Question

If $A=\left[ \begin{matrix} -4 & -1 \\\ 3 & 1 \\\ \end{matrix} \right]$ , then the determinant of the matrix ${{A}^{2016}}-2{{A}^{2015}}-{{A}^{2014}}$ is
A. $2014$
B. $2016$
C. $-175$
D. $-25$

Answer 1

At first, we have to take ${{A}^{2014}}$ common from the expression and get ${{A}^{2014}}\left( {{A}^{2}}-2A-I \right)$ . After that, we find the value of the determinant $\left| A \right|$ and find the value of the matrix ${{A}^{2}}-2A-I$ which come out to be $-1$ and $\left[ \begin{matrix} 20 & 5 \\\ -15 & -5 \\\ \end{matrix} \right]$ respectively. The value of the determinant $\left| {{A}^{2}}-2A-I \right|$ will then be $-25$ . We then apply the property $\left| AB \right|=\left| A \right|\left| B \right|$ to $\left| {{A}^{2014}}\left( {{A}^{2}}-2A-I \right) \right|$ and get $\left| {{A}^{2014}} \right|\left| {{A}^{2}}-2A-I \right|$ . Again, we apply the property $\left| {{A}^{n}} \right|={{\left| A \right|}^{n}}$ and get the value of $\left| {{A}^{2014}} \right|\left| {{A}^{2}}-2A-I \right|={{\left| A \right|}^{2014}}\left| {{A}^{2}}-2A-I \right|$ .

Complete step by step answer:
The matrix that we are given in this problem is,
$A=\left[ \begin{matrix} -4 & -1 \\\ 3 & 1 \\\ \end{matrix} \right]$
We know that for a $2\times 2$ matrix say $\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$ , its determinant will be $\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=ad-bc$ . Applying this to the given matrix, we get,
$\begin{aligned} & \Rightarrow \left| A \right|=\left| \begin{matrix} -4 & -1 \\\ 3 & 1 \\\ \end{matrix} \right|=\left( -4 \right)\times 1-\left( -1 \right)\times 3=-4+3 \\\ & \Rightarrow \left| A \right|=-1....\left( i \right) \\\ \end{aligned}$
The matrix that we need to evaluate is,
${{A}^{2016}}-2{{A}^{2015}}-{{A}^{2014}}$
After taking ${{A}^{2014}}$ common from the expression, this matrix can be rewritten as,
$\Rightarrow {{A}^{2014}}\left( {{A}^{2}}-2A-I \right)$
The determinant of the matrix will then be,
$\Rightarrow \left| {{A}^{2014}}\left( {{A}^{2}}-2A-I \right) \right|$
Now, we know the property of determinant which says that $\left| AB \right|=\left| A \right|\left| B \right|$ . Applying this to the above determinant, the above determinant thus becomes,
$\Rightarrow \left| {{A}^{2014}} \right|\left| {{A}^{2}}-2A-I \right|$
Now, we need to find the value of the matrix ${{A}^{2}}-2A-I$ and then find out its determinant. The matrix will be,
$\begin{aligned} & \Rightarrow {{A}^{2}}-2A-I=\left[ \begin{matrix} -4 & -1 \\\ 3 & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} -4 & -1 \\\ 3 & 1 \\\ \end{matrix} \right]-2\left[ \begin{matrix} -4 & -1 \\\ 3 & 1 \\\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right] \\\ & \Rightarrow {{A}^{2}}-2A-I=\left[ \begin{matrix} 16-3 & 4-1 \\\ -12+3 & -3+1 \\\ \end{matrix} \right]-\left[ \begin{matrix} -8 & -2 \\\ 6 & 2 \\\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right] \\\ & \Rightarrow {{A}^{2}}-2A-I=\left[ \begin{matrix} 20 & 5 \\\ -15 & -5 \\\ \end{matrix} \right] \\\ \end{aligned}$
The determinant of this matrix will then be,
$\begin{aligned} & \Rightarrow \left| {{A}^{2}}-2A-I \right|=\left| \begin{matrix} 20 & 5 \\\ -15 & -5 \\\ \end{matrix} \right|=20\left( -5 \right)-5\left( -15 \right) \\\ & \Rightarrow \left| {{A}^{2}}-2A-I \right|=-25....\left( ii \right) \\\ \end{aligned}$
We know the property of the determinant which says that the determinant of a matrix which is raised to a power is equal to the determinant being raised to the same power. This means,
$\Rightarrow \left| {{A}^{2014}} \right|={{\left| A \right|}^{2014}}$
The value of the determinant $\left| {{A}^{2014}} \right|\left| {{A}^{2}}-2A-1 \right|$ thus becomes,
$\Rightarrow \left| {{A}^{2014}} \right|\left| {{A}^{2}}-2A-1 \right|={{\left( -1 \right)}^{2014}}\times \left( -25 \right)=-25$
Thus, we can conclude that the value of the determinant will be $-25$

So, the correct answer is “Option D”.

Note: In order to solve this problem, we must be accustomed with the various properties of matrices and determinants. Forgetting one important thing will make us unable to solve the problem. Also, we should avoid the common mistake of taking the determinant as $\left| {{A}^{2016}}-2{{A}^{2015}}-{{A}^{2014}} \right|=\left| {{A}^{2016}} \right|-2\left| {{A}^{2015}} \right|-\left| {{A}^{2014}} \right|$ .