Inverse of a Matrix

Given $A=\left[\begin{array}{lll}1& 1& 1\\ 2& 4& 1\\ 2& 3& 1\end{array}\right], B=\left[\begin{array}{ll}2& 3\\ 3& 4\end{array}\right].$ Find P such that $BPA=\left[\begin{array}{lll}1& 0& 1\\ 0& 1& 0\end{array}\right]$.

Let $A^{-1}=\left[\begin{array}{ccc}1& -1& 2\\ -1& 2& 1\\ 2& 1& -1\end{array}\right]$, then -

Matrix $\left[\begin{array}{ccc}a& b& (a\alpha -b)\\ b& c& (b\alpha -c)\\ 2& 1& 0\end{array}\right]$ is non invertible if

If $A$ and $B$ are invertible matrices, which one of the following statement is/are correct -

Let $A$ be an invertible matrix then which of the following is/are true :

If $A=\left[\begin{array}{lll}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]$, then-

If $A=\left[\begin{array}{lll}0& 1& 2\\ 1& 2& 3\\ 3& a& 1\end{array}\right], A^{-1}=\left[\begin{array}{ccc}1/2& -1/2& 1/2\\ -4& 3& c\\ 5/2& -3/2& 1/2\end{array}\right]$, then -

If A is a invertible idempotent matrix of order n, then adj A is equal to

If ${\left[\begin{array}{ll}1& 1\\ 0& 1\end{array}\right]}^{-1}{\left[\begin{array}{ll}1& 2\\ 0& 1\end{array}\right]}^{-1}{\left[\begin{array}{ll}1& 3\\ 0& 1\end{array}\right]}^{-1}\dots \dots ··{\left[\begin{array}{cc}1& 100\\ 0& 1\end{array}\right]}^{-1}=\left[\begin{array}{ll}1& \mathrm{a}\\ \mathrm{b}& 1\end{array}\right]$, then $a+b=$

If $A=\left[\begin{array}{cc}2& 3\\ 5& -2\end{array}\right]$, then $19A^{-1}$ is equal to -

If $P$ is a two-rowed matrix satisfying $P^T=P^{-1}$ , then $P$ can be -

If $F\left(x\right)=\left[\begin{array}{l}\cos x -\sin x 0\\ \sin x \cos x 0\\ 0 0 1\end{array}\right]$ then show that $F\left(x\right).F\left(y\right)=F\left(x+y\right)$. Hence prove that ${\left[F\left(x\right)\right]}^{-1}=F\left(-x\right)$.

Show that $\left[\begin{array}{cc}1& -\tan \frac{\mathrm{\theta }}{2}\\ \tan \frac{\mathrm{\theta }}{2}& 1\end{array}\right]{\left[\begin{array}{cc}1& \tan \frac{\mathrm{\theta }}{2}\\ -\tan \frac{\mathrm{\theta }}{2}& 1\end{array}\right]}^{-1}=\left[\begin{array}{cc}\mathrm{cos\theta }& -\mathrm{sin\theta }\\ \mathrm{sin\theta }& \mathrm{cos\theta }\end{array}\right]$

If $\left[\begin{array}{ll}2& 1\\ 7& 4\end{array}\right]A\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]$, then matrix $A$ equals

Let $A=\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right]$ and $10B=\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& \alpha \\ 1& -2& 3\end{array}\right]$ .If $B$ is the inverse of $A$, then $\alpha$ is