We are given several math problems. We need to find the determinant and inverse of matrices, solve a quadratic equation by completing the square, find the mean of numbers, and calculate the area and volume of figures. Specifically, we need to: 6. (a) Find the determinant of matrix $A = \begin{pmatrix} 3 & 2 & 4 \\ 3 & 1 & 3 \\ 1 & 4 & 1 \end{pmatrix}$ and matrix $B = \begin{pmatrix} 4 & 2 & 3 \\ 1 & 5 & 2 \\ 2 & 1 & 4 \end{pmatrix}$. (b) Find the inverse of matrix A. 7. (a) (i) Use completing the square method to solve the equation $ax^2 + bx + c = 0$. (ii) Use the result in 7.(a)(i) above to find the roots of $4x^2 + 7x - 2 = 0$. (b) The mean of 6 numbers is 40. Three of the numbers sum up to be 90. What is the mean of the remaining 3 numbers? 8. (a) Calculate the area of the shaded portion in Fig. 2.1, where we have a triangle with base 8 cm, side 8 cm and the angle is 54°, and a semi-circle of radius 8 cm. (b) A tin of milk has a radius 6.5 cm and height 13 cm. Find, correct to 2 decimal places, the: (i) Total surface area of the tin. (ii) Volume of liquid in litres that will fill the tin (Take $\pi = \frac{22}{7}$).

1. Problem Description

We are given several math problems. We need to find the determinant and inverse of matrices, solve a quadratic equation by completing the square, find the mean of numbers, and calculate the area and volume of figures.

Specifically, we need to:

6. (a) Find the determinant of matrix $A = \begin{pmatrix} 3 & 2 & 4 \\ 3 & 1 & 3 \\ 1 & 4 & 1 \end{pmatrix}$ and matrix $B = \begin{pmatrix} 4 & 2 & 3 \\ 1 & 5 & 2 \\ 2 & 1 & 4 \end{pmatrix}$.

(b) Find the inverse of matrix A.

7. (a) (i) Use completing the square method to solve the equation $ax^2 + bx + c = 0$.

(ii) Use the result in 7.(a)(i) above to find the roots of $4x^2 + 7x - 2 = 0$.

(b) The mean of 6 numbers is

4

0. Three of the numbers sum up to be

9

0. What is the mean of the remaining 3 numbers?

8. (a) Calculate the area of the shaded portion in Fig. 2.1, where we have a triangle with base 8 cm, side 8 cm and the angle is 54°, and a semi-circle of radius 8 cm.

(b) A tin of milk has a radius 6.5 cm and height 13 cm. Find, correct to 2 decimal places, the:

(i) Total surface area of the tin.

(ii) Volume of liquid in litres that will fill the tin (Take $\pi = \frac{22}{7}$).

2. Solution Steps

6. (a) (i) Determinant of A:

$det(A) = 3(1*1 - 3*4) - 2(3*1 - 3*1) + 4(3*4 - 1*1) = 3(1-12) - 2(3-3) + 4(12-1) = 3(-11) - 2(0) + 4(11) = -33 - 0 + 44 = 11$

(ii) Determinant of B:

$det(B) = 4(5*4 - 2*1) - 2(1*4 - 2*2) + 3(1*1 - 5*2) = 4(20-2) - 2(4-4) + 3(1-10) = 4(18) - 2(0) + 3(-9) = 72 - 0 - 27 = 45$

(b) Inverse of A:

$A = \begin{pmatrix} 3 & 2 & 4 \\ 3 & 1 & 3 \\ 1 & 4 & 1 \end{pmatrix}$

$det(A) = 11$

Cofactor matrix: $C = \begin{pmatrix} -11 & 0 & 11 \\ 14 & -1 & -10 \\ 2 & 3 & -3 \end{pmatrix}$

Adjoint matrix: $adj(A) = C^T = \begin{pmatrix} -11 & 14 & 2 \\ 0 & -1 & 3 \\ 11 & -10 & -3 \end{pmatrix}$

$A^{-1} = \frac{1}{det(A)} adj(A) = \frac{1}{11} \begin{pmatrix} -11 & 14 & 2 \\ 0 & -1 & 3 \\ 11 & -10 & -3 \end{pmatrix} = \begin{pmatrix} -1 & \frac{14}{11} & \frac{2}{11} \\ 0 & -\frac{1}{11} & \frac{3}{11} \\ 1 & -\frac{10}{11} & -\frac{3}{11} \end{pmatrix}$

7. (a) (i) Completing the square:

$ax^2 + bx + c = 0$

$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$

$(x + \frac{b}{2a})^2 - (\frac{b}{2a})^2 + \frac{c}{a} = 0$

$(x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} - \frac{c}{a}$

$(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}$

$x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$

$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

(ii) $4x^2 + 7x - 2 = 0$

$a = 4$, $b = 7$, $c = -2$

$x = \frac{-7 \pm \sqrt{7^2 - 4(4)(-2)}}{2(4)} = \frac{-7 \pm \sqrt{49 + 32}}{8} = \frac{-7 \pm \sqrt{81}}{8} = \frac{-7 \pm 9}{8}$

$x_1 = \frac{-7 + 9}{8} = \frac{2}{8} = \frac{1}{4}$

$x_2 = \frac{-7 - 9}{8} = \frac{-16}{8} = -2$

(b) Mean of 6 numbers is

4

0. Sum of 6 numbers is $6 * 40 = 240$.

Three numbers sum to

9

0. Remaining three numbers sum to $240 - 90 = 150$.

Mean of the remaining 3 numbers is $\frac{150}{3} = 50$.

8. (a) Area of triangle = $\frac{1}{2}ab\sin(C) = \frac{1}{2} * 8 * 8 * \sin(54°) \approx \frac{1}{2} * 64 * 0.809 = 25.888$ cm$^2$

Area of semi-circle = $\frac{1}{2} \pi r^2 = \frac{1}{2} * \frac{22}{7} * 8^2 = \frac{1}{2} * \frac{22}{7} * 64 = \frac{704}{14} \approx 50.286$ cm$^2$

Total area = $25.888 + 50.286 \approx 76.174$ cm$^2$

Area of the shaded portion is approximately $76.17$ cm$^2$.

(b) (i) Total surface area of the tin:

$TSA = 2\pi r (r + h) = 2 * \frac{22}{7} * 6.5 * (6.5 + 13) = 2 * \frac{22}{7} * 6.5 * 19.5 = \frac{5643}{7} \approx 806.14$ cm$^2$

(ii) Volume of the tin:

$V = \pi r^2 h = \frac{22}{7} * (6.5)^2 * 13 = \frac{22}{7} * 42.25 * 13 = \frac{12024.5}{7} \approx 1717.79$ cm$^3$

$1$ litre $= 1000$ cm$^3$

Volume in litres $= \frac{1717.79}{1000} \approx 1.72$ litres.

3. Final Answer

6. (a) (i) $det(A) = 11$

(ii) $det(B) = 45$

(b) $A^{-1} = \begin{pmatrix} -1 & \frac{14}{11} & \frac{2}{11} \\ 0 & -\frac{1}{11} & \frac{3}{11} \\ 1 & -\frac{10}{11} & -\frac{3}{11} \end{pmatrix}$

7. (a) (i) $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

(ii) $x = \frac{1}{4}$ and $x = -2$

(b) The mean of the remaining 3 numbers is

5

0.

8. (a) Area of the shaded portion is approximately $76.17$ cm$^2$.

(b) (i) Total surface area of the tin is approximately $806.14$ cm$^2$.

(ii) Volume of liquid in litres that will fill the tin is approximately $1.72$ litres.