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Maths-

General

Easy

Question

If matrix A = $open square brackets table row 1 0 cell negative 1 end cell row 3 4 5 row 0 6 7 end table close square brackets$ and its inverse is denoted by $A to the power of negative 1 end exponent equals open square brackets table row cell a subscript 11 end subscript end cell cell a subscript 12 end subscript end cell cell a subscript 13 end subscript end cell row cell a subscript 21 end subscript end cell cell a subscript 22 end subscript end cell cell a subscript 23 end subscript end cell row cell a subscript 31 end subscript end cell cell a subscript 32 end subscript end cell cell a subscript 33 end subscript end cell end table close square brackets$ , then the value of a₂₃is equal to-

$\frac{21}{20}$
$\frac{1}{5}$
$- \frac{2}{5}$
$\frac{2}{5}$

The correct answer is: $fraction numerator 1 over denominator 5 end fraction$

Related Questions to study

For the matrix A = $open square brackets table row 1 1 0 row 1 2 1 row 2 1 0 end table close square brackets$ , which of the following is correct

For the matrix A = $open square brackets table row 1 1 0 row 1 2 1 row 2 1 0 end table close square brackets$ , which of the following is correct

If a matrix A is such that3A³+ 2A²+ 5A + I = 0, then A^–1is equal to-

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If M = $open square brackets table row 1 2 row 2 3 end table close square brackets$ and M²– $lambda$ M – $straight I subscript 2$ = 0, then $lambda$ equals -

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If $open square brackets table attributes columnalign center center columnspacing 1em end attributes row 1 cell negative tan theta end cell row cell tan theta end cell 1 end table close square brackets$ $open square brackets table row 1 cell tan invisible function application theta end cell row cell negative tan invisible function application theta end cell 1 end table close square brackets to the power of negative 1 end exponent$ = $open square brackets table row a cell negative b end cell row b a end table close square brackets$ , then -

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parallel

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