Vinod Singh and Shweta Pawar Solutions for Chapter: Matrices, Exercise 3: Competitive Thinking

If $\left[\begin{array}{rrr}1& 1& 1\\ 1& -2& -2\\ 1& 3& 1\end{array}\right]\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}0\\ 3\\ 4\end{array}\right],$ then $\left[\begin{array}{l}x\\ y\\ z\end{array}\right]$ is equal to

If $\left[\begin{array}{lll}1& 3& 3\\ 1& 4& 4\\ 1& 3& 4\end{array}\right]\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}12\\ 15\\ 13\end{array}\right],$ then the values of $x,y,z$ respectively are 

Let $M$ be a $3\times 3$ matrix satisfying $M\left[\begin{array}{l}0\\ 1\\ 0\end{array}\right]=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right], M\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$ and $M \left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 12\end{array}\right]$. Then the sum of the diagonal entries of $M$ is

For a matrix $A=\left[\begin{array}{lll}1& 0& 0\\ 2& 1& 0\\ 3& 2& 1\end{array}\right]$, if $U_1, U_2$ and $U_3$ are $3\times 1$ column matrices satisfying $A{U}_1=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right], A{U}_2=\left[\begin{array}{l}2\\ 3\\ 0\end{array}\right], A{U}_3=\left[\begin{array}{l}2\\ 3\\ 1\end{array}\right]$ and $U$ is $3\times 3$ matrix whose columns are $U_1, U_2$ and $U_3$. Then, the sum of the elements of $U^{-1}$ is

If $A=\left[\begin{array}{rr}1& \tan \frac{\theta }{2} \\ -\tan \frac{\theta }{2}& 1\end{array}\right]$ and $AB=I,$ then $B=$

Given $A=\left[\begin{array}{cc}0& -\tan \alpha \\ \tan \alpha & 0\end{array}\right]$ and $B\left(\alpha \right)=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right],$ then $(I+A)(I-A{)}^{-1}$ is equal to

If $A$ is an $3\times 3$ non-singular matrix such that $A{A}^{\text{'}}=A^{\text{'}}A$ and $B=A^{-1}{A}^{\text{'}},$ then $B{B}^{\text{'}}$ equals

Let $M$ and $N$ be two $3\times 3$ skew-symmetric matrices such that $MN=NM.$ If $P^T$ denotes the transpose of $P$, then  $M^2{N}^2{\left(M^{T}N\right)}^{-1}{\left(M{N}^{-1}\right)}^T$is equal to