Linear Algebra Examples

$x + 2 y = 1$ , $4 x + 5 y = 13$

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

$[\begin{array}{c} 1 & 2 \\ 4 & 5 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 1 \\ 13 \end{array}]$

Find the inverse of the coefficient matrix of $[\begin{array}{c} 1 & 2 \\ 4 & 5 \end{array}]$ .

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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 $[\begin{array}{c} 1 & 2 \\ 4 & 5 \end{array}]$ is $- 3$ .

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

$determinant [\begin{array}{c} 1 & 2 \\ 4 & 5 \end{array}] = | \begin{array}{c} 1 & 2 \\ 4 & 5 \end{array} |$

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

$(1) (5) - 4 \cdot 2$

Simplify the determinant.

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Simplify each term.

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

$5 - 4 \cdot 2$

Multiply $- 4$ by $2$ to get $- 8$ .

$5 - 8$

Subtract $8$ from $5$ to get $- 3$ .

$- 3$

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

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

Simplify each element of the matrix $[\begin{array}{c} 5 & - (2) \\ - (4) & 1 \end{array}]$ .

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

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

Simplify element $1, 0$ by multiplying $- (4)$ to get $- 4$ .

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

Multiply $\frac{1}{- 3}$ by each element of the matrix.

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

Simplify each element of the matrix $[\begin{array}{c} \frac{1}{- 3} \cdot 5 & \frac{1}{- 3} \cdot - 2 \\ \frac{1}{- 3} \cdot - 4 & \frac{1}{- 3} \cdot 1 \end{array}]$ .

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Simplify element $0, 0$ by multiplying $\frac{1}{- 3} \cdot 5$ to get $- \frac{5}{3}$ .

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

Simplify element $0, 1$ by multiplying $\frac{1}{- 3} \cdot - 2$ to get $\frac{2}{3}$ .

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

Simplify element $1, 0$ by multiplying $\frac{1}{- 3} \cdot - 4$ to get $\frac{4}{3}$ .

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

Simplify element $1, 1$ by multiplying $\frac{1}{- 3} \cdot 1$ to get $- \frac{1}{3}$ .

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

Left multiply both sides of the matrix equation $[\begin{array}{c} 1 & 2 \\ 4 & 5 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 1 \\ 13 \end{array}]$ by the inverse matrix $[\begin{array}{c} - \frac{5}{3} & \frac{2}{3} \\ \frac{4}{3} & - \frac{1}{3} \end{array}]$ .

$([\begin{array}{c} - \frac{5}{3} & \frac{2}{3} \\ \frac{4}{3} & - \frac{1}{3} \end{array}] \cdot [\begin{array}{c} 1 & 2 \\ 4 & 5 \end{array}]) \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - \frac{5}{3} & \frac{2}{3} \\ \frac{4}{3} & - \frac{1}{3} \end{array}] \cdot [\begin{array}{c} 1 \\ 13 \end{array}]$

Any matrix multiplied by its inverse is equal to $1$ all the time. $A \cdot A^{- 1} = 1$ . $([\begin{array}{c} - \frac{5}{3} & \frac{2}{3} \\ \frac{4}{3} & - \frac{1}{3} \end{array}] \cdot [\begin{array}{c} 1 & 2 \\ 4 & 5 \end{array}]) = 1$ .

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - \frac{5}{3} & \frac{2}{3} \\ \frac{4}{3} & - \frac{1}{3} \end{array}] \cdot [\begin{array}{c} 1 \\ 13 \end{array}]$

Simplify the right side of the equation.

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Multiply each row in the first matrix $[\begin{array}{c} - \frac{5}{3} & \frac{2}{3} \\ \frac{4}{3} & - \frac{1}{3} \end{array}]$ by each column in the second matrix $[\begin{array}{c} 1 \\ 13 \end{array}]$ .

$[\begin{array}{c} - \frac{5}{3} \cdot 1 + \frac{2}{3} \cdot 13 \\ \frac{4}{3} \cdot 1 - \frac{1}{3} \cdot 13 \end{array}]$

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

$[\begin{array}{c} 7 \\ - 3 \end{array}]$

The equation after simplifying the right and left side of the equation is $[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 7 \\ - 3 \end{array}]$ .

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 7 \\ - 3 \end{array}]$

Find the solution.

$x = 7$

$y = - 3$

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