Question: Mat0900 Mathematics - Academic Year 2025/26, Semester 3
UNIVERSITY OF THE WEST INDIES-OPEN CAMPUS
MATH0900 6 2025/26
SEMESTER 3
Tutorial 6 (Assignment 2) - Individual Assignment
(Based on UNITS 5, 6, 7, 8)
1) If $A = \begin{pmatrix}4 & 2 \\ 5 & 3\end{pmatrix}$ and $B = \begin{pmatrix}-5 & -2 \\ 4 & 1\end{pmatrix}$
Find:- a) $A^2$ b) Adjoint $A$ c) $B^{-1}$ (5, 2, 5mks)
2) a) Using a matrix method solve the following simultaneous equations
$2x + 5y = 13$ and $2x - 3y = 5$ (8 marks)
b) Find the determinant of the following matrix
$\begin{pmatrix} 2 & 3 & -6 \\ 0 & 5 & -1 \\ 1 & -2 & 6 \end{pmatrix}$ (5 marks)
c) If $|A| = -16$ find the possible value(s) for $x$
$A = \begin{pmatrix} x & 4 \\ 3x & 2x \end{pmatrix}$ (6 marks)
3) Given that $f(x) = 2x + 5$, $g(x) = \frac{6 - 2x}{x}$ and $h(x) = 3x + 7$
Find:- i) $f(4)$ = ii) $g[h(-3)]$ = iii) $f[h(-2)]$ =
iv) $h \circ f(x)$ v) $g^{-1}(4)$ = (2, 3, 3, 3, 4 marks)
4) Given the functions $g(x) = \frac{3x + 4}{2x + 1}$ Find $g^{-1}(x)$
(5 marks)
5) a) Two numbers when multiplied equals 96.
If the difference of the two numbers is 4 and $x$ represents the larger number
i) Represent the information given in equation form
ii) Express as a quadratic equation
iii) Solve the quadratic equation to find possible values for $x$
iv) If told that $x$ is a positive numbers, what are the values of the two original numbers?
(8 marks)
b) Solve the following quadratic equations ( hint use the quadratic formula)
i) $2x^2 + 5x - 12 = 0$ ii) $12x^2 + 2x = 4$ (5, 5 Marks)
6) Find the equation of the line joining the points $(3, -5)$ and $(-2, 7)$. (5 marks)
7) Find the equation of the line passing through the point $(6, -4)$ and
perpendicular to the line $6 = 4y + 3x$ (6 marks)
Note Tutorial is out of 80 Marks but will contribute to 40% of your course work marks.
( easy calculations your score out of 80 divided by 2 is your course work score out of 40)
1. **Problem:** Given matrices $A = \begin{pmatrix}4 & 2 \\ 5 & 3\end{pmatrix}$ and $B = \begin{pmatrix}-5 & -2 \\ 4 & 1\end{pmatrix}$, find:
a) $A^2$
b) Adjoint of $A$
c) Inverse of $B$, $B^{-1}$
---
**Step 1: Calculate $A^2 = A \times A$**
$$A^2 = \begin{pmatrix}4 & 2 \\ 5 & 3\end{pmatrix} \times \begin{pmatrix}4 & 2 \\ 5 & 3\end{pmatrix} = \begin{pmatrix}4\times4 + 2\times5 & 4\times2 + 2\times3 \\ 5\times4 + 3\times5 & 5\times2 + 3\times3\end{pmatrix} = \begin{pmatrix}16 + 10 & 8 + 6 \\ 20 + 15 & 10 + 9\end{pmatrix} = \begin{pmatrix}26 & 14 \\ 35 & 19\end{pmatrix}$$
---
**Step 2: Find Adjoint of $A$**
Adjoint of a $2\times2$ matrix $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ is $\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}$.
For $A$, $a=4$, $b=2$, $c=5$, $d=3$.
