Review : Matrices & Vectors — Question 2

Question 2 – Solution

Strategy. Matrix multiplication records the order of geometric operations. Determinants retain only part of that information.

Step 1: Compute both compositions. Direct multiplication gives $$RS=\begin{pmatrix}0&-1\\1&2\end{pmatrix},\qquad SR=\begin{pmatrix}2&-1\\1&0\end{pmatrix}.$$ Thus $RSv=(-1,3)^T$ and $SRv=(1,1)^T$. The first shears then rotates; the second rotates then shears.

Step 2: Test agreement on an arbitrary vector. Subtracting gives $$(RS-SR)w=\begin{pmatrix}-2&0\\0&2\end{pmatrix} \begin{pmatrix}w_1\\w_2\end{pmatrix}.$$ This is zero exactly when $w_1=w_2=0$. Therefore the two compositions agree only on the zero vector, despite each being invertible.

Step 3: Undo the operations in reverse order. Since $S^{-1}=\begin{pmatrix}1&-2\\0&1\end{pmatrix}$ and $R^{-1}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, $$\boxed{(RS)^{-1}=S^{-1}R^{-1}=\begin{pmatrix}2&1\\-1&0\end{pmatrix}.}$$ Multiplication with $RS$ on either side gives $I_2$.

Step 4: Compare the images of the square. For vertices $(0,0),(1,0),(1,1),(0,1)$ in that order, the images are $$\begin{array}{c|cccc} RS&(0,0)&(0,1)&(-1,3)&(-1,2)\\ SR&(0,0)&(2,1)&(1,1)&(-1,0) \end{array}$$ Both determinants are $1$, so both images have area $1$ and preserve the counterclockwise orientation. The parallelograms nevertheless differ. A determinant encodes signed area scaling, not the full image of each vector. The plot uses equal scales on both axes.