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Maths-

General

Easy

Question

If A = $open square brackets table row 1 2 row 3 cell negative 5 end cell end table close square brackets$ , B = $open square brackets table row 1 0 row 0 2 end table close square brackets$ and X is a matrix such that A = BX, then X equals-

$\frac{1}{2} [\begin{matrix} - 2 & 4 \\ 3 & 5 \end{matrix}]$
$\frac{1}{2} [\begin{matrix} 2 & 4 \\ 3 & - 5 \end{matrix}]$
$[\begin{matrix} 2 & 4 \\ 3 & - 5 \end{matrix}]$
None of these

The correct answer is: $fraction numerator 1 over denominator 2 end fraction open square brackets table row 2 4 row 3 cell negative 5 end cell end table close square brackets$

A = BX $rightwards double arrow$ B^–1 A = (B^–1 B) X $rightwards double arrow$ B^–1A = X
$because$ B^–1 = $fraction numerator 1 over denominator 2 end fraction open square brackets table row 2 0 row 0 1 end table close square brackets$ $rightwards double arrow$ X = $fraction numerator 1 over denominator 2 end fraction open square brackets table row 2 4 row 3 cell negative 5 end cell end table close square brackets$

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