2.10.4 Matriks, SPM Praktis (Soalan Panjang)

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

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