Linear Algebra Examples

$[\begin{array}{c} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{array}] \cdot [\begin{array}{c} x \\ y \\ v \\ z \end{array}] = [\begin{array}{c} 110 \\ 160 \\ 140 \\ 130 \end{array}]$

Step 1

Multiply $[\begin{array}{c} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{array}] [\begin{array}{c} x \\ y \\ v \\ z \end{array}]$ .

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Step 1.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is $4 \times 4$ and the second matrix is $4 \times 1$ .

Step 1.2

Multiply each row in the first matrix by each column in the second matrix.

$[\begin{array}{c} 1 x + 1 y + 0 v + 0 z \\ 0 x + 0 y + 1 v + 1 z \\ 1 x + 0 y + 1 v + 0 z \\ 0 x + 1 y + 0 v + 1 z \end{array}] = [\begin{array}{c} 110 \\ 160 \\ 140 \\ 130 \end{array}]$

Step 1.3

Simplify each element of the matrix by multiplying out all the expressions.

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Step 1.3.1

Add $0$ and $y$ .

$[\begin{array}{c} x + y \\ v + z \\ x + v \\ y + 0 + z \end{array}] = [\begin{array}{c} 110 \\ 160 \\ 140 \\ 130 \end{array}]$

Step 1.3.2

Add $y$ and $0$ .

$[\begin{array}{c} x + y \\ v + z \\ x + v \\ y + z \end{array}] = [\begin{array}{c} 110 \\ 160 \\ 140 \\ 130 \end{array}]$

Step 2

The matrix equation can be written as a set of equations.

$x + y = 110$

$v + z = 160$

$x + v = 140$

$y + z = 130$

Step 3

Subtract $y$ from both sides of the equation.

$x = 110 - y$

$v + z = 160$

$x + v = 140$

$y + z = 130$

Step 4

Replace all occurrences of $x$ with $110 - y$ in each equation.

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Step 4.1

Replace all occurrences of $x$ in $x + v = 140$ with $110 - y$ .

$(110 - y) + v = 140$

$x = 110 - y$

$v + z = 160$

$y + z = 130$

Step 4.2

Simplify the left side.

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Step 4.2.1

Remove parentheses.

$110 - y + v = 140$

$x = 110 - y$

$v + z = 160$

$y + z = 130$

$110 - y + v = 140$

$x = 110 - y$

$v + z = 160$

$y + z = 130$

$110 - y + v = 140$

$x = 110 - y$

$v + z = 160$

$y + z = 130$

Step 5

Move all terms not containing $v$ to the right side of the equation.

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Step 5.1

Subtract $110$ from both sides of the equation.

$- y + v = 140 - 110$

$x = 110 - y$

$v + z = 160$

$y + z = 130$

Step 5.2

Add $y$ to both sides of the equation.

$v = 140 - 110 + y$

$x = 110 - y$

$v + z = 160$

$y + z = 130$

Step 5.3

Subtract $110$ from $140$ .

$v = 30 + y$

$x = 110 - y$

$v + z = 160$

$y + z = 130$

$v = 30 + y$

$x = 110 - y$

$v + z = 160$

$y + z = 130$

Step 6

Replace all occurrences of $v$ with $30 + y$ in each equation.

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Step 6.1

Replace all occurrences of $v$ in $v + z = 160$ with $30 + y$ .

$(30 + y) + z = 160$

$v = 30 + y$

$x = 110 - y$

$y + z = 130$

Step 6.2

Simplify the left side.

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Step 6.2.1

Remove parentheses.

$30 + y + z = 160$

$v = 30 + y$

$x = 110 - y$

$y + z = 130$

$30 + y + z = 160$

$v = 30 + y$

$x = 110 - y$

$y + z = 130$

$30 + y + z = 160$

$v = 30 + y$

$x = 110 - y$

$y + z = 130$

Step 7

Move all terms not containing $y$ to the right side of the equation.

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Step 7.1

Subtract $30$ from both sides of the equation.

$y + z = 160 - 30$

$v = 30 + y$

$x = 110 - y$

$y + z = 130$

Step 7.2

Subtract $z$ from both sides of the equation.

$y = 160 - 30 - z$

$v = 30 + y$

$x = 110 - y$

$y + z = 130$

Step 7.3

Subtract $30$ from $160$ .

