2.9.2 Matriks, SPM Practis (Kertas 1)

2.9.2 SPM Practis (Kertas 1)

Soalan 7:

Diberi bahawa

$(3 x) (\begin{array}{l} x \\ - 1 \end{array}) = (18),$ cari nilai x.

Penyelesaian:

$(3 x) (\begin{array}{l} x \\ - 1 \end{array}) = (18)$

[3 × x + x (–1)] = (18)

3x – x = 18

2x = 18

x = 9

Soalan 8:

$(\begin{matrix} 3 & 4 \\ - 2 & 3 \end{matrix}) (\begin{array}{l} 5 \\ - 2 \end{array}) =$

Penyelesaian:

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

Soalan 9:

$(\begin{matrix} 2 & 4 \\ \begin{array}{l} - 3 \\ 4 \end{array} & \begin{array}{l} 0 \\ 1 \end{array} \end{matrix}) (\begin{array}{l} 1 \\ - 3 \end{array}) =$

Penyelesaian:

Peringkat hasil darab dua matriks

$= (3 \times 2) (2 \times 1) = (3 \times 1)$

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

Soalan 10:

$(1 - 1 2) (\begin{matrix} 5 & 1 \\ \begin{array}{l} - 3 \\ 2 \end{array} & \begin{array}{l} 0 \\ 4 \end{array} \end{matrix}) =$

Penyelesaian:

Peringkat hasil darab dua matriks

$= (1 \times 3) (3 \times 2) = (1 \times 2)$

$(1 - 1 2) (\begin{matrix} 5 & 1 \\ \begin{array}{l} - 3 \\ 2 \end{array} & \begin{array}{l} 0 \\ 4 \end{array} \end{matrix}) =$

= (1×5 + (–1)(–3) + (2)(2) 1×1 + (–1)(0) + (2)(4))

= (5 + 3 + 4 1 + 0 + 8)

= (12 9)

2.9.2 Matriks, SPM Practis (Kertas 1)

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