0 & 0 & 0 & - 1 & - 10 \\ 0 & 0 & - 4 & - \frac{19}{3} & - \frac{214}{3} \end{array}]$

Step 10.8

Swap $R_{4}$ with $R_{3}$ to put a nonzero entry at $3, 3$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 1 & \frac{4}{3} & \frac{13}{9} & \frac{100}{9} \\ 0 & 0 & - 4 & - \frac{19}{3} & - \frac{214}{3} \\ 0 & 0 & 0 & - 1 & - 10 \end{array}]$

Step 10.9

Multiply each element of $R_{3}$ by $- \frac{1}{4}$ to make the entry at $3, 3$ a $1$ .

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Step 10.9.1

Multiply each element of $R_{3}$ by $- \frac{1}{4}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 1 & \frac{4}{3} & \frac{13}{9} & \frac{100}{9} \\ - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot - 4 & - \frac{1}{4} (- \frac{19}{3}) & - \frac{1}{4} (- \frac{214}{3}) \\ 0 & 0 & 0 & - 1 & - 10 \end{array}]$

Step 10.9.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 1 & \frac{4}{3} & \frac{13}{9} & \frac{100}{9} \\ 0 & 0 & 1 & \frac{19}{12} & \frac{107}{6} \\ 0 & 0 & 0 & - 1 & - 10 \end{array}]$

Step 10.10

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

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Step 10.10.1

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

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 1 & \frac{4}{3} & \frac{13}{9} & \frac{100}{9} \\ 0 & 0 & 1 & \frac{19}{12} & \frac{107}{6} \\ - 0 & - 0 & - 0 & - - 1 & - - 10 \end{array}]$

Step 10.10.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 1 & \frac{4}{3} & \frac{13}{9} & \frac{100}{9} \\ 0 & 0 & 1 & \frac{19}{12} & \frac{107}{6} \\ 0 & 0 & 0 & 1 & 10 \end{array}]$

Step 10.11

Perform the row operation $R_{3} = R_{3} - \frac{19}{12} R_{4}$ to make the entry at $3, 4$ a $0$ .

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Step 10.11.1

Perform the row operation $R_{3} = R_{3} - \frac{19}{12} R_{4}$ to make the entry at $3, 4$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 1 & \frac{4}{3} & \frac{13}{9} & \frac{100}{9} \\ 0 - \frac{19}{12} \cdot 0 & 0 - \frac{19}{12} \cdot 0 & 1 - \frac{19}{12} \cdot 0 & \frac{19}{12} - \frac{19}{12} \cdot 1 & \frac{107}{6} - \frac{19}{12} \cdot 10 \\ 0 & 0 & 0 & 1 & 10 \end{array}]$

Step 10.11.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 & 1 & \frac{4}{3} & \frac{13}{9} & \frac{100}{9} \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 10 \end{array}]$

Step 10.12

Perform the row operation $R_{2} = R_{2} - \frac{13}{9} R_{4}$ to make the entry at $2, 4$ a $0$ .

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Step 10.12.1

Perform the row operation $R_{2} = R_{2} - \frac{13}{9} R_{4}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 1 & 7 \\ 0 - \frac{13}{9} \cdot 0 & 1 - \frac{13}{9} \cdot 0 & \frac{4}{3} - \frac{13}{9} \cdot 0 & \frac{13}{9} - \frac{13}{9} \cdot 1 & \frac{100}{9} - \frac{13}{9} \cdot 10 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 10 \end{array}]$

Step 10.12.2

Simplify $R_{2}$ .

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

Step 10.13

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

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Step 10.13.1

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

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

Step 10.13.2

Simplify $R_{1}$ .

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

Step 10.14

Perform the row operation $R_{2} = R_{2} - \frac{4}{3} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 10.14.1

Perform the row operation $R_{2} = R_{2} - \frac{4}{3} R_{3}$ to make the entry at $2, 3$ a $0$ .

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

Step 10.14.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 0 & - 3 \\ 0 & 1 & 0 & 0 & - 6 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 10 \end{array}]$

Step 10.15

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

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Step 10.15.1

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

$[\begin{array}{c} 1 - 0 & 1 - 0 & 1 - 1 & 0 - 0 & - 3 - 2 \\ 0 & 1 & 0 & 0 & - 6 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 10 \end{array}]$

Step 10.15.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 0 & 0 & - 5 \\ 0 & 1 & 0 & 0 & - 6 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 10 \end{array}]$

Step 10.16

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

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Step 10.16.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & 0 - 0 & - 5 + 6 \\ 0 & 1 & 0 & 0 & - 6 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 10 \end{array}]$

Step 10.16.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & - 6 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 10 \end{array}]$

Step 11

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

$a = 1$

$b = - 6$

$c = 2$

$d = 10$

Step 12

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

$(1, - 6, 2, 10)$

Step 13

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.

$X = [\begin{array}{c} a \\ b \\ c \\ d \end{array}] = [\begin{array}{c} 1 \\ - 6 \\ 2 \\ 10 \end{array}]$