Question

If \[A = \left[ {\begin{array}{*{20}{c}}2&3\\0&{ - 1}\end{array}} \right]\], then find the value of \[\det \left( {{A^4}} \right) + \det \left( {{A^{10}} - {{\left( {Adj\left( {2A} \right)} \right)}^{10}}} \right)\].

Answer 1

Hint: We find the determinant value of the matrix \[A\]. Then we will find the value of \[\det \left( {{A^4}} \right)\] by using the theorem if the determinate value of the matrix \[A\] is \[k\], then the determinate value of \[{A^n}\] is \[{k^n}\]. After that we will calculate the matrices \[{A^{10}}\] and \[{\left( {Adj\left( {2A} \right)} \right)^{10}}\]. Then we can calculate the value \[\det \left( {{A^4}} \right) + \det \left( {{A^{10}} - {{\left( {Adj\left( {2A} \right)} \right)}^{10}}} \right)\].

Formula used:
If the determinate value of the matrix \[A\] is \[k\], then the determinate value of \[{A^n}\] is \[{k^n}\].

Complete step by step solution:
Given matrix is \[A = \left[ {\begin{array}{*{20}{c}}2&3\\0&{ - 1}\end{array}} \right]\].
The determinate of matrix \[A\] is \[\left| A \right| = \left| {\begin{array}{*{20}{c}}2&3\\0&{ - 1}\end{array}} \right|\]
  \[ = 2 \cdot \left( { - 1} \right) - 0 \cdot 3\]
 \[ = - 2\]
Apply the theorem “if the determinate value of the matrix \[A\] is \[k\], then the determinate value of \[{A^n}\] is \[{k^n}\]. “
Therefore, the value of \[\det \left( {{A^4}} \right)\] is \[{\left( { - 2} \right)^4} = 16\].
Now finding the matrix \[{A^2}\]
\[{A^2} = \left[ {\begin{array}{*{20}{c}}2&3\\0&{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&3\\0&{ - 1}\end{array}} \right]\]
\[ = \left[ {\begin{array}{*{20}{c}}{{2^2} + 0 \cdot 3}&{2 \cdot 3 + 3 \cdot \left( { - 1} \right)}\\{0 \cdot 2 + 0 \cdot \left( { - 1} \right)}&{3 \cdot 0 + \left( { - 1} \right) \cdot \left( { - 1} \right)}\end{array}} \right]\]
\[ = \left[ {\begin{array}{*{20}{c}}{{2^2}}&3\\0&1\end{array}} \right]\]
Now finding the matrix \[{A^4}\]
\[{A^4} = \left[ {\begin{array}{*{20}{c}}{{2^2}}&3\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{2^2}}&3\\0&1\end{array}} \right]\]
 \[ = \left[ {\begin{array}{*{20}{c}}{{2^4} + 0 \cdot 3}&{{2^2} \cdot 3 + 3 \cdot 1}\\{0 \cdot {2^2} + 0 \cdot 1}&{3 \cdot 0 + 1 \cdot 1}\end{array}} \right]\]
\[ = \left[ {\begin{array}{*{20}{c}}{{2^4}}&{15}\\0&1\end{array}} \right]\]
 \[ = \left[ {\begin{array}{*{20}{c}}{{2^4}}&{{2^4} - 1}\\0&1\end{array}} \right]\]
So on.
Similarly, \[{A^n} = \left[ {\begin{array}{*{20}{c}}{{2^n}}&{{2^n} - 1}\\0&{{{\left( { - 1} \right)}^n}}\end{array}} \right]\]
Putting \[n = 10\]
\[{A^{10}} = \left[ {\begin{array}{*{20}{c}}{{2^{10}}}&{{2^{10}} - 1}\\0&{{{\left( { - 1} \right)}^{10}}}\end{array}} \right]\]
\[ = \left[ {\begin{array}{*{20}{c}}{1024}&{1023}\\0&1\end{array}} \right]\]
So, \[2A = \left[ {\begin{array}{*{20}{c}}4&6\\0&{ - 2}\end{array}} \right]\]
Now we will find \[Adj\,2A\]
\[Adj\,\,2A = \left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 6}\\0&4\end{array}} \right]\]
Now we will calculate \[{\left( {Adj\,\,2A} \right)^2}\]
\[{\left( {Adj\,\,2A} \right)^2} = \left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 6}\\0&4\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 6}\\0&4\end{array}} \right]\]
