Matrices

|A| = 2

$\left|\left|A\right|adj\left(5adj{A}^3\right)\right|$

= $\left|A{P}^3\right|\left|adj\left(5adj\left(A^3\right)\right)\right|$

$\begin{array}{l}={\left|A\right|}^{15}.5^6\\ =2^{15}*5^6=2^9*10^6\end{array}$

System of equation is

$\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill \left|\lambda \right|\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left[\begin{array}{l}-2\\ 4\\ 4\lambda -4\end{array}\right]$

R₁ – 2 R₂, R₃ – R₂

$\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -2\hfill & \hfill \left|\lambda \right|-1\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}-10\\ 4\\ 4\lambda -8\end{array}\right)$

System of equation will have no solution for $\lambda$ = 7.

 $A_{3*3},B_{3*3}\&AB=0$

$\ne 0\ne 0\Rightarrow \left|A\right|=0\&\left|B\right|=0$ only .

Sum of all entries of matrix A must be prime p such that 2 < p < 8 then sum of entries may be 3, 5 or 7

If sum is 3 then possible entries are

(0, 5), (0, 1, 4), (0, 2, 3), (0, 1, 3)

(0, 1, 2, 2) and (1, 2)

$\therefore$ Total number of matrices = 4 + 12 + 12 + 12 + 12 + 4 = 56

If sum is 7 then possible entries are

(0, 2, 25), (0, 03, 4), (0, 1, 5), (0, 3, 1), (0, 2, 3), (1, 4), (1, 2, 2), (1, 2, 3) and (0, 1, 2, 4)

Total number of matrices with sum 7 = 104

$\therefore$ total number of required matrices

= 20 + 56 = 104

= 108

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l}True.\\ IfAandBareinvertiblematricesofthesameorder.\\ \therefore {\left(AB\right)}^{-1}={\left(BA\right)}^{-1}\left[?AB=BA\right]\\ But{\left(AB\right)}^{-1}=A^{-1}{B}^{-1}\left[Given\right]\\ \therefore {\left(BA\right)}^{-1}=B^{-1}{A}^{-1}\\ A^{-1}{B}^{-1}=B^{-1}{A}^{-1}\\ \therefore AandBsatisfycommutativepropertyw.r.tmultiplication.\end{array}$

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l}True.\\ {\left(A^2\right)}^{\text{'}}={\left(A^{\text{'}}\right)}^2\\ ={\left[-A\right]}^2\left[?A^{\text{'}}=-A\right]\\ =A^2\\ So,A^{2}isasymmetricmatrix.\end{array}$

This is a True or False Type Questions as classified in NCERT Exemplar

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l}True.\\ LetP=A{A}^{\text{'}}\\ P^{\text{'}}={\left(A{A}^{\text{'}}\right)}^{\text{'}}\\ ={\left(A^{\text{'}}\right)}^{\text{'}}.A^{\text{'}}\left[?{\left(AB\right)}^{\text{'}}=B^{\text{'}}{A}^{\text{'}}\right]\\ =A{A}^{\text{'}}=P\\ So,Pisasymmetricmatrix.\\ Hence,A{A}^{\text{'}}isalwaysasymmetricmatrix.\end{array}$

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l}False.\\ A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right],B=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right]andC=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]\\ \therefore AB=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\\ \therefore AC=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\\ HereAB=AC=0butB\ne C.\end{array}$

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l}True.\\ LetA={\left[a_{ij}\right]}_{m*n}andB={\left[b_{ij}\right]}_{p*q}\\ ABisdefinedwhenn=P\\ \therefore OrderofAB=m*q\\ \Rightarrow Orderof{\left(AB\right)}^{\text{'}}=q*m\\ Orderof{B}^{\text{'}}isq*pandorderof{A}^{\text{'}}isn*m\\ \therefore B^{\text{'}}{A}^{\text{'}}isdefinedwhenP=n\\ andtheorderof{B}^{\text{'}}{A}^{\text{'}}isq*m\\ Hence,Orderof{\left(AB\right)}^{\text{'}}=Orderof{B}^{\text{'}}{A}^{\text{'}}i.e.,q*m.\end{array}$