Linear Algebra Examples

$x + 2 y = 1$ , $5 y = 15$

Find the $A X = B$ from the system of equations.

$[\begin{array}{c} 1 & 2 \\ 0 & 5 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 1 \\ 15 \end{array}]$

Find the inverse of the coefficient matrix of $[\begin{array}{c} 1 & 2 \\ 0 & 5 \end{array}]$ .

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The inverse of a $2 \times 2$ matrix can be found using the formula $\frac{1}{| A |} [\begin{array}{c} d & - b \\ - c & a \end{array}]$ where $| A |$ is the determinant of $A$ .

If $A = [\begin{array}{c} a & b \\ c & d \end{array}]$ then $A^{- 1} = \frac{1}{| A |} [\begin{array}{c} d & - b \\ - c & a \end{array}]$

The determinant of $[\begin{array}{c} 1 & 2 \\ 0 & 5 \end{array}]$ is $5$ .

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These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} 1 & 2 \\ 0 & 5 \end{array}] = | \begin{array}{c} 1 & 2 \\ 0 & 5 \end{array} |$

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$(1) (5) + 0 \cdot 2$

Simplify the determinant.

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Simplify each term.

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Multiply $5$ by $1$ to get $5$ .

$5 + 0 \cdot 2$

Multiply $0$ by $2$ to get $0$ .

$5 + 0$

Add $5$ and $0$ to get $5$ .

$5$

Substitute the known values into the formula for the inverse of a matrix.

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

Simplify each element of the matrix $[\begin{array}{c} 5 & - (2) \\ - (0) & 1 \end{array}]$ .

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Simplify element $0, 1$ by multiplying $- (2)$ to get $- 2$ .

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

Simplify element $1, 0$ by multiplying $- (0)$ to get $0$ .

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

Multiply $\frac{1}{5}$ by each element of the matrix.

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

Simplify each element of the matrix $[\begin{array}{c} \frac{1}{5} \cdot 5 & \frac{1}{5} \cdot - 2 \\ \frac{1}{5} \cdot 0 & \frac{1}{5} \cdot 1 \end{array}]$ .

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Simplify element $0, 0$ by multiplying $\frac{1}{5} \cdot 5$ to get $1$ .

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

Simplify element $0, 1$ by multiplying $\frac{1}{5} \cdot - 2$ to get $- \frac{2}{5}$ .

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

Simplify element $1, 0$ by multiplying $\frac{1}{5} \cdot 0$ to get $0$ .

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

Simplify element $1, 1$ by multiplying $\frac{1}{5} \cdot 1$ to get $\frac{1}{5}$ .

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

Left multiply both sides of the matrix equation $[\begin{array}{c} 1 & 2 \\ 0 & 5 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 1 \\ 15 \end{array}]$ by the inverse matrix $[\begin{array}{c} 1 & - \frac{2}{5} \\ 0 & \frac{1}{5} \end{array}]$ .

$([\begin{array}{c} 1 & - \frac{2}{5} \\ 0 & \frac{1}{5} \end{array}] \cdot [\begin{array}{c} 1 & 2 \\ 0 & 5 \end{array}]) \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 1 & - \frac{2}{5} \\ 0 & \frac{1}{5} \end{array}] \cdot [\begin{array}{c} 1 \\ 15 \end{array}]$

Any matrix multiplied by its inverse is equal to $1$ all the time. $A \cdot A^{- 1} = 1$ . $([\begin{array}{c} 1 & - \frac{2}{5} \\ 0 & \frac{1}{5} \end{array}] \cdot [\begin{array}{c} 1 & 2 \\ 0 & 5 \end{array}]) = 1$ .

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 1 & - \frac{2}{5} \\ 0 & \frac{1}{5} \end{array}] \cdot [\begin{array}{c} 1 \\ 15 \end{array}]$

Simplify the right side of the equation.

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Multiply each row in the first matrix $[\begin{array}{c} 1 & - \frac{2}{5} \\ 0 & \frac{1}{5} \end{array}]$ by each column in the second matrix $[\begin{array}{c} 1 \\ 15 \end{array}]$ .

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

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

$[\begin{array}{c} - 5 \\ 3 \end{array}]$

The equation after simplifying the right and left side of the equation is $[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 5 \\ 3 \end{array}]$ .

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 5 \\ 3 \end{array}]$

Find the solution.

$x = - 5$

$y = 3$

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