Linear Algebra Examples

$3 x + 5 y = 11$ , $x + 3 y = 13$

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

$[\begin{array}{c} 3 & 5 \\ 1 & 3 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 11 \\ 13 \end{array}]$

Find the inverse of the coefficient matrix of $[\begin{array}{c} 3 & 5 \\ 1 & 3 \end{array}]$ .

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

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

$determinant [\begin{array}{c} 3 & 5 \\ 1 & 3 \end{array}] = | \begin{array}{c} 3 & 5 \\ 1 & 3 \end{array} |$

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

$(3) (3) - 1 \cdot 5$

Simplify the determinant.

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Simplify each term.

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Multiply $3$ by $3$ .

$9 - 1 \cdot 5$

Multiply $- 1$ by $5$ .

$9 - 5$

Subtract $5$ from $9$ .

$4$

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

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

Simplify each element of the matrix $[\begin{array}{c} 3 & - (5) \\ - (1) & 3 \end{array}]$ .

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Simplify element $0, 1$ by multiplying $- (5)$ .

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

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

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

Multiply $\frac{1}{4}$ by each element of the matrix.

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

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

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

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

Simplify element $0, 1$ by multiplying $\frac{1}{4} \cdot - 5$ .

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

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

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

Simplify element $1, 1$ by multiplying $\frac{1}{4} \cdot 3$ .

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

Left multiply both sides of the matrix equation $[\begin{array}{c} 3 & 5 \\ 1 & 3 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 11 \\ 13 \end{array}]$ by the inverse matrix $[\begin{array}{c} \frac{3}{4} & - \frac{5}{4} \\ - \frac{1}{4} & \frac{3}{4} \end{array}]$ .

$([\begin{array}{c} \frac{3}{4} & - \frac{5}{4} \\ - \frac{1}{4} & \frac{3}{4} \end{array}] \cdot [\begin{array}{c} 3 & 5 \\ 1 & 3 \end{array}]) \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{3}{4} & - \frac{5}{4} \\ - \frac{1}{4} & \frac{3}{4} \end{array}] \cdot [\begin{array}{c} 11 \\ 13 \end{array}]$

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

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{3}{4} & - \frac{5}{4} \\ - \frac{1}{4} & \frac{3}{4} \end{array}] \cdot [\begin{array}{c} 11 \\ 13 \end{array}]$

Simplify the right side of the equation.

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Multiply each row in the first matrix $[\begin{array}{c} \frac{3}{4} & - \frac{5}{4} \\ - \frac{1}{4} & \frac{3}{4} \end{array}]$ by each column in the second matrix $[\begin{array}{c} 11 \\ 13 \end{array}]$ .

$[\begin{array}{c} \frac{3}{4} \cdot 11 - \frac{5}{4} \cdot 13 \\ - \frac{1}{4} \cdot 11 + \frac{3}{4} \cdot 13 \end{array}]$

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

$[\begin{array}{c} - 8 \\ 7 \end{array}]$

The equation after simplifying the right and left side of the equation is $[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 8 \\ 7 \end{array}]$ .

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - 8 \\ 7 \end{array}]$

Find the solution.

$x = - 8$

$y = 7$

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