Corrected practice final exam in linear algebra and vector geometry (201-SN4-RE)

$\vec{AB}=(2,1,-2)$, $\vec{AC}=(1,4,-1)$, $\vec{AD}=(-1,2,3)$. Norm: $\|\vec{AB}\|=\sqrt{4+1+4}=3$. Angle: the dot product is the tool for angles. $\vec{AB}\cdot\vec{AC}=2+4+2=8$ and $\|\vec{AC}\|=\sqrt{18}=3\sqrt{2}$, so $\cos\theta=\dfrac{8}{3\cdot 3\sqrt{2}}=\dfrac{8}{9\sqrt{2}}\approx 0.6285$ and $\theta\approx 51.1^{\circ}$.

The cross product is the tool for areas. $\vec{AB}\times\vec{AC}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\ 2&1&-2\\ 1&4&-1\end{vmatrix}=(1\cdot(-1)-(-2)\cdot 4)\,\vec{i}-(2\cdot(-1)-(-2)\cdot 1)\,\vec{j}+(2\cdot 4-1\cdot 1)\,\vec{k}=(7,0,7)$. Its norm is $7\sqrt{2}$, which is the area of the parallelogram. The area of the triangle is half of it: $\dfrac{7\sqrt{2}}{2}\approx 4.95$.

The scalar triple product is the tool for volumes. $\vec{AD}\cdot(7,0,7)=-7+0+21=14\neq 0$: the four points are therefore not coplanar, they form a genuine tetrahedron. Volume of the parallelepiped $=|14|=14$, volume of the tetrahedron $=\dfrac{14}{6}=\dfrac{7}{3}\approx 2.33$.

Distance to the plane: from $V=\frac{1}{3}\mathcal{A}h$, where $\mathcal{A}$ is the area of the base $ABC$, we get $h=\dfrac{3V}{\mathcal{A}}=\dfrac{3\cdot\frac{7}{3}}{\frac{7\sqrt{2}}{2}}=\dfrac{7\cdot 2}{7\sqrt{2}}=\sqrt{2}\approx 1.41$. Distance to the line $AB$: $\vec{AD}\times\vec{AB}=(-7,4,-5)$, of norm $\sqrt{49+16+25}=\sqrt{90}=3\sqrt{10}$, so $d=\dfrac{3\sqrt{10}}{3}=\sqrt{10}\approx 3.16$. The two distances measure different things: the first is the distance to a whole plane, the second the distance to a single line lying in that plane. A point is always at least as far from a line as from the plane that contains it, which $\sqrt{10}>\sqrt{2}$ confirms.

The normal of $\pi$ is $\vec{n}=(2,-1,3)$. Since $\vec{n}\cdot\vec{d}=2-1-3=-2\neq 0$, the line is not parallel to the plane (so it is not contained in it either): it crosses it at a single point. Parametrize the line as $(1+t,\,2+t,\,-1-t)$ and substitute: $2(1+t)-(2+t)+3(-1-t)=-2t-3=5$, so $t=-4$ and the point of intersection is $(-3,-2,3)$. Check: $-6+2+9=5$.

The angle $\alpha$ between a line and a plane is the complement of the angle between the line and the normal, hence $\sin\alpha=\frac{|\vec{n}\cdot\vec{d}|}{\|\vec{n}\|\,\|\vec{d}\|}=\frac{2}{\sqrt{14}\sqrt{3}}=\frac{2}{\sqrt{42}}\approx 0.3086$, so $\alpha\approx 17.98^{\circ}$. It really is a sine and not a cosine: that is where most of the marks are lost on the exam.

$d(P,\pi)=\dfrac{|2(4)-1+3(0)-5|}{\sqrt{4+1+9}}=\dfrac{|8-1-5|}{\sqrt{14}}=\dfrac{2}{\sqrt{14}}\approx 0.535$. For the projection, move from $P$ along the normal: $P'=P-\dfrac{2x_{0}-y_{0}+3z_{0}-5}{\|\vec{n}\|^{2}}\,\vec{n}=(4,1,0)-\dfrac{2}{14}(2,-1,3)=\left(\dfrac{26}{7},\dfrac{8}{7},-\dfrac{3}{7}\right)$. Check: $2\cdot\frac{26}{7}-\frac{8}{7}+3\left(-\frac{3}{7}\right)=\frac{52-8-9}{7}=5$.

