Linear Algebra Examples

$15 [\begin{array}{c} 3 & - 6 & 5 \\ 2 & - 1 & 0 \\ - 4 & 7 & 4 \end{array}] - 5 x = 30 [\begin{array}{c} - 1 & - 2 & 1 \\ 5 & 5 & - 4 \\ - 3 & - 2 & 1 \end{array}]$

Step 1

The transformation defines a map from

$ℝ^{3}$ to

$ℝ^{3}$ . To prove the transformation is linear, the transformation must preserve scalar multiplication, addition, and the zero vector.

$ℝ^{3} \to ℝ^{3}$

Step 2

First prove the transform preserves this property.

$M (x + y) = M (x) + M (y)$

Step 3

Set up two matrices to test the addition property is preserved for

$M$ .

$M ([\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}] + [\begin{array}{c} y_{1} \\ y_{2} \\ y_{3} \end{array}])$

Step 4

Add the two matrices.

$M [\begin{array}{c} x_{1} + y_{1} \\ x_{2} + y_{2} \\ x_{3} + y_{3} \end{array}]$

Step 5

Apply the transformation to the vector.

$M (x + y) = [\begin{array}{c} - 1 \\ 5 \\ - 3 \end{array}]$

Step 6

Break the result into two matrices by grouping the variables.

$M (x + y) = [\begin{array}{c} 0 \\ 0 \\ 0 \end{array}] + [\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]$

Step 7

Since the transformation addition property does not hold, this is not a linear transformation.

$M (x + y) \neq M (x) + M (y)$