Question

If $$A=\left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right]$$, 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 $$A=\left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right]$$.
The determinate of matrix $$A$$ is $$\left|A\right|=\left|\begin{array}{cc}2& 3\\ 0& -1\end{array}\right|$$
  $$=2\cdot \left(-1\right)-0\cdot 3$$
 $$=-2$$
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 $$A^2$$
$$A^2=\left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right]\left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right]$$
$$=\left[\begin{array}{cc}{2}^2+0\cdot 3& 2\cdot 3+3\cdot \left(-1\right)\\ 0\cdot 2+0\cdot \left(-1\right)& 3\cdot 0+\left(-1\right)\cdot \left(-1\right)\end{array}\right]$$
$$=\left[\begin{array}{cc}{2}^2& 3\\ 0& 1\end{array}\right]$$
Now finding the matrix $$A^4$$
$$A^4=\left[\begin{array}{cc}{2}^2& 3\\ 0& 1\end{array}\right]\left[\begin{array}{cc}{2}^2& 3\\ 0& 1\end{array}\right]$$
 $$=\left[\begin{array}{cc}{2}^4+0\cdot 3& 2^2\cdot 3+3\cdot 1\\ 0\cdot 2^2+0\cdot 1& 3\cdot 0+1\cdot 1\end{array}\right]$$
$$=\left[\begin{array}{cc}{2}^4& 15\\ 0& 1\end{array}\right]$$
 $$=\left[\begin{array}{cc}{2}^4& 2^4-1\\ 0& 1\end{array}\right]$$
So on.
Similarly, $$A^n=\left[\begin{array}{cc}{2}^n& 2^n-1\\ 0& {\left(-1\right)}^n\end{array}\right]$$
Putting $$n=10$$
$$A^{10}=\left[\begin{array}{cc}{2}^{10}& 2^{10}-1\\ 0& {\left(-1\right)}^{10}\end{array}\right]$$
$$=\left[\begin{array}{cc}1024& 1023\\ 0& 1\end{array}\right]$$
So, $$2A=\left[\begin{array}{cc}4& 6\\ 0& -2\end{array}\right]$$
Now we will find $$Adj2A$$
$$Adj2A=\left[\begin{array}{cc}-2& -6\\ 0& 4\end{array}\right]$$
Now we will calculate $${\left(Adj2A\right)}^2$$
$${\left(Adj2A\right)}^2=\left[\begin{array}{cc}-2& -6\\ 0& 4\end{array}\right]\left[\begin{array}{cc}-2& -6\\ 0& 4\end{array}\right]$$
 $$=\left[\begin{array}{cc}{\left(-2\right)}^2+\left(-6\right)\cdot 0& \left(-6\right)\cdot \left(-2\right)+4\cdot \left(-6\right)\\ 0\cdot \left(-2\right)+4\cdot 0& 0\cdot \left(-6\right)+4\cdot 4\end{array}\right]$$
 $$=\left[\begin{array}{cc}{2}^2& -12\\ 0& 16\end{array}\right]$$
 $$=4\left[\begin{array}{cc}1& -3\\ 0& 4\end{array}\right]$$
$${\left(Adj2A\right)}^4=4\left[\begin{array}{cc}1& -3\\ 0& 4\end{array}\right]\times 4\left[\begin{array}{cc}1& -3\\ 0& 4\end{array}\right]$$
  $$=4^2\left[\begin{array}{cc}1\times 1+\left(-3\right)\times 0& 1\cdot \left(-3\right)+4\cdot \left(-3\right)\\ 0\cdot \left(1\right)+4\cdot 0& 0\cdot \left(-3\right)+4\cdot 4\end{array}\right]$$
$$=2^4\left[\begin{array}{cc}1& -15\\ 0& 16\end{array}\right]$$
 $$=2^4\left[\begin{array}{cc}1& -\left(2^4-1\right)\\ 0& 2^4\end{array}\right]$$
Similarly, $${\left(Adj2A\right)}^{10}=2^{10}\left[\begin{array}{cc}1& -\left(2^{10}-1\right)\\ 0& 2^{10}\end{array}\right]$$
$$=2^{10}\left[\begin{array}{cc}1& -1023\\ 0& 1024\end{array}\right]$$
Now finding the matrix $$A^{10}-{\left(Adj\left(2A\right)\right)}^{10}$$
$$A^{10}-{\left(Adj\left(2A\right)\right)}^{10}=\left[\begin{array}{cc}1024& 1023\\ 0& 1\end{array}\right]-2^{10}\left[\begin{array}{cc}1& -1023\\ 0& 1024\end{array}\right]$$
  $$=\left[\begin{array}{cc}1024-1024& 1023+2^{10}\cdot 1023\\ 0& 1-2^{10}\cdot 1024\end{array}\right]$$
  $$=\left[\begin{array}{cc}0& 1023\left(1+2^{10}\right)\\ 0& 1-2^{10}\cdot 1024\end{array}\right]$$
The determinate of the matrix $$A^{10}-{\left(Adj\left(2A\right)\right)}^{10}$$ is
$$det\left(A^{10}-{\left(Adj\left(2A\right)\right)}^{10}\right)=\left|\begin{array}{cc}0& 1023\left(1+2^{10}\right)\\ 0& 1-2^{10}\cdot 1024\end{array}\right|$$
$$=0$$
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 $${\left(Adj\left(2A\right)\right)}^{10}$$.