A parallel plane has the same normal, so its equation is $2x-y+3z=k$; requiring it to pass through $P$ gives $k=8-1+0=7$. The two planes are $2x-y+3z=5$ and $2x-y+3z=7$, and the distance between two parallel planes is $\dfrac{|7-5|}{\|\vec{n}\|}=\dfrac{2}{\sqrt{14}}$, exactly the distance found in c). This is consistent: the distance from $P$ to $\pi$ is, by definition, the distance between $\pi$ and the parallel plane through $P$.

Expanding along the first row: $\Delta=\begin{vmatrix}1&1&1\\ 2&3&a\\ 1&a&3\end{vmatrix}=(9-a^{2})-(6-a)+(2a-3)=-a^{2}+3a=-a(a-3)$.

The system has a unique solution if and only if $\Delta\neq 0$, that is for $a\neq 0$ and $a\neq 3$. For $a=1$: $R_{3}-R_{1}$ gives $2z=2$, so $z=1$; $R_{2}-2R_{1}$ gives $y-z=1$, so $y=2$; then $R_{1}$ gives $x=1-2-1=-2$. The solution is $(x,y,z)=(-2,2,1)$. Check in $R_{2}$: $-4+6+1=3$, and in $R_{3}$: $-2+2+3=3$.

For $a=3$, $\Delta=0$. Row reduce: $R_{2}-2R_{1}$ gives $y+z=1$ and $R_{3}-R_{1}$ gives $2y+2z=2$, that is again $y+z=1$. The third equation is therefore the same as the second: the coefficient matrix has rank 2, so does the augmented matrix, and the system is consistent with $3-2=1$ free parameter. From $y+z=1$ and $x+y+z=1$ we get $x=0$, then setting $z=t$: $(x,y,z)=(0,\,1-t,\,t),\qquad t\in\mathbb{R}$. This is a line of solutions.

For $a=0$, $\Delta=0$ as well. $R_{2}-2R_{1}$ gives $y-2z=1$ and $R_{3}-R_{1}$ gives $-y+2z=2$, that is $y-2z=-2$. The two contradict each other: the coefficient matrix has rank 2, the augmented matrix has rank 3, and by the consistency theorem the system has no solution. Geometrically: in case c) the three planes meet along a common line; in case d) they form a prism: each pair of planes meets along a line, but the three lines of intersection are parallel and distinct, so no point lies on all three planes.

Expansion along the first column (it contains a zero): $\det A=1\begin{vmatrix}1&4\\ 6&0\end{vmatrix}-0+5\begin{vmatrix}2&3\\ 1&4\end{vmatrix}=1(0-24)+5(8-3)=-24+25=1$.

With cofactors, $A^{-1}=\dfrac{1}{\det A}\operatorname{adj}A=\operatorname{adj}A$ since $\det A=1$. We obtain $A^{-1}=\begin{pmatrix} -24&18&5\\ 20&-15&-4\\ -5&4&1\end{pmatrix}$. Check: $AA^{-1}=I$; for instance the first two entries of the first row of the product are $(1)(-24)+(2)(20)+(3)(-5)=-24+40-15=1$ and $(1)(18)+(2)(-15)+(3)(4)=18-30+12=0$.

$\vec{x}=A^{-1}\vec{b}$: $\vec{x}=\begin{pmatrix} -24&18&5\\ 20&-15&-4\\ -5&4&1\end{pmatrix}\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}=\begin{pmatrix}-24+36+15\\ 20-30-12\\ -5+8+3\end{pmatrix}=\begin{pmatrix}27\\ -22\\ 6\end{pmatrix}$. Direct check: $27+2(-22)+3(6)=27-44+18=1$, $0-22+24=2$, $5(27)+6(-22)=135-132=3$.

