alexandria/2024/documents/by-course/math-4a/selected-solutions/main.typ

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#import "@preview/unequivocal-ams:0.1.1": ams-article, theorem, proof
#show: ams-article.with(title: "Exam Solutions")
Q6.
The general clockwise rotation matrix in $RR^2$ is
$
mat(cos theta, sin theta; -sin theta, cos theta)
$
We have $theta = (3pi)/2$, and
$ cos((3pi) / 2) = 0, sin((3pi) / 2) = -1 $
So our particular rotation matrix is
$
T = mat(0, -1; 1, 0)
$
Clearly, the linear transformation that reflects a vector across the vertical
axis changes the first standard basis vector, $vec(1, 0)$, to $vec(-1, 0)$.
This corresponds to the linear transformation (matrix)
$
S = mat(-1, 0; 0, 1)
$
Matrix composition, $compose$, is an equivalent notion to matrix multiplication. Therefore, we have the two compositions.
+ #[
$T compose S$, the linear transformation corresponding to a reflection followed by rotation:
$
T compose S &= mat(0, -1; 1, 0) mat(-1, 0; 0, 1) \
&#[using _column by coordinate_ rule] \
&= mat((-1 vec(0,1) + 0 vec(-1,0)), (0 vec(0,1) + 1 vec(-1,0))) \
&= mat(0, -1,;-1, 0)
$
]
+ #[
$S compose T$, the linear transformation corresponding to rotation, followed by reflection. Since matrix composition is generally not commutative, we obtain a different matrix.
$
S compose T &= mat(-1,0; 0,1) mat(0,-1; 1,0) \
&#[using _column by coordinate_ rule] \
&= mat((0 vec(-1, 0) + 1 vec(0, 1)), (-1 vec(-1, 0) + 0 vec(0, -1))) \
&= mat(0, 1; 1, 0)
$
]
#pagebreak()
Q7.
#enum(
numbering: "7.1",
[
The matrix $A$ corresponds to a linear transformation
$T: RR^4 |-> RR^3$. $A$ has 3 rows and 4 columns, so its matrix-vector
multiplication is only defined when with vectors in $RR^4$. Accordingly, it
will output a vector in $RR^3$.
So, $p = 4$.
],
[
See above explanation. $q = 3$.
],
[
To find all vectors $arrow(x) in RR^4$ whose image under $T$ is $arrow(b)$, we seek all solutions $arrow(x) = (x_1, x_2, x_3, x_4, x_5)^T$ to the equation
$ T arrow(x) = arrow(b) $
We can do this using our usual row reduction methods.
$
mat(augment: #(-1), -2,3,7,-11,-3;1,0,-2,1,3; 1,-1,-3,4,2) \
mat(augment: #(-1), 1,-1,-3,4,2; 0, 1, 1, -3, 1; 0, 0, 0, 0, 0)
$
We now have the augmented matrix in echelon form. So, $x_4$ and $x_3$ are free. Then, let $s,t in RR$ be free variables
$
x_1 &= 3 - t + 2s \
x_2 &= 1 + 3t - s \
x_3 &= s \
x_4 &= t
$
So, all vectors $arrow(x)$ are of the form
$
arrow(x) = vec(3+2s - t, 1 - s + 3t, s, t)
$
],
)