Problem: The matrix $A$ is given by
\[
A = \begin{bmatrix}
1 & -1 & 0 \\
2 & 0 & 4 \\
3 & -2 & -1 \\
\end{bmatrix}
\]
which represents a transformation $T$. Given that $T$ maps point $P = (x,y,z)$ onto point $P' = (3,7,4)$.
Find the coordinates of the point $P$.
The matrix $B$ represents a transformation $U$. Given that the transformation $T$ followed by the transformation $U$ is equivalent to a rotation through $90^{\circ } $ about the $x$-axis. Find the matrix $B$.
We need to solve the equation
\[
A \begin{bmatrix}
x \\ y \\ z
\end{bmatrix} =
\begin{bmatrix}
3 \\ 7 \\ 4
\end{bmatrix}.
\]
The rotation matrix about the $x$-axis is given by
\[
R_x(90^{\circ }) =
\begin{bmatrix}
1 & 0 & 0\\
0 & 0 & -1\\
0 & 1 & 0
\end{bmatrix}
\]
If the matrix of linear transformations $S \text{ and } T$ are $A$ and $B$, respectively, then the matrix of the compostion $S \circ T$ is given by $AB$.
Solution: The given matrix is
\[
A = \begin{bmatrix}
1 & -1 & 0 \\
2 & 0 & 4 \\
3 & -2 & -1 \\
\end{bmatrix}.
\]
We need to find the coordinates of the point $P$ such that $T(P) = P'$. This means we need to solve the equation
\[
A \begin{bmatrix}
x \\ y \\ z
\end{bmatrix} =
\begin{bmatrix}
3 \\ 7 \\ 4
\end{bmatrix}.
\]
The augmented matrix is
\[
\begin{bmatrix}
1 & -1 & 0 & 3 \\
2 & 0 & 4 & 7 \\
3 & -2 & -1 & 4
\end{bmatrix}.
\]
We will now perform row operations to reduce the augmented matrix to row echelon form and find the solution.
\begin{align*}
\begin{bmatrix}
1 & -1 & 0 & 3 \\
2 & 0 & 4 & 7 \\
3 & -2 & -1 & 4
\end{bmatrix}
& \xrightarrow{
\begin{matrix}
R_2 \rightarrow R_2 - 2R_1 \\
R_3 \rightarrow R_3 - 3R_1
\end{matrix}
}
\begin{bmatrix}
1 & -1 & 0 & 3 \\
0 & 2 & 4 & 1 \\
0 & 1 & -1 & -5
\end{bmatrix} \\[2ex]
& \xrightarrow{
\begin{matrix}
R_2 \rightarrow \frac{1}{2}R_2
\end{matrix}
}
\begin{bmatrix}
1 & -1 & 0 & 3 \\
0 & 1 & 2 & \frac{1}{2} \\
0 & 1 & -1 & -5
\end{bmatrix} \\[2ex]
& \xrightarrow{
\begin{matrix}
R_1 \rightarrow R_1 + R_2 \\
R_3 \rightarrow R_3 - R_2
\end{matrix}
}
\begin{bmatrix}
1 & 0 & 2 & \frac{7}{2} \\[1ex]
0 & 1 & 2 & \frac{1}{2} \\[1ex]
0 & 0 & -3 & -\frac{11}{2}
\end{bmatrix} \\[2ex]
& \xrightarrow{
\begin{matrix}
R_3 \rightarrow -\frac{1}{3}R_3
\end{matrix}
}
\begin{bmatrix}
1 & 0 & 2 & \frac{7}{2} \\[1ex]
0 & 1 & 2 & \frac{1}{2} \\[1ex]
0 & 0 & 1 & \frac{11}{6}
\end{bmatrix} \\[2ex]
& \xrightarrow{
\begin{matrix}
R_1 \rightarrow R_1 - 2R_3\\
R_2 \rightarrow R_2 - 2R_3
\end{matrix}
}
\begin{bmatrix}
1 & 0 & 0 & -\frac{1}{6} \\[1ex]
0 & 1 & 2 & -\frac{19}{6} \\[1ex]
0 & 0 & 1 & \frac{11}{6}
\end{bmatrix}.
\end{align*}
Thus, the coordinate of the point $P$ is
\[
\textcolor{blue}{
\boxed{
P =
\begin{pmatrix}
-\frac{1}{6} &
-\frac{19}{6} &
\frac{11}{6}
\end{pmatrix}
}
}
\]
We could have also solved this by using the inverse of the matrix $A$. The inverse of the matrix $A$ is given by
\[
\begin{bmatrix}
-\frac{4}{3} & \frac{1}{6} & \frac{2}{3} \\[1ex]
-\frac{7}{3} & \frac{1}{6} & \frac{2}{3} \\[1ex]
\frac{2}{3} & \frac{1}{6} & -\frac{1}{3} \\
\end{bmatrix}
\]
and hence, the coordinate of $P$ will be
\[
P = A^{-1}
\begin{bmatrix}
3 \\
7 \\
4 \\
\end{bmatrix}.
\]
The transformation $T$ followed by the transformation $U$ is equivalent to a rotation through $90^{\circ } $ about the $x$-axis. The rotation matrix about the $x$-axis is given by
\[
R_x(90^{\circ }) =
\begin{bmatrix}
1 & 0 & 0\\
0 & 0 & -1\\
0 & 1 & 0
\end{bmatrix}
\]
which is the matrix that represents the transformation $T$ followed by the transformation $U$, that is $T \circ U$ and hence, the corresponding matrix will be $BA$. Thus, we need to find the matrix $B$ such that $BA = R_x (90^{\circ } )$.
Therefore,
\begin{align*}
BA = R_x(90^{\circ} ) & \implies B = R_x(90^{\circ } ) A^{-1} \\
& \implies B =
\begin{bmatrix}
1 & 0 & 0\\
0 & 0 & -1\\
0 & 1 & 0
\end{bmatrix}
\begin{bmatrix}
-\frac{4}{3} & \frac{1}{6} & \frac{2}{3} \\[1ex]
-\frac{7}{3} & \frac{1}{6} & \frac{2}{3} \\[1ex]
\frac{2}{3} & \frac{1}{6} & -\frac{1}{3} \\
\end{bmatrix}
\\[2ex]
& \implies B =
\begin{bmatrix}
-\frac{4}{3} & \frac{1}{6} & \frac{2}{3} \\[1ex]
-\frac{2}{3} & -\frac{1}{6} & \frac{1}{3} \\[1ex]
-\frac{7}{3} & \frac{1}{6} & \frac{2}{3} \\
\end{bmatrix} = \frac{1}{6}
\begin{bmatrix}
-8 & 1 & 4 \\
-4 & -1 & 2 \\
-14 & 1 & 4 \\
\end{bmatrix}.
\end{align*}