2.10.2 Matrices, SPM Paper (Long Questions)

Question 5:
$\text{(a) Given }\frac{1}{14}\left(\begin{array}{cc}2& s\\ -4& t\end{array}\right)\left(\begin{array}{cc}t& -1\\ 4& 2\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),\text{ find the value of }s\text{ and of }t\text{.}$
(b) Write the following simultaneous linear equations as matrix form:
3x – 2y = 5
9x + y = 1
Hence, using matrix method, calculate the value of x and y.

Solution:
$\begin{array}{l}\text{(a)}\\ \frac{1}{14}\left(\begin{array}{cc}2& s\\ -4& t\end{array}\right)\left(\begin{array}{cc}t& -1\\ 4& 2\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\\ \frac{1}{14}\left(\begin{array}{cc}2t+4s& -2+2s\\ -4t+4t& 4+2t\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\\ \frac{-2+2s}{14}=0\\ 2s=2\\ s=1\\ \\ \frac{4+2t}{14}=1\\ 4+2t=14\\ 2t=10\\ t=5\end{array}$

$\begin{array}{l}\text{(b)}\\ \left(\begin{array}{cc}3& -2\\ 9& 1\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}5\\ 1\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{21}\left(\begin{array}{cc}1& 2\\ -9& 3\end{array}\right)\left(\begin{array}{l}5\\ 1\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{21}\left(\begin{array}{l}\left(1\right)\left(5\right)+\left(2\right)\left(1\right)\\ \left(-9\right)\left(5\right)+\left(3\right)\left(1\right)\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{21}\left(\begin{array}{l}7\\ -42\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}\frac{1}{3}\\ -2\end{array}\right)\\ \therefore x=\frac{1}{3},y=-2\end{array}$

Question 6:
$\begin{array}{l}\text{It is given that matrix }P=\left(\begin{array}{cc}6& -3\\ -5& 2\end{array}\right)\text{ and matrix }Q=\frac{1}{m}\left(\begin{array}{cc}2& 3\\ 5& n\end{array}\right)\\ \text{such that }PQ=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\text{.}\end{array}$
(a) Find the value of m and of n.
(b) Write the following simultaneous linear equations as matrix form:
6x – 3y = –24
–5x + 2y = 18
Hence, using matrix method, calculate the value of x and y.

Solution:
$\begin{array}{l}\text{(a)}\\ m=6\left(2\right)-\left(-3\right)\left(-5\right)\\ =12-15\\ m=-3\\ n=6\end{array}$

$\begin{array}{l}\text{(b)}\\ \left(\begin{array}{cc}6& -3\\ -5& 2\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}-24\\ 18\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{12-15}\left(\begin{array}{cc}2& 3\\ 5& 6\end{array}\right)\left(\begin{array}{l}-24\\ 18\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{-3}\left(\begin{array}{l}\left(2\right)\left(-24\right)+\left(3\right)\left(18\right)\\ \left(5\right)\left(-24\right)+\left(6\right)\left(18\right)\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{-3}\left(\begin{array}{l}6\\ -12\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}-2\\ 4\end{array}\right)\\ \therefore x=-2,y=4\end{array}$

Question 7:
$\text{(a) Find the inverse matrix of }\left(\begin{array}{cc}3& 2\\ 5& 4\end{array}\right).$
(b) Ethan and Rahman went to the supermarket to buy cucumbers and carrots. Ethan bought 3 cucumbers and 2 carrots for RM9. Rahman bought 5 cucumbers and 4 carrots for RM16.
By using matrix method, find the price, in RM, of a cucumber and the price of a carrot. 
Solution:
$\begin{array}{l}\text{(a)}\\ \text{Inverse matrix of }\left(\begin{array}{cc}3& 2\\ 5& 4\end{array}\right)\\ =\frac{1}{12-10}\left(\begin{array}{cc}4& -2\\ -5& 3\end{array}\right)\\ =\frac{1}{2}\left(\begin{array}{cc}4& -2\\ -5& 3\end{array}\right)\\ =\left(\begin{array}{cc}2& -1\\ -\frac{5}{2}& \frac{3}{2}\end{array}\right)\end{array}$

$\begin{array}{l}\text{(b)}\\ 3x+2y=\mathrm{9\dots \dots \dots \dots \dots ..}\left(1\right)\\ 5x+4y=\mathrm{16\dots \dots \dots \dots \dots }\left(2\right)\\ \left(\begin{array}{cc}3& 2\\ 5& 4\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}9\\ 16\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{cc}2& -1\\ -\frac{5}{2}& \frac{3}{2}\end{array}\right)\left(\begin{array}{l}9\\ 16\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}\left(2\right)\left(9\right)+\left(-1\right)\left(16\right)\\ \left(-\frac{5}{2}\right)\left(9\right)+\left(\frac{3}{2}\right)\left(16\right)\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}18-16\\ -\frac{45}{2}+24\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}2\\ \frac{3}{2}\end{array}\right)\\ x=2,y=\frac{3}{2}\\ \therefore \text{Price of a cucumber}=\text{RM2}\\ \text{ Price of a carrot}=\text{RM1}\text{.50}\\ \end{array}$