4.10 SPM Practice (Long Questions)

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:
$\begin{array}{l}\text{(a)}\\ \text{Inverse matrix of }\left(\begin{array}{cc}3& 2\\ 5& 4\end{array}\right)\\ =\frac{1}{12-10}\left(\begin{array}{cc}4& -2\\ -5& 3\end{array}\right)\\ =\frac{1}{2}\left(\begin{array}{cc}4& -2\\ -5& 3\end{array}\right)\\ =\left(\begin{array}{cc}2& -1\\ -\frac{5}{2}& \frac{3}{2}\end{array}\right)\end{array}$

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

Question 8:
The inverse matrix of  $\left(\begin{array}{cc}4& -1\\ 2& 5\end{array}\right)\text{ is }t\left(\begin{array}{cc}5& 1\\ -2& n\end{array}\right).$
(a) Find the value of n and of t.
(b) Write the following simultaneous linear equations as matrix equation:
4x – y = 7
2x + 5y = –2
Hence, using matrix method, calculate the value of x and of y.


Solution:
$\begin{array}{l}\text{(a)}\\ t\left(\begin{array}{cc}5& 1\\ -2& n\end{array}\right)={\left(\begin{array}{cc}4& -1\\ 2& 5\end{array}\right)}^{-1}\\ =\frac{1}{\left(4\right)\left(5\right)-\left(-1\right)\left(2\right)}\left(\begin{array}{cc}5& 1\\ -2& 4\end{array}\right)\\ =\frac{1}{22}\left(\begin{array}{cc}5& 1\\ -2& 4\end{array}\right)\\ \therefore t=\frac{1}{22},n=4\end{array}$

$\begin{array}{l}\text{(b)}\\ \left(\begin{array}{cc}4& -1\\ 2& 5\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}7\\ -2\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{22}\left(\begin{array}{cc}5& 1\\ -2& 4\end{array}\right)\left(\begin{array}{l}7\\ -2\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{22}\left(\begin{array}{l}\left(5\right)\left(7\right)+1\left(-2\right)\\ \left(-2\right)\left(7\right)+\left(4\right)\left(-2\right)\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{22}\left(\begin{array}{l}35-2\\ -14-8\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{22}\left(\begin{array}{l}\text{ 33}\\ -22\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}\frac{3}{2}\\ -1\end{array}\right)\\ \therefore x=\frac{3}{2},y=-1\end{array}$