\[\begin{array}{l} g\left( x \right) = \left| {\begin{array}{*{20}{c}} {{a^{ - x}}}&{{e^{x{{\log }_e}a}}}&{{x^2}}\\ {{a^{ - 3x}}}&{{e^{3x{{\log }_e}a}}}&{{x^4}}\\ {{a^{ - 5x}}}&{{e^{5x{{\log }_e}a}}}&1 \end{array}} \right|\\ = \left| {\begin{array}{*{20}{c}} {{a^{ - x}}}&{{e^x}}&{{x^2}}\\ {{a^{ - 3x}}}&{{e^{3x}}}&{{x^4}}\\ {{a^{ - 5x}}}&{{e^{5x}}}&1 \end{array}} \right|\left( {{e^{\log {a^x}}} = {a^x}} \right)\\ \Rightarrow g\left( { - x} \right) = \left| {\begin{array}{*{20}{c}} {{a^x}}&{{a^{ - x}}}&{{x^2}}\\ {{a^{3x}}}&{{a^{ - 3x}}}&{{x^4}}\\ {{a^{5x}}}&{{a^{ - 5x}}}&1 \end{array}} \right|\\ = - \left| {\begin{array}{*{20}{c}} {{a^{ - x}}}&{{a^x}}&{{x^4}}\\ {{a^{ - 3x}}}&{{a^{3x}}}&{{x^4}}\\ {{a^{ - 5x}}}&{{a^{5x}}}&1 \end{array}} \right|\left( \begin{array}{l} {\rm{Interchanging }}\,\,I\\ {\rm{and }}\,\,II\,\,{\rm{ columns}} \end{array} \right) \end{array}\]
$$\eqalign{ & = - g\left( x \right) \cr & \Rightarrow g\left( x \right) + g\left( { - x} \right) = 0 \cr} $$

$$\eqalign{ & \therefore \,\,A' = A,\,\,B' = B \cr & {\text{Now }}\left( {A\left( {BA} \right)} \right)' = \left( {BA} \right)'A' \cr & = \left( {A'B'} \right)A' = \left( {AB} \right)A = A\left( {BA} \right) \cr & {\text{Similarly}}\,\,\left( {\left( {AB} \right)A} \right)' = \left( {AB} \right)A \cr} $$
So, $$A(BA)$$  and $$(AB)A$$  are symmetric matrices.
Again $$(AB)' = B'A’ = BA$$
Now if $$BA = AB,$$   then $$AB$$  is symmetric matrix.

\[f\left( {x + y} \right) = \left[ {\begin{array}{*{20}{c}} {\cos \left( {x + y} \right)}&{ - \sin \left( {x + y} \right)}&0 \\ {\sin \left( {x + y} \right)}&{\cos \left( {x + y} \right)}&0 \\ 0&0&1 \end{array}} \right]\]
\[f\left( {x + y} \right) = \left[ {\begin{array}{*{20}{c}} {\cos x}&{ - \sin x}&0 \\ {\sin x}&{\cos x}&0 \\ 0&0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\cos y}&{ - \sin y}&0 \\ {\sin y}&{\cos y}&0 \\ 0&0&1 \end{array}} \right] = f\left( x \right) \cdot f\left( y \right).\]

