Linear Algebra Examples

$2 x - 2 y - 3 = 0$ , $x - y - 10 = 0$

Step 1

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

$[\begin{array}{c} 2 & - 2 \\ 1 & - 1 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 3 \\ 10 \end{array}]$

Step 2

Find the inverse of the coefficient matrix.

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

Find the determinant of $[\begin{array}{c} 2 & - 2 \\ 1 & - 1 \end{array}]$ .

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

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

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $- 1$ .

$- 2 - 1 \cdot - 2$

Multiply $- 1$ by $- 2$ .

$- 2 + 2$

Add $- 2$ and $2$ .

$0$

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

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

Simplify each element in the matrix.

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Rearrange $- (- 2)$ .

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

Rearrange $- (1)$ .

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

Multiply $\frac{1}{0}$ by each element of the matrix.

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

Rearrange $\frac{1}{0} \cdot - 1$ .

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

Since the matrix is undefined, it cannot be solved.

$Undefined$

Undefined