PPSC Lecturer Math Linear Algebra Solved MCQs & Quiz

Question 1 (Part i): Determine whether the set $W = \{(x, y, z) \in \mathbb{R}^3 \mid 2x - y + 3z = 0\}$ forms a subspace of the real vector space $\mathbb{R}^3$.

Solution:
To verify if $W$ is a subspace of $\mathbb{R}^3$, we test three conditions: zero vector inclusion, additive closure, and scalar closure.

1. Zero Vector: Let $(x,y,z) = (0,0,0)$. Then $2(0) - (0) + 3(0) = 0$. Hence, $(0,0,0) \in W$.

2. Additive Closure: Let $u = (x_1, y_1, z_1) \in W$ and $v = (x_2, y_2, z_2) \in W$. Thus: $$2x_1 - y_1 + 3z_1 = 0 \quad \text{and} \quad 2x_2 - y_2 + 3z_2 = 0$$ Consider $u + v = (x_1 + x_2, y_1 + y_2, z_1 + z_2)$. Substitute into the defining equation: $$2(x_1 + x_2) - (y_1 + y_2) + 3(z_1 + z_2) = (2x_1 - y_1 + 3z_1) + (2x_2 - y_2 + 3z_2) = 0 + 0 = 0$$ Therefore, $u + v \in W$.

3. Scalar Closure: Let $c \in \mathbb{R}$ and $u = (x_1, y_1, z_1) \in W$. Then $c u = (cx_1, cy_1, cz_1)$. Testing the equation: $$2(cx_1) - (cy_1) + 3(cz_1) = c(2x_1 - y_1 + 3z_1) = c(0) = 0$$ Thus, $cu \in W$. All subspace criteria are satisfied.
Answer: $$W \text{ is a valid subspace of } \mathbb{R}^3$$

Question 1 (Part ii): Determine whether the set $W = \{(x, y, z) \in \mathbb{R}^3 \mid x + y + z = 1\}$ forms a subspace of $\mathbb{R}^3$.

Solution:
We first check if the zero vector $0_{\mathbb{R}^3} = (0,0,0)$ lies in $W$.
Substitute $(0,0,0)$ into the condition: $$0 + 0 + 0 = 0 \neq 1$$ Since the zero vector does not satisfy the defining equation of $W$, $W$ fails to contain the additive identity element.
Answer: $$W \text{ is NOT a subspace of } \mathbb{R}^3$$

Question 2 (Part i): Test whether the vectors $v_1 = (1, 2, 3)$, $v_2 = (0, 1, 2)$, and $v_3 = (3, 0, 1)$ in $\mathbb{R}^3$ are linearly independent.

Solution:
Construct a matrix $A$ with these vectors as rows and evaluate its determinant: $$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 0 & 1 \end{pmatrix}$$ Expanding along the first row: $$\det(A) = 1 \cdot \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 0 & 2 \\ 3 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 0 & 1 \\ 3 & 0 \end{vmatrix}$$ $$\det(A) = 1(1 - 0) - 2(0 - 6) + 3(0 - 3)$$ $$\det(A) = 1(1) - 2(-6) + 3(-3) = 1 + 12 - 9 = 4$$ Since $\det(A) = 4 \neq 0$, the row vectors are linearly independent.
Answer: $$\{v_1, v_2, v_3\} \text{ is Linearly Independent}$$

Question 2 (Part ii): Determine if the vectors $v_1 = (1, -1, 0)$, $v_2 = (2, 1, 3)$, and $v_3 = (0, 3, 3)$ are linearly independent in $\mathbb{R}^3$.

