1. **Problem Statement:** Solve the system of linear equations using (i) Cramer's rule and (ii) Gauss elimination, then sketch given functions and express the volume of an open box. --- ## Part 4: System of linear equations Given: $$4x - 5y + 7z = -14$$ $$9x + 2y - 3z = 47$$ $$x - y - 5z = 11$$ ### (i) Cramer's rule 1. Write the coefficient matrix $$A$$ and constants vector $$B$$: $$A = \begin{bmatrix}4 & -5 & 7\\9 & 2 & -3\\1 & -1 & -5\end{bmatrix} , B = \begin{bmatrix}-14 \\ 47 \\ 11\end{bmatrix}$$ 2. Calculate the determinant $$D = \det(A)$$: $$D = 4 \begin{vmatrix}2 & -3 \\ -1 & -5 \end{vmatrix} - (-5) \begin{vmatrix}9 & -3 \\ 1 & -5 \end{vmatrix} + 7 \begin{vmatrix}9 & 2 \\ 1 & -1 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}2 & -3 \\ -1 & -5 \end{vmatrix} = (2)(-5) - (-3)(-1) = -10 - 3 = -13$$ $$\begin{vmatrix}9 & -3 \\ 1 & -5 \end{vmatrix} = (9)(-5) - (-3)(1) = -45 + 3 = -42$$ $$\begin{vmatrix}9 & 2 \\ 1 & -1 \end{vmatrix} = 9(-1) - 2(1) = -9 - 2 = -11$$ Then, $$D = 4(-13) + 5(-42) + 7(-11) = -52 - 210 - 77 = -339$$ 3. Calculate determinants $$D_x, D_y, D_z$$ by replacing corresponding columns in $$A$$ with $$B$$: $$D_x = \det \begin{bmatrix}-14 & -5 & 7 \\ 47 & 2 & -3 \\ 11 & -1 & -5\end{bmatrix}$$ $$D_y = \det \begin{bmatrix}4 & -14 & 7 \\ 9 & 47 & -3 \\ 1 & 11 & -5\end{bmatrix}$$ $$D_z = \det \begin{bmatrix}4 & -5 & -14 \\ 9 & 2 & 47 \\ 1 & -1 & 11\end{bmatrix}$$ Calculate $$D_x$$: $$= -14 \begin{vmatrix}2 & -3 \\ -1 & -5\end{vmatrix} - (-5) \begin{vmatrix}47 & -3 \\ 11 & -5\end{vmatrix} + 7 \begin{vmatrix}47 & 2 \\ 11 & -1\end{vmatrix}$$ $$= -14(-13) + 5(-235 + 33) + 7(-47 - 22)$$ $$= 182 + 5(-202) + 7(-69) = 182 - 1010 - 483 = -1311$$ Calculate $$D_y$$: $$=4 \begin{vmatrix}47 & -3 \\ 11 & -5 \end{vmatrix} - (-14) \begin{vmatrix}9 & -3 \\ 1 & -5 \end{vmatrix} + 7 \begin{vmatrix}9 & 47 \\ 1 & 11 \end{vmatrix}$$ $$= 4(-235 + 33) + 14(-45 + 3) + 7(99 - 47)$$ $$=4(-202) + 14(-42) + 7(52) = -808 - 588 + 364 = -1032$$ Calculate $$D_z$$: $$=4 \begin{vmatrix}2 & -14 \\ -1 & 11 \end{vmatrix} - (-5) \begin{vmatrix}9 & 47 \\ 1 & 11 \end{vmatrix} + (-14) \begin{vmatrix}9 & 2 \\ 1 & -1 \end{vmatrix}$$ $$=4(22 + 14) + 5(99 - 47) - 14(-9 - 2)$$ $$=4(36) + 5(52) + 14(11) = 144 + 260 + 154 = 558$$ 4. Solve for variables: $$x = \frac{D_x}{D} = \frac{-1311}{-339} = \frac{1311}{339} = \frac{1311 ÷ 3}{339 ÷ 3} = \frac{437}{113} \approx 3.87$$ $$y = \frac{D_y}{D} = \frac{-1032}{-339} = \frac{1032}{339} = \frac{1032 ÷ 3}{339 ÷ 3} = \frac{344}{113} \approx 3.04$$ $$z = \frac{D_z}{D} = \frac{558}{-339} = -\frac{558}{339} = -\frac{186}{113} \approx -1.65$$ ### (ii) Gauss elimination method 1. Write augmented matrix $$\left[\begin{array}{ccc|c}4 & -5 & 7 & -14 \\ 9 & 2 & -3 & 47 \\ 1 & -1 & -5 & 11 \end{array}\right]$$ 2. Make leading coefficient of row 1 equal 1 dividing row 1 by 4 (optional, or use elimination directly). 