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(a) Given that $A=$
$$\left(\begin{array}{cc} 3 & x \\ x+1 & 2 \end{array}\right)$$
and $B=$
$$\left(\begin{array}{ll} 1 & 2 \\ 3 & 0 \end{array}\right)$$
find values of $x$ for which $A B$ is a singular matrix.
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\begin{array}{l}
\text { (a) }\left(\begin{array}{cc}
3 & x \\
x+1 & 2
\end{array}\right)\left(\begin{array}{ll}
1 & 2 \\
3 & 0
\end{array}\right)=\left(\begin{array}{cc}
3+3 x & 6 \\
x+7 & 2+2
\end{array}\right)\\
\left(\begin{array}{cc}
3+3 x & 6 \\
x+7 & 2+2
\end{array}\right)=0\\
(3+3 \mathrm{~s})(2 \mathrm{z}+2)-6(x+7)=0\\
6 s+6 x^{2}+6 s-6 s-36=0\\
6 x^{2}+6 s-36=0\\
x^{2}+x-6=0\\
(x+3)(x-2)=0\\
x=2 \text { or }-3
\end{array}

$3 x+5 y=165$
(b) (i) $2 z+4 y=120$
(ii) $\left(\begin{array}{ll}3 & 5 \\ 2 & 4\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{l}165 \\ 120\end{array}\right)$
$$\begin{array}{l} \quad \operatorname{Let} A=\left(\begin{array}{ll} 3 & 5 \\ 2 & 4 \end{array}\right) \\ A^{-1} & \begin{array}{l} 1 \\ 2 \end{array}\left(\begin{array}{cc} 4 & -5 \\ -2 & 3 \end{array}\right) \\ & \frac{1}{2}\left(\begin{array}{cc} 4 & -5 \\ -2 & 3 \end{array}\right)\left(\begin{array}{ll} 3 & 5 \\ 2 & 4 \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)=\frac{1}{2}\left(\begin{array}{cc} 4 & -5 \\ -2 & 3 \end{array}\right)\left(\begin{array}{l} 165 \\ 120 \end{array}\right) \\ \left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{l} 30 \\ 15 \end{array}\right) \\ \left(\begin{array}{l} x \\ y \end{array}\right)=\left(\begin{array}{l} 30 \\ 15 \end{array}\right) \end{array}$$
Cost of an exercise book = Ksh. 30 Cost of a pen Ksh. 15

(iii) $2 \times 36 \times 30+36 \times 15$
$=$ Ksh 2700
by Diamond (74,866 points)

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