Question

If A=diag(-1,-3,2), then |-2A| is equal to: *​

Answer 1

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

$\rm :\longmapsto\:A = diag( - 1, - 3,2)$

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

$\rm :\longmapsto\: | - 2A|$

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

Given that,

$\rm :\longmapsto\:A = diag( - 1, - 3,2)$

We know,

If matrix is given in diagonal form, as

$\rm :\longmapsto\:A = diag(a,b,c)$

then

$\rm :\longmapsto\: |A| = abc$

So, using this

$\rm :\longmapsto\: |A| = ( - 1) \times ( - 3) \times 2 = 6$

Now,

$\rm :\longmapsto\: | - 2A|$

We know,

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

So, using this identity, we get

$\rm \:  =  \:  \: {( - 2)}^{3} |A|$

$\rm \:  =  \:  \: - 8 \times 6$

$\rm \:  =  \:  \: - 48$

Hence,

$\bf :\longmapsto\: | - 2A| = - \: 48$

Additional Information

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

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

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

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

$\boxed{ \bf{ \: {AA}^{ - 1} = {A}^{ - 1}A = I}}$

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