 \[ = \left[ {\begin{array}{*{20}{c}}{{{\left( { - 2} \right)}^2} + \left( { - 6} \right) \cdot 0}&{\left( { - 6} \right) \cdot \left( { - 2} \right) + 4 \cdot \left( { - 6} \right)}\\{0 \cdot \left( { - 2} \right) + 4 \cdot 0}&{0 \cdot \left( { - 6} \right) + 4 \cdot 4}\end{array}} \right]\]
 \[ = \left[ {\begin{array}{*{20}{c}}{{2^2}}&{ - 12}\\0&{16}\end{array}} \right]\]
 \[ = 4\left[ {\begin{array}{*{20}{c}}1&{ - 3}\\0&4\end{array}} \right]\]
\[{\left( {Adj\,\,2A} \right)^4} = 4\left[ {\begin{array}{*{20}{c}}1&{ - 3}\\0&4\end{array}} \right] \times 4\left[ {\begin{array}{*{20}{c}}1&{ - 3}\\0&4\end{array}} \right]\]
  \[ = {4^2}\left[ {\begin{array}{*{20}{c}}{1 \times 1 + \left( { - 3} \right) \times 0}&{1 \cdot \left( { - 3} \right) + 4 \cdot \left( { - 3} \right)}\\{0 \cdot \left( 1 \right) + 4 \cdot 0}&{0 \cdot \left( { - 3} \right) + 4 \cdot 4}\end{array}} \right]\]
\[ = {2^4}\left[ {\begin{array}{*{20}{c}}1&{ - 15}\\0&{16}\end{array}} \right]\]
 \[ = {2^4}\left[ {\begin{array}{*{20}{c}}1&{ - \left( {{2^4} - 1} \right)}\\0&{{2^4}}\end{array}} \right]\]
Similarly, \[{\left( {Adj\,\,2A} \right)^{10}} = {2^{10}}\left[ {\begin{array}{*{20}{c}}1&{ - \left( {{2^{10}} - 1} \right)}\\0&{{2^{10}}}\end{array}} \right]\]
\[ = {2^{10}}\left[ {\begin{array}{*{20}{c}}1&{ - 1023}\\0&{1024}\end{array}} \right]\]
Now finding the matrix \[{A^{10}} - {\left( {Adj\left( {2A} \right)} \right)^{10}}\]
\[{A^{10}} - {\left( {Adj\left( {2A} \right)} \right)^{10}} = \left[ {\begin{array}{*{20}{c}}{1024}&{1023}\\0&1\end{array}} \right] - {2^{10}}\left[ {\begin{array}{*{20}{c}}1&{ - 1023}\\0&{1024}\end{array}} \right]\]
  \[ = \left[ {\begin{array}{*{20}{c}}{1024 - 1024}&{1023 + {2^{10}} \cdot 1023}\\0&{1 - {2^{10}} \cdot 1024}\end{array}} \right]\]
  \[ = \left[ {\begin{array}{*{20}{c}}0&{1023\left( {1 + {2^{10}}} \right)}\\0&{1 - {2^{10}} \cdot 1024}\end{array}} \right]\]
The determinate of the matrix \[{A^{10}} - {\left( {Adj\left( {2A} \right)} \right)^{10}}\] is
\[\det \left( {{A^{10}} - {{\left( {Adj\left( {2A} \right)} \right)}^{10}}} \right) = \left| {\begin{array}{*{20}{c}}0&{1023\left( {1 + {2^{10}}} \right)}\\0&{1 - {2^{10}} \cdot 1024}\end{array}} \right|\]
\[ = 0\]
Now we will put the value of \[\det \left( {{A^{10}} - {{\left( {Adj\left( {2A} \right)} \right)}^{10}}} \right)\] in expression\[\det \left( {{A^4}} \right) + \det \left( {{A^{10}} - {{\left( {Adj\left( {2A} \right)} \right)}^{10}}} \right)\]
\[\det \left( {{A^4}} \right) + \det \left( {{A^{10}} - {{\left( {Adj\left( {2A} \right)} \right)}^{10}}} \right)\]
\[ = 16 + 0\]
\[ = 16\]
Hence the value of \[\det \left( {{A^4}} \right) + \det \left( {{A^{10}} - {{\left( {Adj\left( {2A} \right)} \right)}^{10}}} \right)\] is 16.
Note The theorem” if the determinate value of the matrix \[A\] is \[k\], then the determinate value of \[{A^n}\] is \[{k^n}\]. “does not applicable for calculation \[\det \left( {{A^{10}} - {{\left( {Adj\left( {2A} \right)} \right)}^{10}}} \right)\]. Students often calculate the matrices \[A,{A^2},{A^3}, \cdots ,{A^{10}}\]. But it is long process. For the shortcut we will find the matrices \[{A^2}\] and \[{A^4}\]. After that we will get a pattern and from the pattern, we can get matrix \[{A^{10}}\]. Similar way, we will calculate \[{\left( {Adj\left( {2A} \right)} \right)^{10}}\].