Question 11 (5 marks):
Table 8 shows the information of books purchased by Maslinda.
 
Table 8

Maslinda purchased x history books and y Science books. The total number of books purchased is 5. The total price of the books purchased is RM17.

(a) Write two linear equations in terms of x and y to represent the above information.

(b) Hence, by using matrix method, calculate the value of x and of y.


Solution:
(a)
x + y = 5
4x + 3y = 17

(b)
$\begin{array}{l}\left(\begin{array}{l}\text{1 1}\\ \text{4 3}\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}5\\ 17\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{1\left(3\right)-4\left(1\right)}\left(\begin{array}{l}\text{ 3 }-1\\ -4\text{ 1}\end{array}\right)\left(\begin{array}{l}5\\ 17\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=-1\left(\begin{array}{l}3\left(5\right)+\left(-1\right)\left(17\right)\\ -4\left(5\right)\text{+}\left(1\right)\left(17\right)\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=-1\left(\begin{array}{l}15-17\\ -20+17\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=-1\left(\begin{array}{l}-2\\ -3\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}2\\ 3\end{array}\right)\\ x=2\text{ and }y=3\end{array}$

Question 12 (6 marks):
$\begin{array}{l}\text{Given }A=\left(\begin{array}{l}4-2\\ 3-1\end{array}\right),B=m\left(\begin{array}{l}-1n\\ -34\end{array}\right)\\ \text{and }I=\left(\begin{array}{l}\text{1 0}\\ \text{0 }1\end{array}\right).\end{array}$
(a) Find the value of m and of n if AB = I.
(b) Write the following simultaneous linear equation as matrix equation:
4x – 2y = 3
3x – y = 2
Hence, by using matrix method, calculate the value of x and of y.

Solution:
(a)
$\begin{array}{l}\text{If }AB=I,\text{ then }B=A^{-1}\\ A^{-1}=\frac{1}{4\left(-1\right)-\left(-2\right)\left(3\right)}\left(\begin{array}{l}-12\\ -3\text{ 4}\end{array}\right)\\ A^{-1}=\frac{1}{2}\left(\begin{array}{l}-12\\ -3\text{ 4}\end{array}\right)\\ \\ \text{By comparison:}\\ B=A^{-1}\\ m\left(\begin{array}{l}-1n\\ -34\end{array}\right)=\frac{1}{2}\left(\begin{array}{l}-12\\ -3\text{ 4}\end{array}\right)\\ m=\frac{1}{2};n=2\end{array}$


(b)
$\begin{array}{l}4x-2y=3\\ 3x-y=2\\ \left(\begin{array}{l}4-2\\ 3-1\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}3\\ 2\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{2}\left(\begin{array}{l}-12\\ -3\text{ 4}\end{array}\right)\left(\begin{array}{l}3\\ 2\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{2}\left(\begin{array}{l}\left(-1\right)\left(3\right)\text{+}\left(2\right)\left(2\right)\\ \left(-3\right)\left(3\right)\text{+}\left(4\right)\left(2\right)\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{2}\left(\begin{array}{l}1\\ -1\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}\frac{1}{2}\\ -\frac{1}{2}\end{array}\right)\\ x=\frac{1}{2}\text{ and }y=-\frac{1}{2}\end{array}$

Question 9:
$\text{(a) Given }\frac{1}{s}\left(\begin{array}{cc}-4& 2\\ -5& 3\end{array}\right)\left(\begin{array}{cc}t& -2\\ 5& -4\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),\text{ find the value of }s\text{ and of }t\text{.}$

(b) Using matrices, calculate the value of x and of y that satisfy the following matrix equation:
$\left(\begin{array}{cc}-4& 2\\ -5& 3\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}1\\ 2\end{array}\right)$


Solution:
$\begin{array}{l}\text{(a)}\\ \frac{1}{s}\left(\begin{array}{cc}t& -2\\ 5& -4\end{array}\right)={\left(\begin{array}{cc}-4& 2\\ -5& 3\end{array}\right)}^{-1}\\ =\frac{1}{\left(-4\right)\left(3\right)-\left(2\right)\left(-5\right)}\left(\begin{array}{cc}3& -2\\ 5& -4\end{array}\right)\\ =\frac{1}{-2}\left(\begin{array}{cc}3& -2\\ 5& -4\end{array}\right)\\ \therefore s=-2,t=3\end{array}$

