MATH 1210: Techniques of Classical and Linear Algebra - Matrices, Vectors, Lines and Planes

Practice question-and-answer flashcards covering key problems on matrices, vectors, lines, and planes from MATH 1210.

In Exercise 6.1, why is the matrix subtraction $A - C$ undefined for $A = \begin{pmatrix} 2 & 1 & 3 \\ 1 & 5 & -2 \end{pmatrix}$ and $C = \begin{pmatrix} 1 & 0 & 2 \\ 1 & -1 & 3 \\ 0 & 2 & 1 \end{pmatrix}$?

Matrix subtraction is undefined because $A$ has size $2 \times 3$ and $C$ has size $3 \times 3$, meaning their dimensions are not identical.

In Exercise 6.1(b), what is the transpose $C^T$ of the matrix $C = \begin{pmatrix} 1 & 0 & 2 \\ 1 & -1 & 3 \\ 0 & 2 & 1 \end{pmatrix}$?

$$C^T = \begin{pmatrix} 1 & 1 & 0 \\ 0 & -1 & 2 \\ 2 & 3 & 1 \end{pmatrix}$$

In Exercise 6.1(c), what is the linear combination $3C + 2D$ for matrices $C = \begin{pmatrix} 1 & 0 & 2 \\ 1 & -1 & 3 \\ 0 & 2 & 1 \end{pmatrix}$ and $D = \begin{pmatrix} 5 & 1 & 0 \\ 0 & 3 & 2 \\ 1 & 4 & -1 \end{pmatrix}$?

$$3C + 2D = \begin{pmatrix} 13 & 2 & 6 \\ 3 & 3 & 13 \\ 2 & 14 & 1 \end{pmatrix}$$

In Exercise 6.1(d), what is the evaluated matrix $4A^T - 5B$ given $A = \begin{pmatrix} 2 & 1 & 3 \\ 1 & 5 & -2 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & -1 \\ 1 & 2 \\ 3 & 1 \end{pmatrix}$?

$$4A^T - 5B = \begin{pmatrix} 8 & 9 \\ -1 & 10 \\ -3 & -13 \end{pmatrix}$$

In Exercise 6.2, how do you isolate the matrix $X$ in the equation $\left(2X - \begin{pmatrix} 0 & -2 & 1 \\ -1 & 3 & 4 \end{pmatrix}\right)^T = 3 \begin{pmatrix} 2 & 1 \\ 0 & 5 \\ -3 & 2 \end{pmatrix}$?

Transpose both sides to obtain $2X - \begin{pmatrix} 0 & -2 & 1 \\ -1 & 3 & 4 \end{pmatrix} = 3 \begin{pmatrix} 2 & 0 & -3 \\ 1 & 5 & 2 \end{pmatrix}$, add $\begin{pmatrix} 0 & -2 & 1 \\ -1 & 3 & 4 \end{pmatrix}$ to both sides, and multiply by $\frac{1}{2}$.

In Exercise 6.3(b), why is the matrix product $AB^T$ undefined for $A = \begin{pmatrix} -2 & 5 \\ 3 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -2 & 4 \\ 2 & 1 & 0 \end{pmatrix}$?

Matrix $A$ has dimension $2 \times 2$ and $B^T$ has dimension $3 \times 2$. Since the number of columns of $A$ ($2$) does not equal the number of rows of $B^T$ ($3$), the multiplication cannot be performed.

In Exercise 6.3(a), what is the result of the product $DC$ given $D = \begin{pmatrix} 0 & 1 & 2 \\ -1 & -2 & 0 \\ 3 & 0 & 1 \end{pmatrix}$ and $C = \begin{pmatrix} -3 \\ 1 \\ -1 \end{pmatrix}$?

$$DC = \begin{pmatrix} -1 \\ 1 \\ -10 \end{pmatrix}$$

In Exercise 6.4, what polynomial equation must each diagonal element $d$ satisfy for a diagonal $2 \times 2$ matrix $A$ to fulfill $3A - A^3 = 2I_2$?

