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Maths-

General

Easy

Question

If A = $open square brackets table row 1 2 row cell negative 4 end cell cell negative 1 end cell end table close square brackets$ , then A^–1 =

$\frac{1}{7}$ $[\begin{matrix} 1 & 2 \\ - 4 & - 1 \end{matrix}]$
$[\begin{matrix} - 1 & - 2 \\ 4 & 1 \end{matrix}]$
$\frac{1}{9}$ $[\begin{matrix} 1 & 2 \\ 4 & 1 \end{matrix}]$
$\frac{1}{7}$ $[\begin{matrix} - 1 & - 2 \\ 4 & 1 \end{matrix}]$

hint Hint:

$A equals space open square brackets table row 1 2 row cell negative 4 end cell cell negative 1 end cell end table close square brackets a d j A equals space open square brackets table row cell negative 1 end cell cell negative 2 end cell row 4 1 end table close square brackets vertical line A ∣ equals negative 1 minus left parenthesis negative 8 right parenthesis equals 7 A to the power of negative 1 end exponent equals space 1 over 7 space open square brackets table row cell negative 1 end cell cell negative 2 end cell row 4 1 end table close square brackets$

The correct answer is: $fraction numerator 1 over denominator 7 end fraction$ $open square brackets table row cell negative 1 end cell cell negative 2 end cell row 4 1 end table close square brackets$

we are given matrix A and asked to find A inverse

Therefore the correct option is choice 4

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