So,
$$\text{Adj}(A) = \begin{pmatrix}3 & -2 \\ -5 & 4\end{pmatrix}$$
---
**Step 3: Find $B^{-1}$**
Inverse of $2\times2$ matrix $B = \begin{pmatrix}a & b \\ c & d\end{pmatrix}$ is
$$B^{-1} = \frac{1}{|B|} \begin{pmatrix}d & -b \\ -c & a\end{pmatrix}$$
where determinant $|B| = ad - bc$.
Calculate $|B|$:
$$|B| = (-5)(1) - (-2)(4) = -5 + 8 = 3$$
Calculate adjoint of $B$:
$$\text{Adj}(B) = \begin{pmatrix}1 & 2 \\ -4 & -5\end{pmatrix}$$
Therefore,
$$B^{-1} = \frac{1}{3} \begin{pmatrix}1 & 2 \\ -4 & -5\end{pmatrix} = \begin{pmatrix}\frac{1}{3} & \frac{2}{3} \\ -\frac{4}{3} & -\frac{5}{3}\end{pmatrix}$$
---
2. a) **Problem:** Solve simultaneous equations using matrix method:
$$2x + 5y = 13$$
$$2x - 3y = 5$$
---
**Step 1: Write in matrix form $AX = B$**
$$A = \begin{pmatrix}2 & 5 \\ 2 & -3\end{pmatrix}, \quad X = \begin{pmatrix}x \\ y\end{pmatrix}, \quad B = \begin{pmatrix}13 \\ 5\end{pmatrix}$$
---
**Step 2: Find $A^{-1}$**
Determinant:
$$|A| = 2 \times (-3) - 5 \times 2 = -6 - 10 = -16$$
Adjoint:
$$\text{Adj}(A) = \begin{pmatrix}-3 & -5 \\ -2 & 2\end{pmatrix}$$
Inverse:
$$A^{-1} = \frac{1}{-16} \begin{pmatrix}-3 & -5 \\ -2 & 2\end{pmatrix} = \begin{pmatrix}\frac{3}{16} & \frac{5}{16} \\ \frac{1}{8} & -\frac{1}{8}\end{pmatrix}$$
---
**Step 3: Calculate $X = A^{-1}B$**
$$X = \begin{pmatrix}\frac{3}{16} & \frac{5}{16} \\ \frac{1}{8} & -\frac{1}{8}\end{pmatrix} \begin{pmatrix}13 \\ 5\end{pmatrix} = \begin{pmatrix}\frac{3}{16} \times 13 + \frac{5}{16} \times 5 \\ \frac{1}{8} \times 13 - \frac{1}{8} \times 5\end{pmatrix} = \begin{pmatrix}\frac{39}{16} + \frac{25}{16} \\ \frac{13}{8} - \frac{5}{8}\end{pmatrix} = \begin{pmatrix}\frac{64}{16} \\ \frac{8}{8}\end{pmatrix} = \begin{pmatrix}4 \\ 1\end{pmatrix}$$
So, $x=4$, $y=1$.
---
2. b) **Problem:** Find determinant of
$$M = \begin{pmatrix} 2 & 3 & -6 \\ 0 & 5 & -1 \\ 1 & -2 & 6 \end{pmatrix}$$
---
**Step 1: Use cofactor expansion along first row:**
$$|M| = 2 \times \begin{vmatrix}5 & -1 \\ -2 & 6\end{vmatrix} - 3 \times \begin{vmatrix}0 & -1 \\ 1 & 6\end{vmatrix} + (-6) \times \begin{vmatrix}0 & 5 \\ 1 & -2\end{vmatrix}$$
Calculate minors:
$$\begin{vmatrix}5 & -1 \\ -2 & 6\end{vmatrix} = 5 \times 6 - (-1) \times (-2) = 30 - 2 = 28$$
$$\begin{vmatrix}0 & -1 \\ 1 & 6\end{vmatrix} = 0 \times 6 - (-1) \times 1 = 0 + 1 = 1$$
$$\begin{vmatrix}0 & 5 \\ 1 & -2\end{vmatrix} = 0 \times (-2) - 5 \times 1 = 0 - 5 = -5$$
Substitute:
$$|M| = 2 \times 28 - 3 \times 1 + (-6) \times (-5) = 56 - 3 + 30 = 83$$
---
2. c) **Problem:** Given $|A| = -16$ and
$$A = \begin{pmatrix} x & 4 \\ 3x & 2x \end{pmatrix}$$
Find possible values of $x$.
