If $A = \left[ {\matrix{ 2 & 3 \cr 0 & { - 1} \cr } } \right]$, then the value of det(A4) + det(A10 $-$ (Adj(2A))10) is equal to _____________.
Answer (integer)
16
Solution
$A = \left[ {\matrix{
2 & 3 \cr
0 & { - 1} \cr
} } \right]$
<br><br>$|A|\, = - 2 \Rightarrow |A{|^4} = 16$
<br><br>${A^2} = \left[ {\matrix{
4 & 3 \cr
0 & 1 \cr
} } \right]$
<br><br>$${A^3} = \left[ {\matrix{
8 & 9 \cr
0 & { - 1} \cr
} } \right]$$
<br><br>$\therefore$ $${A^{10}} = \left[ {\matrix{
{{2^{10}}} & {{2^{10}} - 1} \cr
0 & 1 \cr
} } \right] = \left[ {\matrix{
{1024} & {1023} \cr
0 & 1 \cr
} } \right]$$<br><br>$2A = \left[ {\matrix{
4 & 6 \cr
0 & { - 2} \cr
} } \right]$<br><br>$$adj(2A) = \left[ {\matrix{
{ - 2} & { - 6} \cr
0 & 4 \cr
} } \right]$$<br><br>$$adj(2A) = - 2\left[ {\matrix{
1 & 3 \cr
0 & { - 2} \cr
} } \right]$$<br><br>$${(adj(2A))^{10}} = {2^{10}}{\left[ {\matrix{
1 & 3 \cr
0 & { - 2} \cr
} } \right]^{10}}$$<br><br>$$ = {2^{10}}\left[ {\matrix{
1 & { - ({2^{10}} - 1)} \cr
0 & {{2^{10}}} \cr
} } \right]$$<br><br>$$ = {2^{10}}\left[ {\matrix{
1 & { - 1023} \cr
0 & {1024} \cr
} } \right]$$<br><br>$${A^{10}} - {(adj(2A))^{10}} = \left[ {\matrix{
0 & {{2^{11}} \times 1023} \cr
0 & {1 - {{(1024)}^2}} \cr
} } \right]$$<br><br>$|{A^{10}} - adj{(2A)^{10}}| = 0$
<br><br>$\therefore$
det(A<sup>4</sup>) + det(A<sup>10</sup> $-$ (Adj(2A))<sup>10</sup>)
<br><br> = 16 + 0 = 16
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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