Apply the properties, never recomputing a matrix. $\det(2A)=2^{3}\det A=8$. $\det\left(A^{-1}\right)=\dfrac{1}{\det A}=1$. $\det\left(A^{3}\right)=(\det A)^{3}=1$. $\det(\operatorname{adj}A)=(\det A)^{n-1}=1^{2}=1$. Finally $\det\left(A^{-1}B^{2}A^{T}\right)=\frac{1}{\det A}\cdot(\det B)^{2}\cdot\det A=(\det B)^{2}=4$, the factor $\det A$ cancelling with its reciprocal: the result does not depend on $A$.

Replace $R_{2}$ by $R_{2}-2R_{1}$ and $R_{4}$ by $R_{4}-R_{1}$, which does not change the determinant: $\begin{vmatrix} 1&2&0&1\\ 0&-1&1&-2\\ 0&1&2&1\\ 0&-2&1&1\end{vmatrix}=1\cdot\begin{vmatrix} -1&1&-2\\ 1&2&1\\ -2&1&1\end{vmatrix}$. This $3\times 3$ determinant equals $-1(2-1)-1(1+2)+(-2)(1+4)=-1-3-10=-14$. So the required determinant is $-14$.

$\det(2M)=2^{3}\det M=8\cdot 5=40$ (multiplying an $n\times n$ matrix by $k$ multiplies the determinant by $k^{n}$, one row at a time). $\det\left(M^{T}M\right)=\det M^{T}\cdot\det M=(\det M)^{2}=25$ (multiplicativity, and $\det M^{T}=\det M$). $\det\left(M^{-1}\right)=\dfrac{1}{5}$ (because $\det M\cdot\det M^{-1}=\det I=1$). $\det(\operatorname{adj}M)=(\det M)^{n-1}=5^{2}=25$.

Add the last two rows to the first: every entry of $R_{1}$ becomes $\lambda+2$, which factors out. What remains is $\det N=(\lambda+2)\begin{vmatrix} 1&1&1\\ 1&\lambda&1\\ 1&1&\lambda\end{vmatrix}=(\lambda+2)(\lambda-1)^{2}$. The matrix is singular for $\lambda=-2$ and for $\lambda=1$ (a double root). For $\lambda=1$ the three rows are identical (rank 1); for $\lambda=-2$ the sum of the three rows is zero (rank 2).

The scalar triple product decides coplanarity: $\begin{vmatrix} 1&2&3\\ 2&-1&1\\ 4&3&7\end{vmatrix}=1(-7-3)-2(14-4)+3(6+4)=-10-20+30=0$. The determinant is zero, so the three vectors are coplanar (the parallelepiped they span is flat). We then look for $\alpha,\beta$ with $\vec{w}=\alpha\vec{u}+\beta\vec{v}$: the first two components give $\alpha+2\beta=4$ and $2\alpha-\beta=3$, hence $\beta=1$ and $\alpha=2$. The third component confirms it: $3(2)+1=7$. So $\vec{w}=2\vec{u}+\vec{v}$.

$R_{2}-2R_{1}$ gives $(0,0,3,-6)$ and $R_{3}-R_{1}$ gives $(0,0,3,-6)$: the two new rows are identical. The echelon form has only two non-zero rows, so $\operatorname{rank}(A)=2$. In other words, the three rows of $A$ satisfy the relation $R_{1}+R_{3}-R_{2}=\vec{0}$.

The homogeneous system reduces to $x_{1}+2x_{2}-x_{3}+3x_{4}=0$ and $3x_{3}-6x_{4}=0$. The second gives $x_{3}=2x_{4}$; substituting, $x_{1}=-2x_{2}+x_{3}-3x_{4}=-2x_{2}-x_{4}$. The free variables are $x_{2}$ and $x_{4}$: $\vec{x}=x_{2}\begin{pmatrix}-2\\ 1\\ 0\\ 0\end{pmatrix}+x_{4}\begin{pmatrix}-1\\ 0\\ 2\\ 1\end{pmatrix}$. These two vectors are independent (look at components 2 and 4), so they form a basis of the null space, which has dimension 2.

