Let A = ( 1 0 0 0 4 - 1 0 12 - 3 ) . Then the sum of the diagonal elements of the matrix ( A + I ) 11 is equal to: [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let $A = (\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{matrix}) .$ Then the sum of the diagonal elements of the matrix ${(A + I)}^{11}$ is equal to:

1 2050

2 4094

3 6144

4 4097

Ans.
(4)

We have, $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{matrix}] \begin{matrix} \end{matrix}$

$A^{2} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & - 1 \\ 0 & 12 & - 3 \end{matrix}] = A$

$Similarly, A^{n} = A; n \in N$

$Now, {(A + I)}^{11} = {}^{11}C_{0} A^{11} + {}^{11}C_{1} A^{10} I^{1} + {}^{11}C_{2} A^{9} I^{2} + \dots + {}^{11}C_{11} A^{0} I^{11}$

$= A ({}^{11}C_{0} + {}^{11}C_{1} + \dots + {}^{11}C_{10}) + I [∵ A^{n} = A and I^{n} = I]$

$= A (2^{11} - 1) + I$

$Trace of {(A + I)}^{11} = 1 (2^{11} - 1) + 1 + 4 (2^{11} - 1) + 1 + (- 3) (2^{11} - 1) + 1$

$= 2^{11} + 4 (2^{11}) - 3 - 3 (2^{11}) + 4 = 2 \times 2^{11} + 1$

$= 2^{12} + 1 = 4096 + 1 = 4097$

Want Us to Email you About Special Offers & Updates?