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