Each diagonal entry must satisfy $3d - d^3 = 2$, which rearranges to $d^3 - 3d + 2 = 0$ and factors as $(d - 1)^2(d + 2) = 0$, yielding possible values $d = 1$ and $d = -2$.

In Exercise 6.5(a)(iii), what is the evaluated value of $(AD)^T - D^TA^T$?

By the matrix transpose property $(AD)^T = D^TA^T$, the expression simplifies identically to the zero matrix $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.

In Exercise 6.5(a)(iv), why is the matrix expression $3D^T + BC$ defined for $B = \begin{pmatrix} 3 & 1 & 2 \\ -4 & 0 & 5 \end{pmatrix}$, $C = \begin{pmatrix} 1 & 0 \\ -2 & 3 \\ 4 & 2 \end{pmatrix}$, and $D = \begin{pmatrix} 9 & -2 \\ 1 & 10 \end{pmatrix}$?

$3D^T$ is a $2 \times 2$ matrix and $BC$ is the product of a $2 \times 3$ matrix and a $3 \times 2$ matrix, which yields a $2 \times 2$ matrix; since both terms have size $2 \times 2$, their sum is defined.

In Exercise 7.1(a), what is the linear combination $2u + 5v$ for $u = \langle 4, -3, 5 \rangle$ and $v = \langle -2, 1, 2 \rangle$?

$$2u + 5v = 2\langle 4, -3, 5 \rangle + 5\langle -2, 1, 2 \rangle = \langle -2, -1, 20 \rangle$$

In Exercise 7.1(c), what is the unit vector $\hat{u}$ in the direction of $u = \langle 4, -3, 5 \rangle$?

$$\hat{u} = \frac{u}{\|u\|} = \frac{\langle 4, -3, 5 \rangle}{\sqrt{4^2 + (-3)^2 + 5^2}} = \frac{1}{5\sqrt{2}}\langle 4, -3, 5 \rangle = \langle \frac{4}{5\sqrt{2}}, -\frac{3}{5\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle$$

In Exercise 7.1(e), what is the value of the dot product $(u - 2v) \cdot (u + 2v)$ for $u = \langle 4, -3, 5 \rangle$ and $v = \langle -2, 1, 2 \rangle$?

Using algebraic expansion, $(u - 2v) \cdot (u + 2v) = \|u\|^2 - 4\|v\|^2 = 50 - 4(9) = 50 - 36 = 14$.

In Exercise 7.1(h), how is it proven that $\triangle PQR$ is a right triangle for points $P(0, 8, 3)$, $Q(1, 6, 2)$, and $R(-3, 5, 0)$?

By calculating vectors $\vec{QP} = \langle -1, 2, 1 \rangle$ and $\vec{QR} = \langle -4, -1, -2 \rangle$ and showing that their dot product is zero: $\vec{QP} \cdot \vec{QR} = (-1)(-4) + (2)(-1) + (1)(-2) = 4 - 2 - 2 = 0$, proving that the angle at $Q$ is $90^\circ$.

In Exercise 7.2, for what real value of $k$ are the vectors $u = \langle k - 1, 3, -6 \rangle$ and $v = \langle 2, k, 2 - k \rangle$ parallel?

The vectors are parallel when their components are proportional: $\frac{k - 1}{2} = \frac{3}{k} = \frac{-6}{2 - k}$, which holds uniquely when $k = -2$.

In Exercise 7.3(b), what is the value of the scalar triple product $(u \times v) \cdot w$ for $u = \langle 2, 0, -1 \rangle$, $v = \langle 1, -1, 3 \rangle$, and $w = \langle 0, -2, 7 \rangle$?

The scalar triple product equals the determinant of the $3 \times 3$ matrix formed by the vectors: $\det\begin{pmatrix} 2 & 0 & -1 \\ 1 & -1 & 3 \\ 0 & -2 & 7 \end{pmatrix} = 2(-7 + 6) - 1(-2 - 0) = -2 + 2 = 0$.

In Exercise 7.4(a), what is the vector of length $7$ in the opposite direction of $4u - 3v$ where $u = \langle -7, 2, 1 \rangle$ and $v = \langle -9, 2, 1 \rangle$?

