Question

Find the value of \[\det \left( {{A}_{0}}+A_{0}^{2}B_{0}^{2}+A_{0}^{3}+A_{0}^{4}B_{0}^{4}+.......10\text{ terms} \right)\] if the two matrices ${{A}_{0}}$ and ${{B}_{0}}$ are $\left[ \begin{matrix}
   2 & -2 & -4 \\
   -1 & 3 & 4 \\
   1 & -2 & -3 \\
\end{matrix} \right]$ and $\left[ \begin{matrix}
   -4 & -3 & -3 \\
   1 & 0 & 1 \\
   4 & 4 & 3 \\
\end{matrix} \right]$ respectively.
\[\begin{align}
  & A.1000 \\
 & B.-800 \\
 & C.0 \\
 & D.-8000 \\
\end{align}\]

Answer 1

Hint: Students should first find the determinant values using formula, if $X=\left[ \begin{matrix}
   a & d & g \\
   b & e & h \\
   c & f & i \\
\end{matrix} \right]$ then its determinant value or det (X) is $a\left( ei-fh \right)-d\left( bi-ch \right)+g\left( bf-ce \right)$ and then use properties such as,
\[\begin{align}
  & \det \left( A+B \right)=\det \left( A \right)+\det \left( B \right) \\
 & \det \left( AB \right)=\det \left( A \right)\times \det \left( B \right) \\
 & \det \left( {{A}^{4}} \right)={{\left( \det \left( A \right) \right)}^{4}} \\
\end{align}\]
To get the desired answer.

