2.10.2 Matriks, SPM Praktis (Kertas 2 – Soalan 6 – 10)

Soalan 6:
$\begin{array}{l}\text{Diberi bahawa matriks }P=\left(\begin{array}{cc}6& -3\\ -5& 2\end{array}\right)\text{ dan matriks }Q=\frac{1}{m}\left(\begin{array}{cc}2& 3\\ 5& n\end{array}\right)\\ \text{dengan keadaan }PQ=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\text{.}\end{array}$
(a) Cari nilai m dan nilai n.
(b) Tulis persamaan linear serentak berikut dalam bentuk matriks:
6x – 3y = –24
–5x + 2y = 18
Seterusnya, menggunakan kaedah matriks, hitung nilai x dan nilai y.

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

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

Soalan 7:
$\text{(a) Cari matriks songsang bagi }\left(\begin{array}{cc}3& 2\\ 5& 4\end{array}\right).$
(b) Ethan dan Rahman pergi ke pasar raya untuk membeli kiwi dan jambu. Ethan membeli 3 biji kiwi dan 2 biji jambu dengan harga RM9. Rahman membeli 5 biji kiwi dan 4 biji jambu dengan harga RM16.
Dengan menggunakan kaedah matriks, cari harga, dalam RM, bagi sebiji kiwi dan sebiji jambu. 

Penyelesaian:
$\begin{array}{l}\text{(a)}\\ \text{Matriks songsang bagi }\left(\begin{array}{cc}3& 2\\ 5& 4\end{array}\right)\\ =\frac{1}{12-10}\left(\begin{array}{cc}4& -2\\ -5& 3\end{array}\right)\\ =\frac{1}{2}\left(\begin{array}{cc}4& -2\\ -5& 3\end{array}\right)\\ =\left(\begin{array}{cc}2& -1\\ -\frac{5}{2}& \frac{3}{2}\end{array}\right)\end{array}$

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

Soalan 8:
Matriks songsang bagi $\left(\begin{array}{cc}4& -1\\ 2& 5\end{array}\right)\text{ ialah }t\left(\begin{array}{cc}5& 1\\ -2& n\end{array}\right).$
(a) Cari nilai n dan nilai t.
(b) Tulis persamaan linear serentak berikut dalam persamaan matriks:
4x – y = 7
2x + 5y = –2
Seterusnya, menggunakan kaedah matriks, hitung nilai x dan nilai y.


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

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

Soalan 9:
$\text{(a) Diberi }\frac{1}{s}\left(\begin{array}{cc}-4& 2\\ -5& 3\end{array}\right)\left(\begin{array}{cc}t& -2\\ 5& -4\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),\text{ cari nilai }s\text{ dan nilai }t\text{.}$

(b) Menggunakan kaedah matriks, hitung nilai x dan nilai y yang memuaskan persamaan matriks berikut:
$\left(\begin{array}{cc}-4& 2\\ -5& 3\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}1\\ 2\end{array}\right)$


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

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

$\begin{array}{l}\end{array}$