Solution:
Set up the linear combination equation $c_1 v_1 + c_2 v_2 + c_3 v_3 = 0$: $$c_1(1, -1, 0) + c_2(2, 1, 3) + c_3(0, 3, 3) = (0, 0, 0)$$ This yields the system of equations: $$\begin{aligned} c_1 + 2c_2 &= 0 \quad &(1) \\ -c_1 + c_2 + 3c_3 &= 0 \quad &(2) \\ 3c_2 + 3c_3 &= 0 \quad &(3) \end{aligned}$$ From equation (3), $3c_2 = -3c_3 \implies c_2 = -c_3$.
Substitute $c_2 = -c_3$ into (1): $c_1 = -2c_2 = 2c_3$.
Substitute $c_1 = 2c_3$ and $c_2 = -c_3$ into (2): $$-(2c_3) + (-c_3) + 3c_3 = -3c_3 + 3c_3 = 0$$ This identity holds for any non-zero value of $c_3$. Choosing $c_3 = 1$ gives $c_2 = -1$ and $c_1 = 2$.
Since non-trivial coefficients exist ($2v_1 - v_2 + v_3 = 0$), the vectors are linearly dependent.
Answer: $$\{v_1, v_2, v_3\} \text{ is Linearly Dependent}$$

Question 3 (Part i): Find a basis and the dimension for the solution space of the homogeneous linear system: $$x + 2y - z = 0$$ $$2x + 5y + z = 0$$

Solution:
Write the augmented coefficient matrix and perform Row Echelon Form reduction: $$A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 5 & 1 \end{pmatrix} \xrightarrow{R_2 \to R_2 - 2R_1} \begin{pmatrix} 1 & 2 & -1 \\ 0 & 1 & 3 \end{pmatrix}$$ Perform $R_1 \to R_1 - 2R_2$: $$\begin{pmatrix} 1 & 0 & -7 \\ 0 & 1 & 3 \end{pmatrix}$$ Translating back to variables: $$x - 7z = 0 \implies x = 7z$$ $$y + 3z = 0 \implies y = -3z$$ Let $z = t$ be a free parameter ($t \in \mathbb{R}$). Any solution vector $(x,y,z)^T$ is expressed as: $$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 7t \\ -3t \\ t \end{pmatrix} = t \begin{pmatrix} 7 \\ -3 \\ 1 \end{pmatrix}$$ Thus, the solution space is spanned by the single vector $(7, -3, 1)^T$.
Answer: $$\text{Basis} = \left\{ \begin{pmatrix} 7 \\ -3 \\ 1 \end{pmatrix} \right\}, \quad \text{Dimension} = 1$$

Question 3 (Part ii): Find a basis and dimension for the subspace of $\mathbb{R}^3$ spanned by $v_1 = (1, 1, 2)$, $v_2 = (2, 2, 4)$, and $v_3 = (3, 1, 1)$.

Solution:
Place the vectors as rows of a matrix and reduce to Row Echelon Form: $$M = \begin{pmatrix} 1 & 1 & 2 \\ 2 & 2 & 4 \\ 3 & 1 & 1 \end{pmatrix}$$ Apply row operations $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$: $$M \sim \begin{pmatrix} 1 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & -2 & -5 \end{pmatrix}$$ Swap row 2 and row 3 ($R_2 \leftrightarrow R_3$): $$M \sim \begin{pmatrix} 1 & 1 & 2 \\ 0 & -2 & -5 \\ 0 & 0 & 0 \end{pmatrix}$$ The non-zero rows of the reduced echelon matrix form a basis for the spanned subspace.
Answer: $$\text{Basis} = \{(1, 1, 2), (0, -2, -5)\}, \quad \text{Dimension} = 2$$

Question 4 (Part i): Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation defined by $T(x, y, z) = (x + y, y - z)$. Find the kernel $\ker(T)$ and the nullity of $T$.

Solution:
By definition, $\ker(T) = \{(x, y, z) \in \mathbb{R}^3 \mid T(x, y, z) = (0, 0)\}$.
Setting $(x + y, y - z) = (0, 0)$ gives the system: $$\begin{aligned} x + y &= 0 \implies x = -y \\ y - z &= 0 \implies z = y \end{aligned}$$ Expressing in terms of free variable $y = t$: $$(x, y, z) = (-t, t, t) = t(-1, 1, 1)$$ Thus, $\ker(T)$ is the one-dimensional subspace spanned by $(-1, 1, 1)$.
Answer: $$\ker(T) = \text{span}\{(-1, 1, 1)\}, \quad \text{Nullity}(T) = 1$$

Question 4 (Part ii): For the linear transformation $T(x, y, z) = (x + y, y - z)$, find the range space $\text{Im}(T)$ and rank of $T$.