Since, $${A_1}{B_1}{C_1},{A_2}{B_2}{C_2}$$    and $${A_3}{B_3}{C_3}$$  are divisible by $$k,$$ therefore; $$100{A_1} + 10{B_1} + {C_1} = {n_1}k;100{A_2} + 10{B_2} + {C_2} = {n_2}k;100{A_3} + 10{B_3} + {C_3} = {n_3}k$$
(where $${n_1} , {n_2} , {n_3}$$  are integers)
Now, \[\Delta = \left| {\begin{array}{*{20}{c}} {{A_1}}&{{B_1}}&{{C_1}}\\ {{A_2}}&{{B_2}}&{{C_2}}\\ {{A_3}}&{{B_3}}&{{C_3}} \end{array}} \right|\]
\[\begin{array}{l} = \left| {\begin{array}{*{20}{c}} {{A_1}}&{{B_1}}&{100{A_1} + 10{B_1} + {C_1}}\\ {{A_2}}&{{B_2}}&{100{A_2} + 10{B_2} + {C_2}}\\ {{A_3}}&{{B_3}}&{100{A_3} + 10{B_3} + {C_3}} \end{array}} \right|\left[ {{\rm{Applying, }}{C_3} \to {C_3} + 10{C_2} + 100{C_1}} \right]\\ = \left| {\begin{array}{*{20}{c}} {{A_1}}&{{B_1}}&{{n_1}k}\\ {{A_2}}&{{B_2}}&{{n_2}k}\\ {{A_3}}&{{B_3}}&{{n_3}k} \end{array}} \right| = k\left| {\begin{array}{*{20}{c}} {{A_1}}&{{B_1}}&{{n_1}}\\ {{A_2}}&{{B_2}}&{{n_2}}\\ {{A_3}}&{{B_3}}&{{n_3}} \end{array}} \right| = k{\Delta _1} \end{array}\]
⇒ $$\Delta$$ is divisible by $$k$$
[Since, elements of $$\Delta_1$$ are integers ∴ $$\Delta_1$$ is an integer.]

We can write $$\Delta = {\Delta _1} + {y_1}{\Delta _2},{\text{ where}}$$
\[\begin{array}{l} {\Delta _1} = \left| {\begin{array}{*{20}{c}} 1&{1 + {x_1}{y_2}}&{1 + {x_1}{y_3}}\\ 1&{1 + {x_2}{y_2}}&{1 + {x_2}{y_3}}\\ 1&{1 + {x_3}{y_2}}&{1 + {x_3}{y_3}} \end{array}} \right|{\rm{ and}}\\ {\Delta _2} = \left| {\begin{array}{*{20}{c}} {{x_1}}&{1 + {x_1}{y_2}}&{1 + {x_1}{y_3}}\\ {{x_2}}&{1 + {x_2}{y_2}}&{1 + {x_2}{y_3}}\\ {{x_3}}&{1 + {x_3}{y_2}}&{1 + {x_3}{y_3}} \end{array}} \right| \end{array}\]
$$\eqalign{ & {\text{In }}{\Delta _1},{\text{use }}{C_2} \to {C_2} - {C_1}{\text{ and }}{C_3} \to {C_3} - {C_1} \cr & {\text{so that,}} \cr} $$
\[{\Delta _1} = \left| {\begin{array}{*{20}{c}} 1&{{x_1}{y_2}}&{{x_1}{y_3}}\\ 1&{{x_2}{y_2}}&{{x_2}{y_3}}\\ 1&{{x_3}{y_2}}&{{x_3}{y_3}} \end{array}} \right| = 0\]
$$\eqalign{ & \left[ {\because {C_2}{\text{ and }}{C_3}{\text{ are proportional}}} \right] \cr & {\text{In }}{\Delta _2},{\text{us }}{C_2} \to {C_2} - {y_2}{C_1} \cr & {\text{and }}{C_3} \to {C_3} - {y_3}{C_1}{\text{ to get}} \cr} $$
\[{\Delta _2} = \left| {\begin{array}{*{20}{c}} {{x_1}}&1&1 \\ {{x_2}}&1&1 \\ {{x_3}}&1&1 \end{array}} \right| = 0\left[ {\because {C_2}{\text{ and }}{C_3}{\text{ are identical}}} \right]\]
$$\therefore \Delta = 0$$

