Question

if a is square matrix such that a2=1 then find the value of (A-1)^3 + (A+1)^3 -7A​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:A \: is \: square \: matrix \: such \: that \: {A}^{2} = I$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\: {(A - I)}^{3} + {(A + I)}^{3} - 7A$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:A \: is \: square \: matrix \: such \: that \: {A}^{2} = I$

Consider,

$\rm :\longmapsto\: {(A - I)}^{3}$

$\rm \: \: = \:(A - I)(A - I)(A - I)$

$\rm \: \: = \:( {A}^{2} - AI - IA + {I}^{2})(A - I)$

$\rm \: \: = \:(I - A - A + I)(A - I)$

$\red{\bigg \{ \because \:AI = IA = A \: \: and \: {A}^{2} = I \bigg \}}$

$\rm \: \: = \:(2I - 2A)(A - I)$

$\rm \: \: = \:2IA - {2I}^{2} - {2A}^{2} + 2AI$

$\rm \: \: = \:2A - {2I} - {2I} + 2A$

$\rm \: \: = \:4A - {4I}$

$\bf\implies \: {(A - I)}^{3} = 4A - 4I - - (1)$

Consider

$\rm :\longmapsto\: {(A + I)}^{3}$

$\rm \: \: = \:(A + I)(A + I)(A + I)$

$\rm \: \: = \:( {A}^{2} + AI + IA + {I}^{2})(A + I)$

$\rm \: \: = \:(I + A + A + I)(A + I)$

$\rm \: \: = \:(2I + 2A)(A + I)$

$\rm \: \: = \:2IA + {2I}^{2} + {2A}^{2} + 2AI$

$\rm \: \: = \:2A + {2I} + {2I} + 2A$

$\rm \: \: = \:4A + {4I}$

$\bf\implies \: {(A + I)}^{3} = 4A + 4I - - (2)$

Now, Consider

$\rm :\longmapsto\: {(A - I)}^{3} + {(A + I)}^{3} - 7A$

On substituting the values from equation (1) and (2), we get

$\rm \: \: = \:4A - 4I + 4A + 4I - 7A$

$\rm \: \: = \:8A - 7A$

$\rm \: \: = \:A$

Hence,

The value of

$\: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf{\: {(A - I)}^{3} + {(A + I)}^{3} - 7A = A}}}$