Question

if A is an investible matrix of order 3 and detA=2, then del (A-1) equals​

Answer 1

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

Given that

A is a square matrix of order 3 such that | A | = 3

Now,

We know that

$\rm :\longmapsto\: {AA}^{ - 1} = I$

Taking determinant om both sides, we get

$\rm :\longmapsto\: | {AA}^{ - 1} | = |I|$

We know,

$\boxed{ \bf{ \: |AB| = |A| \: |B|}}$

and

$\boxed{ \bf{ \: |I| = 1}}$

So, using this property, we get

$\rm :\longmapsto\: |A| \: | {A}^{ - 1} | = 1$

$\bf\implies \: | {A}^{ - 1} | = \dfrac{1}{ |A| }$

$\bf\implies \: | {A}^{ - 1} | = \dfrac{1}{2}$

Additional Information -

$\boxed{ \bf{ \: |adjA| = { |A| }^{n - 1}}}$

$\boxed{ \bf{ \: |AadjA| = { |A| }^{n}}}$

$\boxed{ \bf{ \: |kA| \: = \: {k}^{n} \: |A| }}$

$\boxed{ \bf{ \: |kAB| \: = \: {k}^{n} \: |A| \: |B| }}$

$\boxed{ \bf{ \: |A| = | {A}^{T} | }}$