Solution:
Using the Rank-Nullity Theorem: $$\dim(\text{Domain}) = \text{Rank}(T) + \text{Nullity}(T)$$ Here, domain is $\mathbb{R}^3$, so $\dim(\mathbb{R}^3) = 3$. From Part (i), $\text{Nullity}(T) = 1$.
$$\text{Rank}(T) = 3 - 1 = 2$$ Since $\text{Im}(T) \subseteq \mathbb{R}^2$ and $\dim(\text{Im}(T)) = 2 = \dim(\mathbb{R}^2)$, the range space is the entirety of $\mathbb{R}^2$.
Answer: $$\text{Im}(T) = \mathbb{R}^2, \quad \text{Rank}(T) = 2$$

Question 5 (Part i): Find all eigenvalues of the matrix $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$.

Solution:
Solve the characteristic equation $\det(A - \lambda I) = 0$: $$\det \begin{pmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{pmatrix} = 0$$ $$(2 - \lambda)^2 - 1 = 0$$ $$\lambda^2 - 4\lambda + 4 - 1 = 0 \implies \lambda^2 - 4\lambda + 3 = 0$$ Factoring the quadratic equation: $$(\lambda - 3)(\lambda - 1) = 0 \implies \lambda_1 = 3, \quad \lambda_2 = 1$$ Answer: $$\lambda = 1, 3$$

Question 5 (Part ii): Find the eigenvectors corresponding to each eigenvalue of $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$.

Solution:
1. For $\lambda_1 = 3$: Solve $(A - 3I)v = 0$: $$\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies -x + y = 0 \implies x = y$$ Eigenvector $v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

2. For $\lambda_2 = 1$: Solve $(A - 1I)v = 0$: $$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies x + y = 0 \implies x = -y$$ Eigenvector $v_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
Answer: $$v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \text{ for } \lambda=3, \quad v_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \text{ for } \lambda=1$$

Question 6 (Part i): Verify the Cayley-Hamilton Theorem for the matrix $M = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$.

Solution:
The characteristic equation of $M$ is $\det(M - \lambda I) = 0$: $$\det \begin{pmatrix} 1 - \lambda & 2 \\ 3 & 4 - \lambda \end{pmatrix} = (1 - \lambda)(4 - \lambda) - 6 = \lambda^2 - 5\lambda - 2 = 0$$ By Cayley-Hamilton Theorem, $M$ must satisfy its own characteristic equation: $M^2 - 5M - 2I = 0$.
Calculate $M^2$: $$M^2 = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 1+6 & 2+8 \\ 3+12 & 6+16 \end{pmatrix} = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix}$$ Substitute $M^2$, $M$, and $I$ into the polynomial expression: $$M^2 - 5M - 2I = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix} - 5\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - 2\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ $$= \begin{pmatrix} 7 - 5 - 2 & 10 - 10 - 0 \\ 15 - 15 - 0 & 22 - 20 - 2 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$ Answer: $$M^2 - 5M - 2I = \mathbf{0} \quad (\text{Theorem Verified})$$

Question 6 (Part ii): Use the Cayley-Hamilton equation verified in Part (i) to find the matrix inverse $M^{-1}$.

Solution:
From Cayley-Hamilton: $M^2 - 5M - 2I = 0 \implies 2I = M^2 - 5M$.
Multiply both sides by $M^{-1}$: $$2 M^{-1} = M - 5I \implies M^{-1} = \frac{1}{2}(M - 5I)$$ Substitute $M$ and $I$: $$M^{-1} = \frac{1}{2} \left[ \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \right] = \frac{1}{2} \begin{pmatrix} -4 & 2 \\ 3 & -1 \end{pmatrix}$$ $$M^{-1} = \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}$$ Answer: $$M^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}$$