If A=−4 −13 1 then the determinant line matrix A2016−2A2015−A2014, is

A
2014
B
2016
C
-175
D
-25

Solution:

Let $B = A^{2016} - 2 A^{2015} - A^{2014}$ Then ,

$\begin{array}{l} B = A^{2014} (A^{2} - 2 A - I) \\ \Rightarrow | B | = | A |^{2014} |A^{2} - 2 A - I| \end{array}$

Now, $A - [\begin{array}{rr} - 4 - 1 \\ 3 1 \end{array}]$

$\begin{array}{l} ∴ A^{2} - 2 A - I = [\begin{array}{rr} - 4 & - 1 \\ 3 & 1 \end{array}] [\begin{array}{rr} - 4 & - 1 \\ 3 & 1 \end{array}] - 2 [\begin{array}{rr} - 4 & - 1 \\ 3 & 1 \end{array}] - [\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}] \\ = [\begin{array}{rr} 13 & 3 \\ - 9 & - 2 \end{array}] + [\begin{array}{rr} 8 & 2 \\ - 6 & - 2 \end{array}] + [\begin{array}{rr} - 1 & 0 \\ 0 & - 1 \end{array}] = [\begin{array}{rr} 20 & 5 \\ - 15 & - 5 \end{array}] \\ \Rightarrow |A^{2} - 2 A - I| = |\begin{array}{rr} 20 & 5 \\ - 15 & - 5 \end{array}| = - 25 \end{array}$

and $| A | = |\begin{array}{rr} - 4 - 1 \\ 3 1 \end{array}| = - 1$

Hence, $| B | = (- 1)^{2014} = - 25$

If $A = [\begin{array}{rr} - 4 - 1 \\ 3 1 \end{array}]$ then the determinant line
matrix $(A^{2016} - 2 A^{2015} - A^{2014})$ , is

Solution:

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