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Hint- $\left| {3A} \right|$means first A matrix is multiplied with 3 and then it’s determinant is to be found. Evaluate each LHS and RHS separately, to prove.

We have given that $A = \left[ {\begin{array}{*{20}{c}}

1&0&1 \\

0&1&2 \\

0&0&4

\end{array}} \right]$

Now we show that $\left| {3A} \right| = 27\left| A \right|$

First let’s calculate the LHS part so $3A = 3\left[ {\begin{array}{*{20}{c}}

1&0&1 \\

0&1&2 \\

0&0&4

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

3&0&3 \\

0&3&6 \\

0&0&{12}

\end{array}} \right]$

Now the determinant of 3A that is $\left| {3A} \right|$

$\left| {\begin{array}{*{20}{c}}

3&0&3 \\

0&3&6 \\

0&0&{12}

\end{array}} \right|$=$\left[ {3\left( {3 \times 12 - 0 \times 6} \right) - 0\left( {0 \times 12 - 0 \times 6} \right) + 3\left( {0 \times 0 - 0 \times 3} \right)} \right]$

On simplifying we get

$\left| {\begin{array}{*{20}{c}}

3&0&3 \\

0&3&6 \\

0&0&{12}

\end{array}} \right| = 3 \times 36 = 108$………………………………….. (1)

Now we have to find $27\left| A \right|$

That is $27\left| {\begin{array}{*{20}{c}}

1&0&1 \\

0&1&2 \\

0&0&4

\end{array}} \right| = 27\left[ {1 \times \left( {1 \times 4 - 0 \times 2} \right) - 0\left( {0 \times 4 - 0 \times 2} \right) + 1\left( {0 \times 0 - 0 \times 1} \right)} \right]$

On simplifying we get

$27\left| {\begin{array}{*{20}{c}}

1&0&1 \\

0&1&2 \\

0&0&4

\end{array}} \right| = 27 \times 4 = 108$……………………………… (2)

Clearly equation (1) is equal to equation (2) thus we can say that $\left| {3A} \right| = 27\left| A \right|$

Hence proved.

Note- The key concept involved here is that we need to understand the basics of determinant evaluation: the quantity inside the determinant resembles a matrix , if it is multiplied with a scalar then the determinant of that scalar multiplied matrix is to be found. However if a scalar is multiplied with a determinant then simply the product of determinant and scalar number is to be evaluated.

We have given that $A = \left[ {\begin{array}{*{20}{c}}

1&0&1 \\

0&1&2 \\

0&0&4

\end{array}} \right]$

Now we show that $\left| {3A} \right| = 27\left| A \right|$

First let’s calculate the LHS part so $3A = 3\left[ {\begin{array}{*{20}{c}}

1&0&1 \\

0&1&2 \\

0&0&4

\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}

3&0&3 \\

0&3&6 \\

0&0&{12}

\end{array}} \right]$

Now the determinant of 3A that is $\left| {3A} \right|$

$\left| {\begin{array}{*{20}{c}}

3&0&3 \\

0&3&6 \\

0&0&{12}

\end{array}} \right|$=$\left[ {3\left( {3 \times 12 - 0 \times 6} \right) - 0\left( {0 \times 12 - 0 \times 6} \right) + 3\left( {0 \times 0 - 0 \times 3} \right)} \right]$

On simplifying we get

$\left| {\begin{array}{*{20}{c}}

3&0&3 \\

0&3&6 \\

0&0&{12}

\end{array}} \right| = 3 \times 36 = 108$………………………………….. (1)

Now we have to find $27\left| A \right|$

That is $27\left| {\begin{array}{*{20}{c}}

1&0&1 \\

0&1&2 \\

0&0&4

\end{array}} \right| = 27\left[ {1 \times \left( {1 \times 4 - 0 \times 2} \right) - 0\left( {0 \times 4 - 0 \times 2} \right) + 1\left( {0 \times 0 - 0 \times 1} \right)} \right]$

On simplifying we get

$27\left| {\begin{array}{*{20}{c}}

1&0&1 \\

0&1&2 \\

0&0&4

\end{array}} \right| = 27 \times 4 = 108$……………………………… (2)

Clearly equation (1) is equal to equation (2) thus we can say that $\left| {3A} \right| = 27\left| A \right|$

Hence proved.

Note- The key concept involved here is that we need to understand the basics of determinant evaluation: the quantity inside the determinant resembles a matrix , if it is multiplied with a scalar then the determinant of that scalar multiplied matrix is to be found. However if a scalar is multiplied with a determinant then simply the product of determinant and scalar number is to be evaluated.

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