Since $4u - 3v = \langle -1, 2, 1 \rangle$ with norm $\sqrt{(-1)^2 + 2^2 + 1^2} = \sqrt{6}$, the required vector is $-7\frac{4u - 3v}{\|4u - 3v\|} = -\frac{7}{\sqrt{6}}\langle -1, 2, 1 \rangle = \langle \frac{7}{\sqrt{6}}, -\frac{14}{\sqrt{6}}, -\frac{7}{\sqrt{6}} \rangle$.

In Exercise 8.1(a), what is an equation of the plane passing through point $P(0, 5, -2)$ and perpendicular to $\vec{QR}$, where $Q(-3, 1, 4)$ and $R(-1, 2, 6)$?

The normal vector is $\vec{QR} = \langle 2, 1, 2 \rangle$, giving the plane equation $2(x - 0) + 1(y - 5) + 2(z + 2) = 0$, which simplifies to $2x + y + 2z = 1$.

In Exercise 8.1(b), what are parametric equations of the line passing through points $P(0, 5, -2)$ and $Q(-3, 1, 4)$?

Using $P$ and direction vector $\vec{PQ} = \langle -3, -4, 6 \rangle$, the equations are $x = -3t$, $y = 5 - 4t$, $z = -2 + 6t$, with $t \in \mathbb{R}$.

In Exercise 8.2(a), why are the lines $\ell_1 : x = 1 - t, y = -3, z = 5 + 2t$ and $\ell_2 : x = 3 + 2t, y = 0, z = 7 - 4t$ parallel?

Their direction vectors are $d_1 = \langle -1, 0, 2 \rangle$ and $d_2 = \langle 2, 0, -4 \rangle$, which satisfy $d_2 = -2d_1$, meaning they have parallel direction vectors.

In Exercise 8.2(b), what are the coordinates of the intersection point $P$ of the lines $\ell_1 : x = 1 - t, y = -3, z = 5 + 2t$ and $\ell_3 : x = 4 + t, y = 1 + 2t, z = 1 - t$?

The intersection point is $P(2, -3, 3)$, found by setting parameter $t = -1$ in $\ell_1$ and parameter value $-2$ in $\ell_3$.

In Exercise 8.3, what is the geometric relationship between the line $\ell : x = 3 + 2t, y = -1 + t, z = 3t$ and the plane $\pi : 5x + 2y - 4z = 9$?

The line $\ell$ is parallel to the plane $\pi$ and does not intersect it, because the direction vector $\langle 2, 1, 3 \rangle$ is orthogonal to the normal vector $\langle 5, 2, -4 \rangle$ ($5(2) + 2(1) - 4(3) = 0$), but substituting the line into the plane yields the false equality $13 = 9$.

In Exercise 8.3, what is an equation of the plane that is parallel to $\pi : 5x + 2y - 4z = 9$ and contains the line $\ell : x = 3 + 2t, y = -1 + t, z = 3t$?

Using the normal vector $\langle 5, 2, -4 \rangle$ and the point on the line $(3, -1, 0)$, the plane equation is $5(x - 3) + 2(y + 1) - 4(z - 0) = 0$, which simplifies to $5x + 2y - 4z = 13$.

In Exercise 8.4(a), what is the point of intersection $Q$ between the lines $\ell_1 : x = 3 + t, y = 1 - t, z = 3t$ and $\ell_2 : \frac{x}{2} = y + 2 = \frac{z - 5}{-1}$?

The point of intersection is $Q(4, 0, 3)$, corresponding to $t = 1$ for line $\ell_1$ and parameter value $2$ for line $\ell_2$.

In Exercise 8.4(d), what are the parametric equations of the line that passes through the origin and is parallel to line $\ell_1 : x = 3 + t, y = 1 - t, z = 3t$?

The direction vector of $\ell_1$ is $\langle 1, -1, 3 \rangle$, so the parametric equations passing through $(0, 0, 0)$ are $x = t$, $y = -t$, $z = 3t$, with $t \in \mathbb{R}$.