Question

rotate the given triangle $90^{circ}$ counterclockwise about the origin.
$\

$$\begin{bmatrix}-1&2&2\\\\-1&-1&3\\end{bmatrix}$$

$
$\

$$\begin{bmatrix}1&1&-3\\\\-1&2&?\\end{bmatrix}$$

$

Explanation:

Step1: Recall the rotation matrix

The rotation matrix for a \(90^{\circ}\) counter - clockwise rotation about the origin is \(

$$\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$

\). If a point \((x,y)\) is rotated \(90^{\circ}\) counter - clockwise about the origin, the new point \((x',y')\) is given by \(x'=-y\) and \(y' = x\).

Step2: Apply the rotation rule to the third column

For the third column of the first matrix (representing a point \((x = 2,y = 3)\)):
When we rotate a point \((x,y)\) \(90^{\circ}\) counter - clockwise about the origin, using the formula \(x'=-y\) and \(y'=x\).
Substitute \(x = 2\) and \(y = 3\) into the formulas.
For the \(x\) - coordinate of the new point: \(x'=-y=-3\).
For the \(y\) - coordinate of the new point: \(y'=x = 2\).

Answer:

\(2\)