Sample Mathematics Solutions

Q.1 [14 marks]

Fill in the blanks using appropriate choice from the given options

Q1.1 [1 mark]

If $A_{2×3}$ and $B_{3×4}$ are two matrices then find order of $AB$ = ______

a. $4×2$ b. $2×4$ c. $3×3$ d. AB is not possible

Answer: a. $2×4$

Solution: For matrix multiplication $AB$ to be possible, the number of columns in matrix $A$ must equal the number of rows in matrix $B$ .

Given:

Matrix $A$ has order $2×3$ (2 rows, 3 columns)
Matrix $B$ has order $3×4$ (3 rows, 4 columns)

Since the number of columns in $A$ (3) equals the number of rows in $B$ (3), multiplication is possible.

The order of the resultant matrix $AB$ will be: $AB_{(2×3)} \times B_{(3×4)} = (AB)_{(2×4)}$

Therefore, the order of $AB$ is $2×4$ .

Q1.2 [1 mark]

If $A = [1\quad 3\quad 2]$ and $B = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$ then find $AB$ = ______

a. Not possible b. $9$ c. $[1\quad 1]$ d. $[1\quad 6\quad 2]$

Answer: b. $9$

Solution: Given: $A = [1\quad 3\quad 2] \text{ (order: } 1×3\text{)}$ $B = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} \text{ (order: } 3×1\text{)}$

Since $A$ is $1×3$ and $B$ is $3×1$ , multiplication is possible and the result will be $1×1$ (a scalar).

$AB = [1\quad 3\quad 2] \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$

$AB = (1)(1) + (3)(2) + (2)(1)$ $AB = 1 + 6 + 2 = 9$

Q1.3 [1 mark]

If $A \cdot I_2 = A$ then $I_2$ = ______

a. $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ b. $\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$ c. $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ d. $\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$

Answer: c. $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Solution: The condition $A \cdot I_2 = A$ means that $I_2$ is the identity matrix of order $2×2$ .

The identity matrix has the property that when any matrix is multiplied by it, the original matrix remains unchanged.

The $2×2$ identity matrix is: $I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Verification: For any $2×2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ :

$A \cdot I_2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = A$

Calculus Problems

Q1.4 [1 mark]

If $\frac{d}{dx}(\sin^2 x + \cos^2 x) =$ ______

a. $1$ b. $0$ c. $-1$ d. $x$

Answer: b. $0$

Solution: We know that $\sin^2 x + \cos^2 x = 1$ (fundamental trigonometric identity).

Since the derivative of a constant is zero: $\frac{d}{dx}(\sin^2 x + \cos^2 x) = \frac{d}{dx}(1) = 0$

Alternative approach (step by step): $\frac{d}{dx}(\sin^2 x + \cos^2 x)$ $= \frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(\cos^2 x)$ $= 2\sin x \cos x + 2\cos x(-\sin x)$ $= 2\sin x \cos x - 2\sin x \cos x = 0$

Integration Example

Evaluate: $\int x^5 dx$

Solution: Using the power rule for integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ where $n \neq -1$

$\int x^5 dx = \frac{x^{5+1}}{5+1} + C = \frac{x^6}{6} + C$

Complex Numbers

Sample Problem

Simplify: $(3+4i)(4-5i)$

Solution: $(3+4i)(4-5i)$ $= 3(4) + 3(-5i) + 4i(4) + 4i(-5i)$ $= 12 - 15i + 16i - 20i^2$

Since $i^2 = -1$ : $= 12 - 15i + 16i - 20(-1)$ $= 12 - 15i + 16i + 20$ $= 32 + i$

Differential Equations

Sample Problem

Solve: $x\frac{dy}{dx} + y = 0$

Solution: This is a separable differential equation.

Rearranging: $x\frac{dy}{dx} = -y$

Separating variables: $\frac{dy}{y} = -\frac{dx}{x}$

Integrating both sides: $\int \frac{dy}{y} = \int -\frac{dx}{x}$ $\ln|y| = -\ln|x| + C_1$ $\ln|y| = \ln|x^{-1}| + C_1$ $|y| = e^{\ln|x^{-1}| + C_1} = e^{C_1} \cdot |x^{-1}|$

Let $C = \pm e^{C_1}$ , then: $y = \frac{C}{x}$

This is the general solution.