Linear Algebra: Exam-Focused Concepts and Problem Solving: 19 MCQs with answers

A · $2x + 3y = 5$

A linear equation has the form $ax + by = c$ with each variable to the first power and no products. $2x+3y=5$ fits; the others contain $y^2$, $xy$, or $\sin x$.

B · $\begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$

The determinant is $1\cdot4 - 2\cdot3 = -2$. The inverse is $\frac{1}{-2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$.

D · $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

An elementary matrix is obtained by one elementary row operation on an identity matrix. Swapping rows of $I_2$ gives $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. The $3\times3$ matrix has a zero row and duplicate row, requiring multiple operations; the others are not from a single row operation on $I$.

C · $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

The matrix is a reflection; its square is the identity. Since 30 is even, the 30th power is the identity.

B · $15$

Determinants multiply: $\det(AB) = \det(A)\det(B) = 3 \cdot 5 = 15$.

A · True

A square matrix is invertible if and only if its determinant is nonzero.

B · $k = 0$ and $k = 4$

Set determinant to zero: $(k-2)^2 - 4 = 0 \Rightarrow (k-2)^2 = 4 \Rightarrow k-2 = \pm 2 \Rightarrow k = 0$ or $k = 4$.

C · $k = \pm \frac{5}{7}$

$\|v\| = \sqrt{4+9+0+36} = 7$. Then $\|kv\| = |k|\|v\| = 7|k| = 5$, so $|k| = 5/7$ and $k = \pm 5/7$.

A · Yes, with $a=2$, $b=-1$.

Solve $a(1,1,1)+b(1,0,1)=(1,2,1)$: $a+b=1$, $a=2$, $a+b=1$. With $a=2$, $b=-1$ works: $2(1,1,1)-1(1,0,1)=(1,2,1)$.

A · Yes, because their determinant is $-3 \neq 0$.

Form the matrix with these as columns: determinant $= 2\cdot0 - 3\cdot1 = -3 \neq 0$, so they are independent and span $\mathbb{R}^2$, hence a basis.

B · $\begin{pmatrix} 1 & i \\ -i & 2 \end{pmatrix}$

A Hermitian matrix satisfies $A^* = A$. For $\begin{pmatrix} 1 & i \\ -i & 2 \end{pmatrix}$, the conjugate transpose is $\begin{pmatrix} 1 & i \\ -i & 2 \end{pmatrix}$, so it is Hermitian. The others are not equal to their conjugate transpose.

A · $\sqrt{23}$

$\|u\| = \sqrt{|2-i|^2 + |4i|^2 + |1+i|^2} = \sqrt{5 + 16 + 2} = \sqrt{23}$.

B · $\{(a,0,0) : a \in \mathbb{R}\}$

A subspace must contain the zero vector. Only $\{(a,0,0)\}$ contains $(0,0,0)$; the others have a fixed nonzero component.

D · No, because $u \cdot v = 1 \neq 0$.

Orthogonal means dot product zero. $u \cdot v = 1\cdot0 + 2\cdot1 + (-1)\cdot1 = 1 \neq 0$, so they are not orthogonal.

B · Commutativity of scalar multiplication

The ten vector space axioms include closure, zero vector, and distributivity, but do not require commutativity of scalar multiplication (scalars multiply vectors, not each other in that way).

C · Infinitely many solutions forming a line

Row reduction shows the last two equations are negatives, so the system is consistent with one free variable, giving a line of solutions.

D · Only \( k = 0 \) and \( k = 4 \)

The matrix is singular when its determinant is zero. Calculating the determinant gives \( (k-2)^2 - 4 = 0 \), leading to \( k - 2 = \pm 2 \), thus \( k = 0 \) or \( k = 4 \).

A · 2, 3, 3

The eigenvalues of a triangular matrix are the entries on its diagonal. Hence, the eigenvalues of \( A \) are \( 2, 3, 3 \).

D · c_1 = 1, c_2 = -2, c_3 = 2

By equating coefficients from \( 6 + 11x + 6x^2 = c_1(2+x+4x^2) + c_2(1-x+3x^2) + c_3(3+2x+5x^2) \) and solving the resulting system, we find \( c_1 = 1, c_2 = -2, c_3 = 2 \).