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Mathematics

Grade9

Easy

Question

Simplify the product
$negative 3 open square brackets table row cell negative 4 end cell 1 0 row 9 cell negative 5 end cell cell negative 7 end cell end table close square brackets$

$[\begin{matrix} 12 & 3 & 0 \\ - 27 & 15 & 21 \end{matrix}]$
$[\begin{matrix} 12 & - 3 & 0 \\ - 27 & - 15 & - 21 \end{matrix}]$
$[\begin{matrix} - 12 & - 3 & 0 \\ 27 & 15 & 21 \end{matrix}]$
$[\begin{matrix} 12 & - 3 & 0 \\ - 27 & 15 & 21 \end{matrix}]$

hint Hint:

The multiplication rule of matrices is applicable here. Multiply the digit outside the matrix with all the number inside the matrix.

The correct answer is: $open square brackets table row 12 cell negative 3 end cell 0 row cell negative 27 end cell 15 21 end table close square brackets$

$negative 3 open square brackets table row cell negative 4 end cell 1 0 row 9 cell negative 5 end cell cell negative 7 end cell end table close square brackets equals open square brackets table row cell open parentheses negative 3 close parentheses open parentheses negative 4 close parentheses end cell cell open parentheses negative 3 close parentheses open parentheses 1 close parentheses end cell cell open parentheses negative 3 close parentheses open parentheses 0 close parentheses end cell row cell open parentheses negative 3 close parentheses open parentheses 9 close parentheses end cell cell open parentheses negative 3 close parentheses open parentheses negative 5 close parentheses end cell cell open parentheses negative 3 close parentheses open parentheses negative 7 close parentheses end cell end table close square brackets equals open square brackets table row 12 cell negative 3 end cell 0 row cell negative 27 end cell 15 21 end table close square brackets$

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