Question

(If A is a 3rd order matrix, such that det. A is 2, then det. (3A^2) is​

Answer 1

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

A is a square matrix of order 3 × 3 such that |A| = 2

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

$\rm \: |3 {A}^{2} |$

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

Given that,

A is a square matrix of order 3 × 3 such that |A| = 2

Now, Consider

$\rm \: |3 {A}^{2} |$

We know,

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

So, using this result, we get

$\rm \: = \: {3}^{3} \: | {A}^{2} |$

can be rewritten as

$\rm \: = \: 27 |A.A|$

We know

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

So, using this result, we get

$\rm \: = \: 27 \: |A| \: |A|$

$\rm \: = \: 27 \times 2 \times 2$

$\rm \: = \: 108$

Hence,

$\rm\implies \:\boxed{\tt{ \: |3 \: {A}^{2} | \: = \: 108 \: \: }}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ \: |adj \: A| = { |A| }^{n - 1} \: }}$

$\boxed{\tt{ \: |adj(adjA| = { |A| }^{ {(n - 1)}^{2} } \: \: }}$

$\boxed{\tt{ \: | {A}^{ - 1} | \: = \: \dfrac{1}{ |A| } \: }}$

$\boxed{\tt{ \: |A'| = |A| \: }}$

$\boxed{\tt{ \: A \: is \: singular \: matrix \: iff \: |A| = 0 \: }}$

$\boxed{\tt{ \: {A}^{ - 1} \: doesnot \: exist \: if \: |A| = 0 \: }}$

$\boxed{\tt{ \: adj(adjA) \: = \: { |A| }^{n - 2} \: A \: }}$

$\boxed{\tt{ \: adj(AB) = (adjB) \: (adjA) \: }}$

Answer 2

$\green{ \underline{\sf \: Given -} } \\ \\ \sf \: \: A \: is \: A \: square \: mathrix \: of \: order \: 3 \times 3 \\ \sf \: such \: that \: |A| \\ \\ \purple { \sf \underline{ \: To \: Find}} \\ \\ \tt \: |3 {A}^{2} | \\ \\ \red{ \sf \underline{solution}} \\ \\ \sf \: Given \: that \\ \sf \: A\: is \: A \: square \: matrix \: of \: order \\ \sf \: 3 \times 3 \: such \: that |a| = 2 \\ \\ \sf \: Now ,\: consider \\ \\ \tt |3 {a}^{2} | \\ \\ \sf We \: know \\ \\ \boxed{ \tt \: |KA| = {K}^{3} |A| } \\ \\ \sf \: so \: \bold{using \: this \: result} \: \: we \: get \\ \\ = \sf \: {3}^{3} | {A}^{2} | \\ \\ \sf \: can \bold{ \: Be \: rewritten} \: as \\ = \sf \: 27 |A.A| \\ \sf \: we \: know \: \\ \boxed{ \tt |AB| = |A| |B| } \\ \\ \sf \: so \: \bold{Using \: the \: result} \: we \: get \\ \\ = \sf27 |A| |A| \\ = 27 \times 2 \times 2 \\ = 108 \\ \\ \bold{ \tt \: Hence} \\ \\ : \to \boxed{ \tt \: |3 {A}^{2} | = 108 }$