---
**Step 1: Calculate determinant:**
$$|A| = x \times 2x - 4 \times 3x = 2x^2 - 12x$$
Set equal to $-16$:
$$2x^2 - 12x = -16$$
Rewrite:
$$2x^2 - 12x + 16 = 0$$
Divide both sides by 2:
$$\cancel{2}x^2 - \cancel{2}6x + \cancel{2}8 = 0 \Rightarrow x^2 - 6x + 8 = 0$$
---
**Step 2: Solve quadratic equation:**
$$x = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 1 \times 8}}{2} = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm 2}{2}$$
So,
$$x = \frac{6 + 2}{2} = 4 \quad \text{or} \quad x = \frac{6 - 2}{2} = 2$$
---
3. **Problem:** Given functions $f(x) = 2x + 5$, $g(x) = \frac{6 - 2x}{x}$, $h(x) = 3x + 7$, find:
i) $f(4)$
ii) $g[h(-3)]$
iii) $f[h(-2)]$
iv) $h \circ f(x)$
v) $g^{-1}(4)$
---
**Step 1: Calculate $f(4)$:**
$$f(4) = 2 \times 4 + 5 = 8 + 5 = 13$$
---
**Step 2: Calculate $h(-3)$:**
$$h(-3) = 3 \times (-3) + 7 = -9 + 7 = -2$$
Then,
$$g[h(-3)] = g(-2) = \frac{6 - 2(-2)}{-2} = \frac{6 + 4}{-2} = \frac{10}{-2} = -5$$
---
**Step 3: Calculate $h(-2)$:**
$$h(-2) = 3 \times (-2) + 7 = -6 + 7 = 1$$
Then,
$$f[h(-2)] = f(1) = 2 \times 1 + 5 = 7$$
---
**Step 4: Find $h \circ f(x) = h(f(x))$:**
$$f(x) = 2x + 5$$
$$h(f(x)) = 3(2x + 5) + 7 = 6x + 15 + 7 = 6x + 22$$
---
**Step 5: Find $g^{-1}(4)$:**
Given $g(x) = \frac{6 - 2x}{x}$, find $x$ such that $g(x) = 4$.
Set:
$$4 = \frac{6 - 2x}{x}$$
Multiply both sides by $x$:
$$4x = 6 - 2x$$
Add $2x$ to both sides:
$$4x + 2x = 6$$
$$6x = 6$$
Divide both sides by 6:
$$x = 1$$
So,
$$g^{-1}(4) = 1$$
---
4. **Problem:** Given $g(x) = \frac{3x + 4}{2x + 1}$, find $g^{-1}(x)$.
---
**Step 1: Set $y = g(x)$:**
$$y = \frac{3x + 4}{2x + 1}$$
**Step 2: Solve for $x$ in terms of $y$:**
Multiply both sides by $2x + 1$:
$$y(2x + 1) = 3x + 4$$
Expand:
$$2xy + y = 3x + 4$$
Group $x$ terms on one side:
$$2xy - 3x = 4 - y$$
Factor $x$:
$$x(2y - 3) = 4 - y$$
Divide both sides:
$$x = \frac{4 - y}{2y - 3}$$
**Step 3: Replace $y$ by $x$ to write inverse function:**
$$g^{-1}(x) = \frac{4 - x}{2x - 3}$$
---
5. a) **Problem:** Two numbers multiply to 96, difference is 4, $x$ is larger number.
