Self-check: 201-SN4-RE, Linear Algebra and Vector Geometry

b) We have $\vec{u}\times\vec{v}=(-10, 7, 9)$ so $\|\vec{u}\times\vec{v}\|^{2}=230$, and $\vec{u}\cdot\vec{v}=-8$ so $(\vec{u}\cdot\vec{v})^{2}=64$. The sum is $230+64=294$. On the other side, $\|\vec{u}\|^{2}=14$ and $\|\vec{v}\|^{2}=21$, so $\|\vec{u}\|^{2}\|\vec{v}\|^{2}=14\times 21=294$. Both sides are equal: the identity checks out.

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a) $\vec{n_{1}}=(2, -1, 2)$ and $\vec{n_{2}}=(1, 2, -2)$ have norms $\|\vec{n_{1}}\|=\sqrt{4+1+4}=3$ and $\|\vec{n_{2}}\|=\sqrt{1+4+4}=3$: they are equal. For $Q(3, 1, 0)$: $\text{dist}(Q,\pi_{1})=\dfrac{|2(3)-1+0|}{3}=\dfrac{5}{3}$ and $\text{dist}(Q,\pi_{2})=\dfrac{|3+2(1)-0|}{3}=\dfrac{5}{3}$. The point $Q$ is equidistant from the two planes.

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a) $A+B=\begin{pmatrix} 2+1 & -1+4 \\ 3-2 & 0+5 \end{pmatrix}=\begin{pmatrix} 3 & 3 \\ 1 & 5 \end{pmatrix}$. For $3A-2B$: $3A=\begin{pmatrix} 6 & -3 \\ 9 & 0 \end{pmatrix}$ and $2B=\begin{pmatrix} 2 & 8 \\ -4 & 10 \end{pmatrix}$, so $3A-2B=\begin{pmatrix} 4 & -11 \\ 13 & -10 \end{pmatrix}$.

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b) Augmented matrix $\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 1 & 2 & 3 & 14 \\ 2 & 3 & 4 & 20 \end{array}\right]$. Apply $R_{2}\to R_{2}-R_{1}$, which gives $(0, 1, 2\,|\,8)$, then $R_{3}\to R_{3}-2R_{1}$, which gives $(0, 1, 2\,|\,8)$.

The two new rows are identical, so $R_{3}\to R_{3}-R_{2}$ gives $(0, 0, 0\,|\,0)$. The row echelon form is $\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 8 \\ 0 & 0 & 0 & 0 \end{array}\right]$.

Here the zero row has a ZERO right-hand side, which expresses no contradiction at all: it is the equation $0=0$, always true, which simply signals a redundant equation. The system is consistent. It has two pivots (columns of $x$ and $y$) for three unknowns, so $3-2=1$ free parameter. The unknown $z$, whose column carries no pivot, becomes the parameter.

Set $z=t$. The second row gives $y+2t=8$, so $y=8-2t$. The first gives $x+(8-2t)+t=6$, that is $x=6-8+2t-t=t-2$.

The solution set is $\{(t-2, 8-2t, t)\ :\ t\in\mathbb{R}\}$, INFINITELY many solutions. The whole set must be given: answering with a single particular solution, even a correct one, does not earn the marks.

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d) $\det(A^{2})=\det(A)\det(A)=6\times 6=36$, by the property $\det(AB)=\det(A)\det(B)$. Since $\det(A)=6\neq 0$, the matrix $A$ is invertible and $\det(A^{-1})=\dfrac{1}{\det(A)}=\dfrac{1}{6}$. Finally $\det(A^{T})=\det(A)=6$.

For $\det(A+A)$: the trap is to write $\det(A)+\det(A)=12$. That is wrong, because the determinant is NOT additive. You have to see that $A+A=2A$, so $\det(A+A)=\det(2A)=2^{3}\times 6=48$. Remember the formula that does not exist: $\det(A+B)\neq\det(A)+\det(B)$ in general. The determinant is multiplicative, never additive.

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a) We look for $a$ and $b$ such that $a\vec{u}+b\vec{v}=\vec{w}$, that is $\begin{cases} a+2b=4 \\ 2a-b=3 \\ -a+3b=1 \end{cases}$. The first two give $a=4-2b$, then $2(4-2b)-b=3$, that is $8-5b=3$, so $b=1$ and $a=2$. Test of the third equation: $-2+3(1)=1$, it holds. So $\vec{w}=2\vec{u}+\vec{v}$. Since one of the three vectors is expressed in terms of the other two, the set is linearly DEPENDENT. The dependence relation can also be written in its standard form: $2\vec{u}+\vec{v}-\vec{w}=\vec{0}$, a zero combination whose coefficients are not all zero.

