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Mathematics

Algebra

Easy

Question

$L e t space M equals open square brackets table attributes columnalign left end attributes row 0 1 a row 1 2 3 row 3 b 1 end table close square brackets space a n d space adj M equals open square brackets table row cell negative 1 end cell 1 cell negative 1 end cell row 8 cell negative 6 end cell 2 row cell negative 5 end cell 3 cell negative 1 end cell end table close square brackets space text where a and b are real numbers. Which of the end text text following options is/are correct? end text$

$(a d j M)^{- 1} + a d j M^{- 1} = - M$
$det (adj M^{2}) = 81$
$a + b = 3$
$I f M = [\begin{array}{l} α \\ β \\ γ \end{array}] = [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}], then α - β + γ = 3$

The correct answers are: $I f space M equals open square brackets table attributes columnalign left end attributes row alpha row beta row gamma end table close square brackets equals open square brackets table attributes columnalign left end attributes row 1 row 2 row 3 end table close square brackets comma text then end text alpha minus beta plus gamma equals 3$ , $a plus b equals 3$ , $left parenthesis a d j M right parenthesis to the power of negative 1 end exponent plus a d j M to the power of negative 1 end exponent equals negative M$

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$L e t space M equals open square brackets table attributes columnalign center center columnspacing 1em rowspacing 4 pt end attributes row cell sin to the power of 4 theta end cell cell negative 1 minus sin squared theta end cell row cell 1 plus cos squared theta end cell cell cos to the power of 4 theta end cell end table close square brackets equals alpha I plus beta M to the power of negative 1 end exponent comma text where end text alpha equals alpha left parenthesis theta right parenthesis text and end text beta equals beta left parenthesis theta right parenthesis space text are real numbers, and I is the end text 2 cross times 2 text identity matrix. If end text alpha to the power of ∗ text is the minimum of the set end text left curly bracket alpha left parenthesis theta right parenthesis colon theta element of left square bracket 0 comma 2 pi right parenthesis right curly bracket text and end text beta to the power of ∗ space text is the minimum of the end text set left curly bracket beta left parenthesis theta right parenthesis colon theta element of left square bracket 0 comma 2 pi right parenthesis right curly bracket space comma space text then value of end text alpha to the power of ∗ plus beta to the power of ∗ text is - end text$

$L e t space M equals open square brackets table attributes columnalign center center columnspacing 1em rowspacing 4 pt end attributes row cell sin to the power of 4 theta end cell cell negative 1 minus sin squared theta end cell row cell 1 plus cos squared theta end cell cell cos to the power of 4 theta end cell end table close square brackets equals alpha I plus beta M to the power of negative 1 end exponent comma text where end text alpha equals alpha left parenthesis theta right parenthesis text and end text beta equals beta left parenthesis theta right parenthesis space text are real numbers, and I is the end text 2 cross times 2 text identity matrix. If end text alpha to the power of ∗ text is the minimum of the set end text left curly bracket alpha left parenthesis theta right parenthesis colon theta element of left square bracket 0 comma 2 pi right parenthesis right curly bracket text and end text beta to the power of ∗ space text is the minimum of the end text set left curly bracket beta left parenthesis theta right parenthesis colon theta element of left square bracket 0 comma 2 pi right parenthesis right curly bracket space comma space text then value of end text alpha to the power of ∗ plus beta to the power of ∗ text is - end text$

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$text The total number of matrices end text A equals open parentheses table attributes columnalign center center center columnspacing 1em rowspacing 4 pt end attributes row 0 cell 2 y end cell 1 row cell 2 x end cell y cell negative 1 end cell row cell 2 x end cell cell negative y end cell 1 end table close parentheses comma left parenthesis x comma y element of R comma x not equal to y right parenthesis space text for which end text straight A to the power of straight T straight A equals 3 straight I subscript 3 is colon$

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If $left square bracket table attributes columnalign left left columnspacing 1em rowspacing 4 pt end attributes row 1 1 row 0 1 end table right square bracket left square bracket table attributes columnalign left left columnspacing 1em rowspacing 4 pt end attributes row 1 2 row 0 1 end table right square bracket left square bracket table attributes columnalign left left columnspacing 1em rowspacing 4 pt end attributes row 1 3 row 0 1 end table right square bracket horizontal ellipsis horizontal ellipsis. left square bracket table attributes columnalign center center columnspacing 1em rowspacing 4 pt end attributes row 1 cell n minus 1 end cell row 0 1 end table right square bracket equals left square bracket table attributes columnalign center center columnspacing 1em rowspacing 4 pt end attributes row 1 78 row 0 1 end table right square bracket$ then the inverse of $left square bracket table attributes columnalign left left columnspacing 1em rowspacing 4 pt end attributes row 1 n row 0 1 end table right square bracket$ is

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