Latih Kendiri 2.2d – Buku Teks Matematik Tingkatan 5 Bab 2 – Matriks


Soalan 1:
Antara matriks berikut, yang manakah matriks identiti? Jika bukan, berikan sebab anda.

Penyelesaian:
(a)
[0   1 ]
Bukan. Ini bukan matriks segi empat sama.

(b)

Bukan. Unsur di pepenjuru utama bukan 1.

(c)

Bukan.
Unsur di pepenjuru utama bukan 1.

(d)

Ya
, matriks identiti.

(e)

Ya, matriks identiti.

(f)

Bukan. Unsur di pepenjuru utama bukan 1.


Soalan 2:
$$ \text { Diberi matriks } C=\left[\begin{array}{cc} -1 & 3 \\ 2 & 5 \end{array}\right] \text { dan matriks } D=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text {. } $$
Tunjukkan matriks D ialah matriks identiti.

Penyelesaian:
$$ \begin{aligned} C D & =\left[\begin{array}{cc} -1 & 3 \\ 2 & 5 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{cc} -1(1)+3(0) & (-1) 0+3(1) \\ 2(1)+5(0) & 2(0)+5(1) \end{array}\right] \\ & =\left[\begin{array}{cc} -1 & 3 \\ 2 & 5 \end{array}\right] \end{aligned} $$
$$ \begin{aligned} D C & =\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 3 \\ 2 & 5 \end{array}\right] \\ & =\left[\begin{array}{cc} 1(-1)+0(2) & 1(3)+0(5) \\ 0(-1)+1(2) & 0(3)+1(5) \end{array}\right] \\ & =\left[\begin{array}{cc} -1 & 3 \\ 2 & 5 \end{array}\right] \end{aligned} $$
$$ C D=D C=C \text {. Maka, } D \text { ialah matriks identiti. } $$


Soalan 3:
$$ \text { Diberi matriks } S=\left[\begin{array}{ll} 7 & 2 \\ 6 & 3 \end{array}\right] \text { dan matriks } T=\left[\begin{array}{cc} 3 & 1 \\ -5 & 4 \end{array}\right] \text {. Hitung } $$
(a) SI + TI
(b) (IS)T
(c) 4ITI2
(d) (S I)I

Penyelesaian:
(a)
$$ \begin{aligned} & S I+T I \\ & =\left[\begin{array}{ll} 7 & 2 \\ 6 & 3 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\left[\begin{array}{cc} 3 & 1 \\ -5 & 4 \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{ll} 7 & 2 \\ 6 & 3 \end{array}\right]+\left[\begin{array}{cc} 3 & 1 \\ -5 & 4 \end{array}\right] \\ & =\left[\begin{array}{cc} 7+3 & 2+1 \\ 6+(-5) & 3+4 \end{array}\right] \\ & =\left[\begin{array}{cc} 10 & 3 \\ 1 & 7 \end{array}\right] \end{aligned} $$

(b)
$$ \begin{aligned} & (I S) T \\ & =\left(\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 7 & 2 \\ 6 & 3 \end{array}\right]\right)\left[\begin{array}{cc} 3 & 1 \\ -5 & 4 \end{array}\right] \\ & =\left[\begin{array}{ll} 7 & 2 \\ 6 & 3 \end{array}\right]\left[\begin{array}{cc} 3 & 1 \\ -5 & 4 \end{array}\right] \\ & =\left[\begin{array}{cc} 7(3)+2(-5) & 7(1)+2(4) \\ 6(3)+3(-5) & 6(1)+3(4) \end{array}\right] \\ & =\left[\begin{array}{cc} 11 & 15 \\ 3 & 18 \end{array}\right] \end{aligned} $$

(c)
$$ \begin{aligned} 4 I T-I^2 & =4 T-I \\ & =4\left[\begin{array}{cc} 3 & 1 \\ -5 & 4 \end{array}\right]-\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{cc} 12 & 4 \\ -20 & 16 \end{array}\right]-\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{cc} 11 & 4 \\ -20 & 15 \end{array}\right] \end{aligned} $$
(d)
$$ \begin{aligned} (S-I) I & =S-I \\ & =\left[\begin{array}{ll} 7 & 2 \\ 6 & 3 \end{array}\right]-\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ & =\left[\begin{array}{ll} 6 & 2 \\ 6 & 2 \end{array}\right] \end{aligned} $$

Leave a Comment