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b) Standard basis of $P_{2}$: $\left\{1,\;x,\;x^{2}\right\}$. It SPANS $P_{2}$, since every polynomial $ax^{2}+bx+c$ can be written $c\cdot 1+b\cdot x+a\cdot x^{2}$; it is LINEARLY INDEPENDENT, since $c+bx+ax^{2}=0$ for every $x$ forces $a=b=c=0$ (the zero polynomial has all its coefficients equal to zero). Hence $\dim P_{2}=3$. For $M_{2\times 2}$, the standard basis consists of the four matrices with a single 1 and 0 everywhere else, so $\dim M_{2\times 2}=4$. General rule to remember: $\dim P_{n}=n+1$ (watch the $+1$, forgotten very often) and $\dim M_{m\times n}=mn$.

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b) $R_{60}\begin{pmatrix} 2 \\ 0 \end{pmatrix}=\begin{pmatrix} \dfrac{1}{2}(2)-\dfrac{\sqrt{3}}{2}(0) \\[4pt] \dfrac{\sqrt{3}}{2}(2)+\dfrac{1}{2}(0) \end{pmatrix}=\begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix}$. The image is $A'(1, \sqrt{3})$, whose distance to the origin is $\sqrt{1+3}=2$, exactly that of $A$. A rotation preserves lengths, and this check should become a reflex.

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c) If you swap the columns of $P$ to get $P'=\begin{pmatrix}2&-1\\1&1\end{pmatrix}$ while keeping $D=\begin{pmatrix}2&0\\0&5\end{pmatrix}$, the equality fails: you would be claiming that $\begin{pmatrix}2\\1\end{pmatrix}$ goes with the eigenvalue 2, when it goes with 5. The calculation confirms it, $P'DP'^{-1}=\begin{pmatrix}3&-2\\-1&4\end{pmatrix}\neq A$. The rule is therefore strict: the $i$-th column of $P$ must be an eigenvector associated with the $i$-th diagonal entry of $D$. On the other hand, swapping the columns of $P$ AND the entries of $D$ together is perfectly legal and gives another diagonalization, just as valid: diagonalization is not unique, only the PAIRING counts.

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c) $p_{12}=0.3$ is the entry in row 1 and column 2: it is the probability of moving FROM state 2 TO state 1, in other words the share of B that joins A. The order of the indices is the most frequent source of error in the chapter: in the column convention, $p_{ij}$ reads “from $j$ to $i$”.

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a) $|z|=\sqrt{(-2)^{2}+(2\sqrt{3})^{2}}=\sqrt{4+12}=\sqrt{16}=4$. For the argument, impose BOTH conditions: $\cos\theta=\dfrac{-2}{4}=-\dfrac{1}{2}$ and $\sin\theta=\dfrac{2\sqrt{3}}{4}=\dfrac{\sqrt{3}}{2}$. A negative cosine and a positive sine put the angle in the second quadrant, and the reference angle for $\dfrac{1}{2}$ and $\dfrac{\sqrt{3}}{2}$ is $\dfrac{\pi}{3}$. So $\theta=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}$.

The classic pitfall: using the calculator for $\arctan\left(\dfrac{2\sqrt{3}}{-2}\right)=\arctan(-\sqrt{3})=-\dfrac{\pi}{3}$, and stopping there. That value is in the fourth quadrant, while $z$ is in the second. The arctangent function only ever returns angles between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$: it cannot tell opposite quadrants apart. Always locate the point before concluding.

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b) $A=\begin{pmatrix} -\cos 30^{\circ} & \cos 60^{\circ} \\ \sin 30^{\circ} & \sin 60^{\circ} \end{pmatrix}$ and $B=\begin{pmatrix} 0 \\ 200 \end{pmatrix}$. Then $\det A=-\cos 30^{\circ}\sin 60^{\circ}-\cos 60^{\circ}\sin 30^{\circ}=-(\sin 60^{\circ}\cos 30^{\circ}+\cos 60^{\circ}\sin 30^{\circ})=-\sin(60^{\circ}+30^{\circ})=-\sin 90^{\circ}=-1$. The addition formula avoids every decimal, and the exact result $-1$ comes from the two cables being PERPENDICULAR, $30^{\circ}+60^{\circ}=90^{\circ}$.

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