Matrices

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Totalnumberofpossiblematricesoforder3*3witheachentry0or2=2^{3*3}=2^9=512.\\ Hence,thecorrectoptionis(d).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}GiventhatA=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\\ Here,numberofcolumnsandthenumberofrowsareequali.e.,3.So,Aisasquarematrix.\\ Hence,thecorrectoptionis(a).\end{array}$

This is a Long Answers Type Questions as classified in NCERT Exemplar

Sol. 

This is a Long Answers Type Questions as classified in NCERT Exemplar

Sol. 

This is a Long Answers Type Questions as classified in NCERT Exemplar

Sol. 

$\begin{array}{l}LetP\left(n\right):{\left(AB\right)}^n=A^n{B}^n\\ Step1:Putn=1,P\left(1\right):AB=ABwhichistrueforn=1\\ Step2:Putn=k,P\left(k\right):{\left(AB\right)}^k=A^k{B}^{k}Letitbetrueforanyk\in N\\ Step3:Putn=k+1,P\left(k+1\right):{\left(AB\right)}^{k+1}=A^{k+1}{B}^{k+1}\\ L.H.S.{\left(AB\right)}^{k+1}={\left(AB\right)}^k.AB\\ =A^k{B}^k.AB\left[Fromstep2\right]\\ =A^{k+1}{A}^{k+1}R.H.S.\\ Hence,ifP\left(n\right)istrueforP\left(k\right)thenitistrueforP\left(k+1\right).\end{array}$

Given, A²= A.

(E) we need to calculate,

(I + A)³- 7A = I³ + A³ + 3IA (I + A) - 7A { (x + y)³ = x³ + y³ + 3xy (a + y)}

Given, A is both symmetric and skew-symmetric.

(E) Then, A' = A ____ (1) and A' = -A ____ (2)

So using (2), A' = -A.

A = -A {eqⁿ (I)}

A + A = 0

2A = 0

A = 0.

A is a zero matrix

So, option B is correct.

Kindly go through the solution