$y = 130 - z$

$v = 30 + y$

$x = 110 - y$

$y + z = 130$

$y = 130 - z$

$v = 30 + y$

$x = 110 - y$

$y + z = 130$

Step 8

Replace all occurrences of $y$ with $130 - z$ in each equation.

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Step 8.1

Replace all occurrences of $y$ in $v = 30 + y$ with $130 - z$ .

$v = 30 + 130 - z$

$y = 130 - z$

$x = 110 - y$

$y + z = 130$

Step 8.2

Simplify $v = 30 + 130 - z$ .

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Step 8.2.1

Simplify the left side.

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Step 8.2.1.1

Remove parentheses.

$v = 30 + 130 - z$

$y = 130 - z$

$x = 110 - y$

$y + z = 130$

$v = 30 + 130 - z$

$y = 130 - z$

$x = 110 - y$

$y + z = 130$

Step 8.2.2

Simplify the right side.

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Step 8.2.2.1

Add $30$ and $130$ .

$v = 160 - z$

$y = 130 - z$

$x = 110 - y$

$y + z = 130$

$v = 160 - z$

$y = 130 - z$

$x = 110 - y$

$y + z = 130$

$v = 160 - z$

$y = 130 - z$

$x = 110 - y$

$y + z = 130$

Step 8.3

Replace all occurrences of $y$ in $x = 110 - y$ with $130 - z$ .

$x = 110 - (130 - z)$

$v = 160 - z$

$y = 130 - z$

$y + z = 130$

Step 8.4

Simplify the right side.

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Step 8.4.1

Simplify $110 - (130 - z)$ .

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Step 8.4.1.1

Simplify each term.

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Step 8.4.1.1.1

Apply the distributive property.

$x = 110 - 1 \cdot 130 + z$

$v = 160 - z$

$y = 130 - z$

$y + z = 130$

Step 8.4.1.1.2

Multiply $- 1$ by $130$ .

$x = 110 - 130 + z$

$v = 160 - z$

$y = 130 - z$

$y + z = 130$

Step 8.4.1.1.3

Multiply $- - z$ .

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Step 8.4.1.1.3.1

Multiply $- 1$ by $- 1$ .

$x = 110 - 130 + 1 z$

$v = 160 - z$

$y = 130 - z$

$y + z = 130$

Step 8.4.1.1.3.2

Multiply $z$ by $1$ .

$x = 110 - 130 + z$

$v = 160 - z$

$y = 130 - z$

$y + z = 130$

$x = 110 - 130 + z$

$v = 160 - z$

$y = 130 - z$

$y + z = 130$

$x = 110 - 130 + z$

$v = 160 - z$

$y = 130 - z$

$y + z = 130$

Step 8.4.1.2

Subtract $130$ from $110$ .

$x = - 20 + z$

$v = 160 - z$

$y = 130 - z$

$y + z = 130$

$x = - 20 + z$

$v = 160 - z$

$y = 130 - z$

$y + z = 130$

$x = - 20 + z$

$v = 160 - z$

$y = 130 - z$

$y + z = 130$

Step 8.5

Replace all occurrences of $y$ in $y + z = 130$ with $130 - z$ .

$(130 - z) + z = 130$

$x = - 20 + z$

$v = 160 - z$

$y = 130 - z$

Step 8.6

Simplify the left side.

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Step 8.6.1

Combine the opposite terms in $(130 - z) + z$ .

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Step 8.6.1.1

Add $- z$ and $z$ .

$130 + 0 = 130$

$x = - 20 + z$

$v = 160 - z$

$y = 130 - z$

Step 8.6.1.2

Add $130$ and $0$ .

$130 = 130$

$x = - 20 + z$

$v = 160 - z$

$y = 130 - z$

$130 = 130$

$x = - 20 + z$

$v = 160 - z$

$y = 130 - z$

$130 = 130$

$x = - 20 + z$

$v = 160 - z$

$y = 130 - z$

$130 = 130$

$x = - 20 + z$

$v = 160 - z$

$y = 130 - z$

Step 9

Remove any equations from the system that are always true.

$x = - 20 + z$

$v = 160 - z$

$y = 130 - z$