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

Question 10 (5 marks):
During the sport day, students used coupons to buy food and drink. Ali and Larry spent RM31 and RM27 respectively. Ali bought 2 food coupons and 5 drinks coupons while Larry bought 3 coupons and 1 drinks coupon.
Using the matrix method, calculate the price, in RM, of a food coupon and of a drinks coupon.

Solution:
Ali spent RM31. He bought 2 food coupons and 5 drinks coupons.
Larry spent RM27. He bought 3 food coupons and 1 drinks coupons.
x = price of one food coupon
y = price of one drinks coupon

$\begin{array}{l}\left(\begin{array}{l}2\text{ 5}\\ 31\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}31\\ 27\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{2\left(1\right)-5\left(3\right)}\left(\begin{array}{l}1-5\\ -3\text{ 2}\end{array}\right)\left(\begin{array}{l}31\\ 27\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{2-15}\left(\begin{array}{l}1\left(31\right)+\left(-5\right)\left(27\right)\\ -3\left(31\right)\text{+}2\left(27\right)\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{-13}\left(\begin{array}{l}-104\\ -39\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}8\\ 3\end{array}\right)\\ x=8\text{ and }y=3\\ \\ \text{Thus, the price for a food coupon is RM8}\\ \text{and the price for drink coupon is RM3}\text{.}\end{array}$

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

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

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

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

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 3:
(a) Find Q.

Solution:
(a) 
$\begin{array}{l}Q={\left(\begin{array}{cc}3& 2\\ 6& 5\end{array}\right)}^{-1}\\ Q=\frac{1}{3\left(5\right)-2\left(6\right)}\left(\begin{array}{cc}5& -2\\ -6& 3\end{array}\right)\\ Q=\frac{1}{3}\left(\begin{array}{cc}5& -2\\ -6& 3\end{array}\right)\\ Q=\left(\begin{array}{cc}\frac{5}{3}& -\frac{2}{3}\\ -2& 1\end{array}\right)\end{array}$

(b)
$\begin{array}{l}\left(\begin{array}{cc}3& 2\\ 6& 5\end{array}\right)\left(\begin{array}{l}u\\ v\end{array}\right)=\left(\begin{array}{l}5\\ 2\end{array}\right)\\ \left(\begin{array}{l}u\\ v\end{array}\right)=\frac{1}{3}\left(\begin{array}{cc}5& -2\\ -6& 3\end{array}\right)\left(\begin{array}{l}5\\ 2\end{array}\right)\\ \left(\begin{array}{l}u\\ v\end{array}\right)=\frac{1}{3}\left(\begin{array}{l}\left(5\right)\left(5\right)+\left(-2\right)\left(2\right)\\ \left(-6\right)\left(5\right)+\left(3\right)\left(2\right)\end{array}\right)\\ \left(\begin{array}{l}u\\ v\end{array}\right)=\frac{1}{3}\left(\begin{array}{l}21\\ -24\end{array}\right)\\ \left(\begin{array}{l}u\\ v\end{array}\right)=\left(\begin{array}{l}7\\ -8\end{array}\right)\\ \therefore u=7,v=-8\end{array}$

Question 4:
Hence, using matrix method, calculate the value of x and y.

