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Question

Answers

A) 1

B) 3

C) 24

D) 50

Answer
Verified

Combination formula:- $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$

${\left( I \right)^n} = \left( I \right)$

$\left( A \right)\left( I \right) = \left( A \right)$

${(x + y)^n} = \sum\limits_{k = 0}^n {^n{C_k}.{x^k}{y^{n - k}} = {[^n}{C_0}{x^n}{y^0}{ + ^n}{C_1}{x^{n - 1}}{y^1}{ + ^n}{C_2}{x^{n - 2}}{y^2}......{ + ^n}{C_n}{x^0}{y^n}} $

B is a $3 \times 3$ matrix such that ${B^2} = 0$ so if we multiply B matrix on both sides then we will get higher power B values.

$ \Rightarrow \left( B \right){\left( B \right)^2} = \left( B \right)0$

$ \Rightarrow {\left( B \right)^3} = 0$

Similarly if we multiply $\left( B \right)$ again then we will get

$ \Rightarrow \left( B \right){\left( B \right)^3} = \left( B \right)0$

$ \Rightarrow {\left( B \right)^4} = 0$

So we can conclude that ${\left( B \right)^n} = 0$ for $n \ge 2$

Now we will apply binomial expansion inside the given matrix $\det [{(I + B)^{50}} - 50B]$

$ \Rightarrow {(I + B)^{50}}{ = ^{50}}{C_0}.{I^{50}}{ + ^{50}}{C_1}.{I^{49}}.B{ + ^{50}}{C_2}.{I^{48}}.{B^2}.......{ + ^{50}}{C_{50}}.{B^{50}}$

As we have proved above, ${\left( B \right)^n} = 0$ for $n \ge 2$ so all matrices B with having power higher than equal to 2 will become zero.

$ \Rightarrow {(I + B)^{50}}{ = ^{50}}{C_0}.{I^{50}}{ + ^{50}}{C_1}.{I^{49}}.B$

Now we know that identity matrix multiplied n times gives an identity matrix so we will use this property.

$ \Rightarrow {(I + B)^{50}}{ = ^{50}}{C_0}.I{ + ^{50}}{C_1}.I.B$

Now we will apply combination formula $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} = \dfrac{{50!}}{{0!(50 - 0)!}} = 1$

$^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} = \dfrac{{50!}}{{1!(50 - 1)!}} = \dfrac{{50 \times 49!}}{{49!}} = 50$

$ \Rightarrow {(I + B)^{50}} = I + 50I.B$

Using identity $\left( A \right)\left( I \right) = \left( A \right)$we will get,

$ \Rightarrow {(I + B)^{50}} = I + 50B$

Now we will put this value in the main equation given in question.

$ \Rightarrow \det [I + 50B - 50B]$

Now we will cancel equal and opposite in sign terms.

$ \Rightarrow \det [I]$ And determinant of identity matrix is 1

$\therefore \det [{(I + B)^{50}} - 50B] = 1$

Determinant will be |ad-cb|