1. (a) Given matrices: $$A = \begin{pmatrix}2 & 4 \\ 1 & 3\end{pmatrix},\quad B = \begin{pmatrix}3 & 2 \\ 5 & 1\end{pmatrix}$$ To find values of $x$ such that $AB$ is singular means $\det(AB) = 0$. 1. Compute $AB$: $$AB = \begin{pmatrix}2 & 4 \\ 1 & 3\end{pmatrix} \begin{pmatrix}3 & 2 \\ 5 & 1\end{pmatrix} = \begin{pmatrix}2 \times 3 + 4 \times 5 & 2 \times 2 + 4 \times 1 \\ 1 \times 3 + 3 \times 5 & 1 \times 2 + 3 \times 1\end{pmatrix} = \begin{pmatrix}26 & 8 \\ 18 & 5 \end{pmatrix}$$ 2. Calculate $\det(AB)$: $$\det(AB) = (26)(5) - (8)(18) = 130 - 144 = -14 \neq 0$$ Since $AB$ does not depend on $x$ and $\det(AB) \neq 0$, $AB$ is nonsingular for given matrices with no $x$ parameter to solve. Possibly $x$ missing or misread from problem. (b) Given $$A=\begin{pmatrix}30 & -6 & -19 \\ -16 & 4 & 10 \\ -10 & 2 & 7\end{pmatrix}, \quad B=\begin{pmatrix}2 & 1 & 4 \\ 3 & 5 & 1 \\ 2 & 0 & 6\end{pmatrix}$$ (i) Find $AB$: Calculate each element: 1st row: $$AB_{11} = 30 \times 2 + (-6) \times 3 + (-19) \times 2 = 60 - 18 - 38 = 4$$ $$AB_{12} = 30 \times 1 + (-6) \times 5 + (-19) \times 0 = 30 - 30 + 0 = 0$$ $$AB_{13} = 30 \times 4 + (-6) \times 1 + (-19) \times 6 = 120 - 6 - 114 = 0$$ 2nd row: $$AB_{21} = (-16) \times 2 + 4 \times 3 + 10 \times 2 = -32 + 12 + 20 = 0$$ $$AB_{22} = (-16) \times 1 + 4 \times 5 + 10 \times 0 = -16 + 20 + 0 = 4$$ $$AB_{23} = (-16) \times 4 + 4 \times 1 + 10 \times 6 = -64 + 4 + 60 = 0$$ 3rd row: $$AB_{31} = (-10) \times 2 + 2 \times 3 + 7 \times 2 = -20 + 6 + 14 = 0$$ $$AB_{32} = (-10) \times 1 + 2 \times 5 + 7 \times 0 = -10 + 10 + 0 = 0$$ $$AB_{33} = (-10) \times 4 + 2 \times 1 + 7 \times 6 = -40 + 2 + 42 = 4$$ Thus $$AB = \begin{pmatrix}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{pmatrix} = 4I$$ (ii) Find $A^{-1}$ and $B^{-1}$: Notice from (i) that $AB = 4I$ so multiply both sides by $B^{-1}$: $$A = 4 B^{-1} \\ \Rightarrow B^{-1} = \frac{1}{4} A$$ Similarly multiply by $A^{-1}$: $$B = 4 A^{-1} \\ \Rightarrow A^{-1} = \frac{1}{4} B$$ Thus $$A^{-1} = \frac{1}{4} \begin{pmatrix}2 & 1 & 4 \\ 3 & 5 & 1 \\ 2 & 0 & 6\end{pmatrix} = \begin{pmatrix}0.5 & 0.25 & 1 \\ 0.75 & 1.25 & 0.25 \\ 0.5 & 0 & 1.5\end{pmatrix}$$ $$B^{-1} = \frac{1}{4} \begin{pmatrix}30 & -6 & -19 \\ -16 & 4 & 10 \\ -10 & 2 & 7\end{pmatrix} = \begin{pmatrix}7.5 & -1.5 & -4.75 \\ -4 & 1 & 2.5 \\ -2.5 & 0.5 & 1.75\end{pmatrix}$$ (iii) Solve system: $$2x + y + z = 8$$ $$3x + 5y + z = 0$$ $$2x + 6z = 12$$ Express in matrix form: $AX = D$, where $$A = \begin{pmatrix}2 & 1 & 1 \\ 3 & 5 & 1 \\ 2 & 0 & 6\end{pmatrix},\quad X = \begin{pmatrix}x \\ y \\ z\end{pmatrix},\quad D = \begin{pmatrix}8 \\ 0 \\ 12\end{pmatrix}$$ Using matrix inverse, $$X = A^{-1} D$$ Calculate $A^{-1}$ (not same as previous $A^{-1}$) by standard method or using given matrices. Alternatively, solve manually: From third equation: $$2x + 6z = 12 \Rightarrow x + 3z = 6 \Rightarrow x = 6 - 3z$$ Substitute $x$ into first equation: $$2(6 - 3z) + y + z = 8 \Rightarrow 12 - 6z + y + z = 8 \Rightarrow y - 5z = -4 \Rightarrow y = -4 + 5z$$ Substitute $x$ and $y$ into second equation: $$3(6 - 3z) + 5(-4 + 5z) + z = 0 \\ 18 - 9z - 20 + 25z + z = 0 \\ -2 + 17z = 0 \\ 17z = 2 \Rightarrow z = \frac{2}{17}$$ Then $$x = 6 - 3 \times \frac{2}{17} = 6 - \frac{6}{17} = \frac{102}{17} - \frac{6}{17} = \frac{96}{17}$$ $$y = -4 + 5 \times \frac{2}{17} = -4 + \frac{10}{17} = -\frac{68}{17} + \frac{10}{17} = -\frac{58}{17}$$ Final solutions: $$x = \frac{96}{17},\quad y = -\frac{58}{17},\quad z = \frac{2}{17}$$