Solution:
(a)
$\begin{array}{l}Q=\frac{1}{3\left(-4\right)-\left(5\right)\left(-2\right)}\left(\begin{array}{cc}-4& 2\\ -5& 3\end{array}\right)\\ =-\frac{1}{2}\left(\begin{array}{cc}-4& 2\\ -5& 3\end{array}\right)\\ =\left(\begin{array}{cc}2& -1\\ \frac{5}{2}& -\frac{3}{2}\end{array}\right)\end{array}$

(b)
$\begin{array}{l}\left(\begin{array}{cc}3& -2\\ 5& -4\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}7\\ 9\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=-\frac{1}{2}\left(\begin{array}{cc}-4& 2\\ -5& 3\end{array}\right)\left(\begin{array}{l}7\\ 9\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=-\frac{1}{2}\left(\begin{array}{l}\left(-4\right)\left(7\right)+\left(2\right)\left(9\right)\\ \left(-5\right)\left(7\right)+\left(3\right)\left(9\right)\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=-\frac{1}{2}\left(\begin{array}{l}-10\\ -8\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}5\\ 4\end{array}\right)\\ \therefore x=5,y=4\end{array}$

$\begin{array}{l}\left(\begin{array}{cc}3& 4\\ -2& 3\end{array}\right)\left(\begin{array}{l}5\\ -2\end{array}\right)=\left(\begin{array}{l}\left(3\right)\left(5\right)+\left(4\right)\left(-2\right)\\ \left(-2\right)\left(5\right)+\left(3\right)\left(-2\right)\end{array}\right)\\ =\left(\begin{array}{l}15-8\\ -10-6\end{array}\right)\\ =\left(\begin{array}{l}7\\ -16\end{array}\right)\end{array}$

Order of the product of the two matrices
$=\left(\boxed{3}\times 2\right)\left(2\times \boxed{1}\right)=\left(\boxed{3\times 1}\right)$

$\begin{array}{l}\left(\begin{array}{cc}2& 4\\ \begin{array}{l}-3\\ 4\end{array}& \begin{array}{l}0\\ 1\end{array}\end{array}\right)\left(\begin{array}{l}1\\ -3\end{array}\right)=\left(\begin{array}{l}2\left(1\right)+4\left(-3\right)\\ \left(-3\right)\left(1\right)+0\left(-3\right)\\ \left(4\right)\left(1\right)+1\left(-3\right)\end{array}\right)\\ =\left(\begin{array}{l}2-12\\ -3+0\\ 4-3\end{array}\right)\\ =\left(\begin{array}{l}-10\\ -3\\ 1\end{array}\right)\end{array}$

$\left(1-12\right)\left(\begin{array}{cc}5& 1\\ \begin{array}{l}-3\\ 2\end{array}& \begin{array}{l}0\\ 4\end{array}\end{array}\right)=$ = (12   9)

Find the values of m and n in each of the following matrix equations.
Solution:
$\begin{array}{l}\text{(a)}\\ \left(\begin{array}{l}3\\ m\end{array}\right)(1n)=\left(\begin{array}{cc}3& 12\\ -2& -8\end{array}\right)\\ \left(\begin{array}{cc}3& 3n\\ m& mn\end{array}\right)=\left(\begin{array}{cc}3& 12\\ -2& -8\end{array}\right)\\ m=-2,\\ 3n=12\\ n=4\end{array}$

$\begin{array}{l}\text{(b)}\\ \left(\begin{array}{cc}m& 2\\ -3& 1\end{array}\right)\left(\begin{array}{l}2\\ n\end{array}\right)=\left(\begin{array}{l}12\\ 4+2n\end{array}\right)\\ \left(\begin{array}{l}2m+2n\\ -6+n\end{array}\right)=\left(\begin{array}{l}12\\ 4+2n\end{array}\right)\\ -6+n=4+2n\\ n=-10\\ 2m+2n=12\\ 2m+2(-10)=12\\ 2m-20=12\\ 2m=32\\ m=16\end{array}$

$\begin{array}{l}\text{(c)}\\ \left(\begin{array}{cc}m& -3\\ -1& 1\end{array}\right)\left(\begin{array}{cc}-1& 2\\ 4& n\end{array}\right)=\left(\begin{array}{cc}-14& -11\\ 5& 3\end{array}\right)\\ \left(\begin{array}{cc}-m+(-12)& 2m+(-3n)\\ 1+4& -2+n\end{array}\right)=\left(\begin{array}{cc}-14& -11\\ 5& 3\end{array}\right)\\ -m-12=-14\\ -m=-2\\ m=2\\ -2+n=3\\ n=5\end{array}$