Let A - begin bmatrix 48111-98 x-2end bmatrix - and - A -100- -beginbmatrix a- 18 a- 3 1a2 - a- 4lendbmatrix - then value of a 1- a 4 is

A = B + I where B = nbsp;[ 3 amp; 1; nbsp; 9 amp; 3 ] B^2 = nbsp;[ 3 amp; 1; nbsp; 9 amp; 3 ][ 3 amp; 1; nbsp; 9 amp; 3 ...

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Byju's Answer

Standard XII

Mathematics

Equality of Matrices

Let A = begin...

Question

Let $A = [\begin{matrix} 4 & 1 - 9 & - 2 \end{matrix}] a n d A^{100} = [\begin{matrix} a_{1} & a_{3} a_{2} & a_{4} \end{matrix}],$ then value of $a_{1} + a_{4}$ is

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Solution

A = B + I
where $B = [\begin{matrix} 3 & 1 - 9 & - 3 \end{matrix}]$
$B^{2} = [\begin{matrix} 3 & 1 - 9 & - 3 \end{matrix}] [\begin{matrix} 3 & 1 - 9 & - 3 \end{matrix}]$
$[\begin{matrix} 0 & 0 0 & 0 \end{matrix}]$
$A^{100} = (B + I)^{100} = I + 100 . B + 0....$
$= [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] + [\begin{matrix} 300 & 100 - 900 & - 300 \end{matrix}]$
$= [\begin{matrix} 301 & 100 - 900 & - 299 \end{matrix}]$
So $a_{1} + a_{4} = 2$

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Let A=[41−9−2] and A100=[a1a3a2a4], then value of a1+a4 is

A = B + I where B=[31−9−3] B2=[31−9−3][31−9−3] [0000] A100=(B+I)100=I+100.B+0.... =[1001]+[300100−900−300] =[301100−900−299] So a1+a4=2

Let $A = [\begin{matrix} 4 & 1 - 9 & - 2 \end{matrix}] a n d A^{100} = [\begin{matrix} a_{1} & a_{3} a_{2} & a_{4} \end{matrix}],$ then value of $a_{1} + a_{4}$ is