The rank theorem states $\operatorname{rank}(A)+\dim(\text{null space})=n$, the number of columns. Here $2+2=4$, and $A$ does have 4 columns. The count checks out.

Since $R_{1}+R_{3}-R_{2}=\vec{0}$ on the coefficient side, the same combination must cancel the right-hand side: the consistency condition is $b_{1}+b_{3}-b_{2}=0$. For $\vec{b}=(1,3,2)^{T}$: $1+2-3=0$, so the system is consistent. Solving: $3x_{3}-6x_{4}=b_{2}-2b_{1}=1$, so $x_{3}=\frac{1}{3}+2x_{4}$, then $x_{1}=1-2x_{2}+x_{3}-3x_{4}=\frac{4}{3}-2x_{2}-x_{4}$. The general solution is $\vec{x}=\begin{pmatrix}4/3\\ 0\\ 1/3\\ 0\end{pmatrix}+x_{2}\begin{pmatrix}-2\\ 1\\ 0\\ 0\end{pmatrix}+x_{4}\begin{pmatrix}-1\\ 0\\ 2\\ 1\end{pmatrix}$, that is, one particular solution plus the whole null space. This is the general structure of the solution set of a consistent system.

Yes. $W$ is the solution set of a homogeneous linear equation: it contains $\vec{0}$, and if $\vec{u},\vec{v}\in W$ and $\alpha,\beta\in\mathbb{R}$, then $\alpha\vec{u}+\beta\vec{v}$ still satisfies the equation by linearity. It is therefore a subspace. Setting $x=-2y+z$, we get $(x,y,z)=y(-2,1,0)+z(1,0,1)$: a basis is $\{(-2,1,0),(1,0,1)\}$ and $\dim W=2$ (it is a plane through the origin).

No. The zero vector $(0,0,0)$ does not satisfy $x+y=1$, and every subspace must contain the zero vector: $U$ is not a subspace. (It is an affine plane, parallel to a subspace but shifted.) The zero vector test is the first reflex, and it settles the question at once.

The two generators of $F$ are not proportional, so $\dim F=2$; likewise $\dim G=2$. For $F+G$, it is enough to find three independent vectors among the four: $\begin{vmatrix} 1&1&0\\ 0&1&1\\ 1&0&1\end{vmatrix}=1(1)-1(-1)+0=2\neq 0$, so $F+G$ contains three independent vectors of $\mathbb{R}^{3}$: $F+G=\mathbb{R}^{3}$ and $\dim(F+G)=3$. Grassmann's formula then gives $\dim(F\cap G)=\dim F+\dim G-\dim(F+G)=2+2-3=1$. For a direction vector, write $a(1,1,0)+b(0,1,1)=c(1,0,1)+d(1,1,1)$, which gives $a=c+d$, $a+b=d$, $b=c+d$. So $b=a$, then $d=2a$ and $c=-a$. With $a=1$: the vector $(1,2,1)$, which is indeed in $F$ (the sum of the two generators) and in $G$ (since $-\left(1,0,1\right)+2\left(1,1,1\right)=(1,2,1)$).

In the standard basis $\{1,x,x^{2}\}$, the three polynomials have coordinate vectors $(1,1,0)$, $(0,1,1)$ and $(1,0,1)$. The determinant of these three columns is 2, which is non-zero: the set is linearly independent, and since $\dim P_{2}=3$, it is a basis. For the coordinates of $p$, solve $\alpha(1+x)+\beta(x+x^{2})+\gamma(1+x^{2})=2+3x+4x^{2}$, that is $\alpha+\gamma=2$, $\alpha+\beta=3$, $\beta+\gamma=4$. Adding all three: $2(\alpha+\beta+\gamma)=9$, so $\alpha+\beta+\gamma=\frac{9}{2}$, hence $\gamma=\frac{9}{2}-3=\frac{3}{2}$, $\alpha=\frac{9}{2}-4=\frac{1}{2}$ and $\beta=\frac{9}{2}-2=\frac{5}{2}$. The coordinates are $\left(\frac{1}{2},\frac{5}{2},\frac{3}{2}\right)$.