3. Eliminate $$x$$ terms from rows 2 and 3: row 2 - 9 * row 1 $$R2 = R2 - \frac{9}{4} R1:$$ $$9 - 9 = 0, 2 - (9/4)(-5) = 2 + 11.25 = 13.25, -3 - (9/4)(7) = -3 - 15.75 = -18.75, 47 - (9/4)(-14) = 47 + 31.5 = 78.5$$ row 3 - 1 * row 1 $$R3 = R3 - R1:$$ $$1 - 1 = 0, -1 - (-5) = 4, -5 - 7 = -12, 11 - (-14) = 25$$ New matrix: $$\left[\begin{array}{ccc|c}4 & -5 & 7 & -14 \\ 0 & 13.25 & -18.75 & 78.5 \\ 0 & 4 & -12 & 25 \end{array}\right]$$ 4. Make leading coefficient of row 2 in $$y$$ equal to 1 dividing row 2 by 13.25: $$R2 = R2 / 13.25$$ Approximate: $$\left[0, 1, -1.415, 5.925 \right]$$ 5. Eliminate $$y$$ from row 3: $$R3 = R3 - 4 * R2$$ $$0, 4 - 4 * 1 = 0, -12 - 4*(-1.415) = -12 + 5.66 = -6.34, 25 - 4*5.925 = 25 - 23.7 = 1.3$$ New matrix: $$\left[\begin{array}{ccc|c}4 & -5 & 7 & -14 \\ 0 & 1 & -1.415 & 5.925 \\ 0 & 0 & -6.34 & 1.3 \end{array}\right]$$ 6. Solve for $$z$$: $$-6.34 z = 1.3 \implies z = -\frac{1.3}{6.34} \approx -0.205$$ 7. Back substitute into row 2: $$y - 1.415 * (-0.205) = 5.925 \implies y + 0.29 = 5.925 \implies y = 5.635$$ 8. Back substitute into row 1: $$4x - 5y + 7z = -14$$ $$4x - 5(5.635) + 7(-0.205) = -14$$ $$4x - 28.175 - 1.435 = -14$$ $$4x - 29.61 = -14$$ $$4x = 15.61 \implies x = 3.90$$ --- ## Part 5(a): Sketch graphs (i) $$f(x) = \frac{x}{|x|}$$ - Defined for all $$x \ne 0$$ - $$f(x) = 1$$ when $$x > 0$$ - $$f(x) = -1$$ when $$x < 0$$ - Step function with a jump at $$x=0$$ (ii) $$f(x) = \sqrt{4 - x^2}$$ - Domain: $$-2 \leq x \leq 2$$ - Graph is the upper half of a circle centered at origin with radius 2 (iii) Piecewise function $$f(x) = \begin{cases}x^2, & x > 1 \\ 2, & x \leq 1 \end{cases}$$ - For $$x>1$$, parabola - For $$x\leq 1$$, constant value 2 - Jump discontinuity at $$x=1$$ because $$1^2=1\ne 2$$ --- ## Part 5(b): Open box volume Given a sheet of size 8cm by 15cm, cut squares of side $$x$$ from corners and fold up edges. 1. Dimensions after cutting: - Length: $$15 - 2x$$ - Width: $$8 - 2x$$ - Height: $$x$$ 2. Volume function: $$V(x) = x (15 - 2x)(8 - 2x)$$ Expand: $$V = x (120 - 30x - 16x + 4x^2) = x(120 - 46x + 4x^2) = 120x - 46x^2 + 4x^3$$ 3. Domain: $$x > 0$$ since positive cut, and lengths must remain positive: $$15 - 2x > 0 \Rightarrow x < 7.5$$ $$8 - 2x > 0 \Rightarrow x < 4$$ Therefore, $$0 < x < 4$$ 4. Range: Volume is $$0$$ at both ends $$x=0$$ and $$x=4$$, and is positive in between; the maximum volume is found by calculus but range is $$[0, V_{max}]$$ with $$V_{max}$$ at some $$x \in (0,4)$$. --- Final answers: - (4)(i) $$x=\frac{437}{113} \approx 3.87, y=\frac{344}{113} \approx 3.04, z=-\frac{186}{113} \approx -1.65$$ - (4)(ii) $$x \approx 3.90, y \approx 5.64, z \approx -0.205$$ - (5)(a) Graphs described as above - (5)(b) $$V(x) = 120x - 46x^2 + 4x^3$$, domain $$0 < x < 4$$, range $$[0,V_{max}]$$