Answer
Verified
481.5k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE