
If $A = \left[ {\begin{array}{*{20}{c}}
1&0&1 \\
0&1&2 \\
0&0&4
\end{array}} \right]$ then show that$\left| {3A} \right| = 27\left| A \right|$.
Answer
608.4k+ views
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.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

When was the first election held in India a 194748 class 12 sst CBSE

