First slide

Operations of matrix

Question

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

Moderate

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

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App