If a is a complex cube root 2, find
\[\vartriangle = \left( {\begin{array}{*{20}{c}}
1&{2a}&1 \\
{{a^2}}&1&{4{a^2}} \\
3&{2a}&1
\end{array}} \right)\]
Answer
637.5k+ views
Hint: It is given in the question that a is cube root infinity. Therefore,
$ \Rightarrow a = \sqrt[3]{2}$
Find the determinant of the matrix given in the question and after that put the value of a in the determinant.
Determinant of a matrix can be calculated by expanding the matrix either row wise or column wise. It is a single number and determinants are always calculated from a singular matrix.
Complete step-by-step answer:
We are given with a value of a i.e. equal to cube root of 2.
$ \Rightarrow a = \sqrt[3]{2}$
We have to find the value of matrix given below
\[ \Rightarrow \vartriangle = \left( {\begin{array}{*{20}{c}}
1&{2a}&1 \\
{{a^2}}&1&{4{a^2}} \\
3&{2a}&1
\end{array}} \right)\]
To find the determinant expand it row wise, you can also expand it column wise as follows
\[
\Rightarrow \vartriangle = \left( {\begin{array}{*{20}{c}}
1&{2a}&1 \\
{{a^2}}&1&{4{a^2}} \\
3&{2a}&1
\end{array}} \right) \\
\Rightarrow \vartriangle = 1(1 - (4{a^2})(2a)) - 2a({a^2} - 3(4{a^2})) + 1(({a^2})(2a) - 3) \\
\Rightarrow \vartriangle = 1 - 8{a^3} - 2a({a^2} - 12{a^2}) + 2{a^3} - 3 \\
\Rightarrow \vartriangle = 1 - 8{a^3} - 2{a^3} + 24{a^3} + 2{a^3} - 3 \\
\Rightarrow \vartriangle = 16{a^3} - 2 \\
\]
Now, we have found the determinant of the matrix and the value of a is given and put that value in above equation, we get
$
\Rightarrow \vartriangle = 16{a^3} - 2 \\
\Rightarrow \vartriangle = 16{(\sqrt[3]{2})^3} - 2 \\
\Rightarrow \vartriangle = 16(2) - 2 \\
\Rightarrow \vartriangle = 32 - 2 \\
\Rightarrow \vartriangle = 30 \\
$
So, the required value of the matrix is 30.
Note: During calculating the determinant of the matrix take care of the sign. Many students make mistakes in multiplying the sign i.e. they write minus while multiplying minus sign with minus sign which is wrong. The signs of numbers are multiplied as follows:
$ \Rightarrow a = \sqrt[3]{2}$
Find the determinant of the matrix given in the question and after that put the value of a in the determinant.
Determinant of a matrix can be calculated by expanding the matrix either row wise or column wise. It is a single number and determinants are always calculated from a singular matrix.
Complete step-by-step answer:
We are given with a value of a i.e. equal to cube root of 2.
$ \Rightarrow a = \sqrt[3]{2}$
We have to find the value of matrix given below
\[ \Rightarrow \vartriangle = \left( {\begin{array}{*{20}{c}}
1&{2a}&1 \\
{{a^2}}&1&{4{a^2}} \\
3&{2a}&1
\end{array}} \right)\]
To find the determinant expand it row wise, you can also expand it column wise as follows
\[
\Rightarrow \vartriangle = \left( {\begin{array}{*{20}{c}}
1&{2a}&1 \\
{{a^2}}&1&{4{a^2}} \\
3&{2a}&1
\end{array}} \right) \\
\Rightarrow \vartriangle = 1(1 - (4{a^2})(2a)) - 2a({a^2} - 3(4{a^2})) + 1(({a^2})(2a) - 3) \\
\Rightarrow \vartriangle = 1 - 8{a^3} - 2a({a^2} - 12{a^2}) + 2{a^3} - 3 \\
\Rightarrow \vartriangle = 1 - 8{a^3} - 2{a^3} + 24{a^3} + 2{a^3} - 3 \\
\Rightarrow \vartriangle = 16{a^3} - 2 \\
\]
Now, we have found the determinant of the matrix and the value of a is given and put that value in above equation, we get
$
\Rightarrow \vartriangle = 16{a^3} - 2 \\
\Rightarrow \vartriangle = 16{(\sqrt[3]{2})^3} - 2 \\
\Rightarrow \vartriangle = 16(2) - 2 \\
\Rightarrow \vartriangle = 32 - 2 \\
\Rightarrow \vartriangle = 30 \\
$
So, the required value of the matrix is 30.
Note: During calculating the determinant of the matrix take care of the sign. Many students make mistakes in multiplying the sign i.e. they write minus while multiplying minus sign with minus sign which is wrong. The signs of numbers are multiplied as follows:
| Sign of A | Sign of B | Sign of $A \times B$ |
| Plus (+) | Plus (+) | Plus (+) |
| Plus (+) | Minus (-) | Minus (-) |
| Minus (-) | Plus (+) | Minus (-) |
| Minus (-) | Minus (-) | Plus (+) |
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