Complete step-by-step solution:
In the above question, we are given two matrices ${{A}_{0}}$ and ${{B}_{0}}$ which are $\left[ \begin{matrix}
   2 & -2 & -4 \\
   -1 & 3 & 4 \\
   1 & -2 & -3 \\
\end{matrix} \right]$ and $\left[ \begin{matrix}
   -4 & -3 & -3 \\
   1 & 0 & 1 \\
   4 & 4 & 3 \\
\end{matrix} \right]$ and from this we have to find the value of \[\det \left( {{A}_{0}}+A_{0}^{2}B_{0}^{2}+A_{0}^{3}+A_{0}^{4}B_{0}^{4}+.......10\text{ terms} \right)\]
So, as we know that,
\[\begin{align}
  & \det \left( A+B \right)=\det \left( A \right)+\det \left( B \right) \\
 & \det \left( AB \right)=\det \left( A \right)\times \det \left( B \right) \\
 & \det \left( {{A}^{4}} \right)={{\left( \det \left( A \right) \right)}^{4}} \\
\end{align}\]
We can rewrite the given statement \[\det \left( {{A}_{0}}+A_{0}^{2}B_{0}^{2}+A_{0}^{3}+A_{0}^{4}B_{0}^{4}+.......10\text{ terms} \right)\] as \[\det \left( {{A}_{0}} \right)+\det \left( A_{0}^{2} \right)\times \det \left( B_{0}^{2} \right)+\det \left( A_{0}^{3} \right).......10\text{ terms}\], which can be further written as,
\[\det \left( {{A}_{0}} \right)+{{\left( \det \left( {{A}_{0}} \right) \right)}^{2}}\times {{\left( \det \left( {{B}_{0}} \right) \right)}^{2}}+{{\left( \det \left( {{A}_{0}} \right) \right)}^{3}}+.......10\text{ terms}\]
Now, as we know that,
\[{{A}_{0}}=\left[ \begin{matrix}
   2 & -2 & -4 \\
   -1 & 3 & 4 \\
   1 & -2 & -3 \\
\end{matrix} \right]\]
Then we will find the determinant using formula. Let us consider a matrix of the form
$X=\left[ \begin{matrix}
   a & d & g \\
   b & e & h \\
   c & f & i \\
\end{matrix} \right]$
Then, we know that the determinant of this matrix can be written as \[\det \left( X \right)=a\left( ei-fh \right)-d\left( bi-ch \right)+g\left( bf-ce \right)\]
Now, we will apply this for our matrix. So we get the determinant of the matrix, $\det \left( {{A}_{0}} \right)$ as \[\Rightarrow \left\{ 2\left( -9+8 \right)+2\left( 3-4 \right)-4\left( 2-3 \right) \right\}\Rightarrow \left( -2-2+4 \right)\Rightarrow 0\]
Now, in a similar manner, we will find the determinant of matrix B.
For, the matrix \[{{B}_{0}}=\left[ \begin{matrix}
   -4 & -3 & -3 \\
   1 & 0 & 1 \\
   4 & 4 & 3 \\
\end{matrix} \right]\] again using the above formula we can get its determinant $\det \left( {{B}_{0}} \right)$ as
\[\Rightarrow \left\{ -4\left( 0-4 \right)+3\left( 3-4 \right)+\left( -3 \right)\left( 4-0 \right) \right\}\Rightarrow \left( 16-3-12 \right)\Rightarrow 1\]
Thus, from the above calculations, we have got \[\det \left( {{A}_{0}} \right)=0\text{ and }\det \left( {{B}_{0}} \right)=1\]
Now, we can evaluate the value of the given expression in the question. So, we have the value of the given expression as,
\[\begin{align}
  & \det \left( {{A}_{0}} \right)+{{\left( \det \left( {{A}_{0}} \right) \right)}^{2}}\times {{\left( \det \left( {{B}_{0}} \right) \right)}^{2}}+{{\left( \det \left( {{A}_{0}} \right) \right)}^{3}}+{{\left( \det \left( {{A}_{0}} \right) \right)}^{4}}\times {{\left( \det \left( {{B}_{0}} \right) \right)}^{4}}+ \\
 & {{\left( \det \left( {{A}_{0}} \right) \right)}^{5}}\times {{\left( \det \left( {{B}_{0}} \right) \right)}^{5}}+{{\left( \det \left( {{A}_{0}} \right) \right)}^{6}}\times {{\left( \det \left( {{B}_{0}} \right) \right)}^{6}}+{{\left( \det \left( {{A}_{0}} \right) \right)}^{7}}\times {{\left( \det \left( {{B}_{0}} \right) \right)}^{7}}+ \\
 & {{\left( \det \left( {{A}_{0}} \right) \right)}^{8}}\times {{\left( \det \left( {{B}_{0}} \right) \right)}^{8}}+{{\left( \det \left( {{A}_{0}} \right) \right)}^{9}}\times {{\left( \det \left( {{B}_{0}} \right) \right)}^{9}}+{{\left( \det \left( {{A}_{0}} \right) \right)}^{10}}\times {{\left( \det \left( {{B}_{0}} \right) \right)}^{10}} \\
\end{align}\]
Now on substituting $\det \left( {{A}_{0}} \right)=0\text{ and }\det \left( {{B}_{0}} \right)=1$ in the above expression, we get:
\[\begin{align}
  & 0+{{\left( 0 \right)}^{2}}\times {{\left( 1 \right)}^{2}}+{{\left( 0 \right)}^{3}}+{{\left( 0 \right)}^{4}}\times {{\left( 1 \right)}^{4}}+{{\left( 0 \right)}^{5}}\times {{\left( 1 \right)}^{5}}+{{\left( 0 \right)}^{6}}\times {{\left( 1 \right)}^{6}}+ \\
 & {{\left( 0 \right)}^{7}}\times {{\left( 1 \right)}^{7}}+{{\left( 0 \right)}^{8}}\times {{\left( 1 \right)}^{8}}+{{\left( 0 \right)}^{9}}\times {{\left( 1 \right)}^{9}}+{{\left( 0 \right)}^{10}}\times {{\left( 1 \right)}^{10}} \\
\end{align}\]
Thus, on simplification of all the above terms, we get:
\[0+0+0+0+0+0+0+0+0+0\Rightarrow 0\]
Thus the correct option is C.

Note: Students while finding determinant values, generally do calculation mistakes which can make their answer wrong. Thus, they should be careful about that. Students must note that in the given expression whose value is to be found, we have to find the determinant of matrix A and B and then take the square, cube, or higher powers. They might by mistake find the square of the matrix by multiplying it and then take its determinant. So, this must be avoided.