Find:
i) Equation form
ii) Quadratic equation
iii) Solve quadratic
iv) Positive $x$ values and original numbers
---
**Step 1: Let smaller number be $x - 4$**
Product:
$$x(x - 4) = 96$$
---
**Step 2: Expand and form quadratic:**
$$x^2 - 4x = 96$$
Rewrite:
$$x^2 - 4x - 96 = 0$$
---
**Step 3: Solve quadratic using formula:**
$$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times (-96)}}{2} = \frac{4 \pm \sqrt{16 + 384}}{2} = \frac{4 \pm \sqrt{400}}{2} = \frac{4 \pm 20}{2}$$
So,
$$x = \frac{4 + 20}{2} = 12 \quad \text{or} \quad x = \frac{4 - 20}{2} = -8$$
---
**Step 4: Since $x$ is positive, $x=12$**
Smaller number:
$$12 - 4 = 8$$
Original numbers are 12 and 8.
---
5. b) Solve quadratic equations:
i) $2x^2 + 5x - 12 = 0$
**Step 1: Use quadratic formula:**
$$x = \frac{-5 \pm \sqrt{5^2 - 4 \times 2 \times (-12)}}{2 \times 2} = \frac{-5 \pm \sqrt{25 + 96}}{4} = \frac{-5 \pm \sqrt{121}}{4} = \frac{-5 \pm 11}{4}$$
Solutions:
$$x = \frac{-5 + 11}{4} = \frac{6}{4} = 1.5$$
$$x = \frac{-5 - 11}{4} = \frac{-16}{4} = -4$$
---
ii) $12x^2 + 2x = 4$
Rewrite:
$$12x^2 + 2x - 4 = 0$$
Divide by 2:
$$6x^2 + x - 2 = 0$$
Use quadratic formula:
$$x = \frac{-1 \pm \sqrt{1^2 - 4 \times 6 \times (-2)}}{2 \times 6} = \frac{-1 \pm \sqrt{1 + 48}}{12} = \frac{-1 \pm \sqrt{49}}{12} = \frac{-1 \pm 7}{12}$$
Solutions:
$$x = \frac{-1 + 7}{12} = \frac{6}{12} = 0.5$$
$$x = \frac{-1 - 7}{12} = \frac{-8}{12} = -\frac{2}{3}$$
---
6. **Problem:** Find equation of line through points $(3, -5)$ and $(-2, 7)$.
---
**Step 1: Calculate slope $m$:**
$$m = \frac{7 - (-5)}{-2 - 3} = \frac{12}{-5} = -\frac{12}{5}$$
---
**Step 2: Use point-slope form with point $(3, -5)$:**
$$y - (-5) = -\frac{12}{5}(x - 3)$$
Simplify:
$$y + 5 = -\frac{12}{5}x + \frac{36}{5}$$
$$y = -\frac{12}{5}x + \frac{36}{5} - 5 = -\frac{12}{5}x + \frac{36}{5} - \frac{25}{5} = -\frac{12}{5}x + \frac{11}{5}$$
Equation:
$$y = -\frac{12}{5}x + \frac{11}{5}$$
---
7. **Problem:** Find equation of line through $(6, -4)$ perpendicular to line $6 = 4y + 3x$.
---
**Step 1: Rewrite given line in slope-intercept form:**
$$6 = 4y + 3x \Rightarrow 4y = 6 - 3x \Rightarrow y = \frac{6 - 3x}{4} = -\frac{3}{4}x + \frac{3}{2}$$
Slope of given line:
$$m_1 = -\frac{3}{4}$$
---
**Step 2: Slope of perpendicular line:**
$$m_2 = -\frac{1}{m_1} = -\frac{1}{-3/4} = \frac{4}{3}$$
---
**Step 3: Use point-slope form with point $(6, -4)$:**
$$y - (-4) = \frac{4}{3}(x - 6)$$
Simplify:
$$y + 4 = \frac{4}{3}x - 8$$
$$y = \frac{4}{3}x - 8 - 4 = \frac{4}{3}x - 12$$