Matrix A is a matrix of order 2.

Identity matrix of second order is $\left[\begin{array}{cc}1 & 0\\ 0 & 1 \end{array}\right]$

For A to be an identity matrix,

$$\begin{gathered} \left[ \begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] = \left[ \begin{array}{cc}\mathrm{cos\alpha }& - \mathrm{sin\alpha }\\ \mathrm{sin\alpha }& \mathrm{cos\alpha }\end{array} \right] \\ \Rightarrow \mathrm{cos\alpha } = 1 \mathrm{and} \mathrm{sin\alpha } = 0 \\ \Rightarrow \mathrm{cos\alpha } = \cos 0° \mathrm{and} \mathrm{sin\alpha } = \sin 0° \\ \Rightarrow \mathrm{\alpha } = 0° \\ \mathrm{Thus}, \mathrm{for} \mathrm{\alpha } = 0°, \mathrm{A} \mathrm{is} \mathrm{an} \mathrm{identity} \mathrm{matrix}. \end{gathered}$$

$$\begin{gathered} \mathrm{Let} {\cos }^{-1} \left(-\frac{\sqrt{3}}{2}\right) = \mathrm{x} \\ \Rightarrow \cos \mathrm{x} = -\frac{\sqrt{3}}{2} \\ \Rightarrow \cos \mathrm{x} = - \cos \left( \frac{\mathrm{\pi }}{6} \right) \\ \Rightarrow \cos \mathrm{x} = \cos \left( \mathrm{\pi } - \frac{\mathrm{\pi }}{6} \right) \\ \Rightarrow \cos \mathrm{x} = \cos \left( \frac{5\mathrm{\pi }}{6} \right) \\ \Rightarrow \mathrm{x} = \frac{5\mathrm{\pi }}{6} \end{gathered}$$

Therefore, the principal value of  ${\cos }^{-1} \left( - \frac{\sqrt{3}}{2} \right) \mathrm{is} \frac{5\mathrm{\pi }}{2}$

f ( x ) is not defined at x = 1.

$$\begin{gathered} \mathrm{for} \mathrm{x} \ge 1, \mathrm{f} ( \mathrm{x} ) = \frac{\left| \mathrm{x} - 1 \right|}{\left( \mathrm{x} - 1 \right)} = \frac{\left( \mathrm{x} - 1 \right)}{\left( \mathrm{x} - 1 \right)} = 1 \\ \mathrm{for} \mathrm{x} \le 1, \mathrm{f} ( \mathrm{x} ) = \frac{\left| \mathrm{x} - 1 \right|}{\left( \mathrm{x} - 1 \right)} = \frac{- \left( \mathrm{x} - 1 \right)}{\left( \mathrm{x} - 1 \right)} = \frac{\left( 1 - \mathrm{x} \right)}{\left( \mathrm{x} - 1 \right)} = -1 \end{gathered}$$

Thus, range of the function is either -1 or 1 at all the points and is undefined  at x = 1

Given determinant = $\left| \begin{array}{ccc}2& -3 & 5\\ 6& 0& 4\\ 1& 5& -7\end{array} \right|$

Minor of the element  a₂₃  is  M₂₃

Obtained by deleting III column and II row

$$\begin{gathered} {\mathrm{M}}_{23} = \left| \begin{array}{cc}2& -3\\ 1& 5\end{array} \right| \\ = 10 - ( -3 ) \\ = 13 \end{gathered}$$

$\mathrm{Given} \left[ \begin{array}{cc}1 & 2\\ 3 & 4\end{array} \right] = \left[ \begin{array}{cc}3 & 1\\ 2 & 5\end{array} \right] = \left[ \begin{array}{cc}7& 11\\ \mathrm{k} & 23 \end{array} \right]$

Now using matrix multiplication in L.H.S., we get

$$\begin{gathered} \left[ \begin{array}{cc}1 \mathrm{x} 3 + 2 \mathrm{x} 2& 1 \mathrm{x} 1 + 2 \mathrm{x} 5\\ 3 \mathrm{x} 3 + 4 \mathrm{x} 2& 3 \mathrm{x} 1 + 4 \mathrm{x} 5\end{array} \right] = \left[ \begin{array}{cc}7& 11\\ \mathrm{k} & 23\end{array} \right] \\ \Rightarrow \left[ \begin{array}{cc}3 + 4& 1 + 10\\ 9 + 8& 3 + 20\end{array} \right] = \left[ \begin{array}{cc}7& 11\\ \mathrm{k} & 23\end{array} \right] \\ \Rightarrow \left[ \begin{array}{cc}7& 11\\ 17 & 23\end{array} \right] = \left[ \begin{array}{cc}7& 11\\ \mathrm{k} & 23\end{array} \right] \end{gathered}$$

Now on equating the corresponding elements we get the value of k = 17