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Maths-

General

Easy

Question

If the product of the matrix B = $open square brackets table row 2 6 4 row 1 0 1 row cell negative 1 end cell 1 cell negative 1 end cell end table close square brackets$ with a matrix A has inverse C = $open square brackets table row cell negative 1 end cell 0 1 row 1 1 3 row 2 0 2 end table close square brackets$ , then A^–1 equals-

$[\begin{matrix} - 3 & - 5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 6 \end{matrix}]$
$[\begin{matrix} - 3 & 5 & 5 \\ 0 & 0 & 9 \\ 2 & 14 & 16 \end{matrix}]$
$[\begin{matrix} - 3 & - 5 & - 5 \\ 0 & 0 & 2 \\ 2 & 14 & 6 \end{matrix}]$
$[\begin{matrix} - 3 & - 3 & - 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6 \end{matrix}]$

The correct answer is: $open square brackets table row cell negative 3 end cell cell negative 5 end cell cell negative 5 end cell row 0 0 2 row 2 14 6 end table close square brackets$

We have (BA)^–1 = C $rightwards double arrow$ A^–1 B^–1 = C $rightwards double arrow$ A^–1 = CB
$therefore$ A^–1 = $open square brackets table row cell negative 1 end cell 0 1 row 1 1 3 row 2 0 2 end table close square brackets$ $open square brackets table row 2 6 4 row 1 0 1 row cell negative 1 end cell 1 cell negative 1 end cell end table close square brackets$
= $open square brackets table row cell negative 3 end cell cell negative 5 end cell cell negative 5 end cell row 0 9 2 row 2 14 6 end table close square brackets$

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$MeC identical to straight C minus COCl not stretchy ⟶ with straight H subscript 2 plus text Lindlar's Catalyst end text on top$ (A) The product (A) is:

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chemistry-General

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