Given \[\left| {\begin{array}{*{20}{c}} p&b&c\\ a&q&c\\ a&b&r \end{array}} \right| = 0\]
$${R_1} \to {R_1} - {R_2},{R_2} \to {R_2} - {R_3}$$      reduces the determinant to
\[\left| {\begin{array}{*{20}{c}} {p - a}&{b - q}&0\\ 0&{q - b}&{c - r}\\ a&b&r \end{array}} \right| = 0\]
$$ \Rightarrow \,\left( {p - a} \right)\left( {q - b} \right)r + a\left( {b - q} \right)\left( {c - r} \right) - b\left( {p - a} \right)\left( {c - r} \right) = 0$$
$$ \Rightarrow $$ Dividing throughout by $$\left( {p - a} \right)\left( {q - b} \right)\left( {r - c} \right),$$     we get
$$\eqalign{ & \Rightarrow \,\frac{r}{{r - c}} + \frac{a}{{p - a}} + \frac{b}{{q - b}} = 0 \cr & \Rightarrow \,\frac{r}{{r - c}} + \frac{a}{{p - a}} + \frac{b}{{q - b}} = 2 \cr} $$

\[\begin{array}{l} \frac{{dy}}{{dx}} = \left| {\begin{array}{*{20}{c}} {\cos x}&{ - \sin x}&{\cos x}\\ {\cos x}&{ - \sin x}&{\cos x}\\ x&1&1 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {\sin x}&{\cos x}&{\sin x}\\ { - \sin x}&{ - \cos x}&{ - \sin x}\\ x&1&1 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {\sin x}&{\cos x}&{\sin x}\\ {\cos x}&{ - \sin x}&{\cos x}\\ 1&0&0 \end{array}} \right|\\ = 0 - \left| {\begin{array}{*{20}{c}} {\sin x}&{\cos x}&{\sin x}\\ {\sin x}&{\cos x}&{\sin x}\\ x&1&1 \end{array}} \right| + 1\left| {\begin{array}{*{20}{c}} {\cos x}&{\sin x}\\ { - \sin x}&{\cos x} \end{array}} \right| \end{array}\]
$$ = 0 + \left( {{{\cos }^2}x + {{\sin }^2}x} \right)$$

Given that, \[A = \left[ {\begin{array}{*{20}{c}} 1&2\\ 0&3 \end{array}} \right]\]
\[{A^2} = \left[ {\begin{array}{*{20}{c}} 1&2\\ 0&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&2\\ 0&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{2 + 6}\\ 0&9 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&8\\ 0&9 \end{array}} \right]\]
Since, $$f\left( x \right) = {x^2} - x + 2$$
Putting $$A$$ in place of $$x$$
$$f\left( A \right) = {A^2} - A + 2I$$
\[\begin{array}{l} = \left[ {\begin{array}{*{20}{c}} 1&8\\ 0&9 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 1&2\\ 0&3 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 2&0\\ 0&2 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {1 - 1 + 2}&{8 - 2 + 0}\\ {0 - 0 + 0}&{9 - 3 + 2} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 2&6\\ 0&8 \end{array}} \right] \end{array}\]

$${c_{ij}} = \sum\limits_{k = 1}^n {{a_{ik}}{b_{kj}}} \,\,\,\left( {{\text{In general}}} \right)$$
and in a diagonal matrix non-diagonal elements are zero
i.e., \[{a_{ij}} = \left\{ \begin{gathered} 0,\,\,{\text{if }}i \ne j \hfill \\ {a_{ii}}{\text{, if }}i = j \hfill \\ \end{gathered} \right.\]
So, $${c_{ij}} = {a_{ii}}{b_{ij}}$$

Let, \[A = \left[ {\begin{array}{*{20}{c}} 0&{2b}&c\\ a&b&{ - c}\\ a&{ - b}&c \end{array}} \right]\]
Now, \[{A^T} = \left[ {\begin{array}{*{20}{c}} 0&a&a\\ {2b}&b&{ - b}\\ c&{ - c}&c \end{array}} \right]\]
∵ $$A$$ is orthogonal
∴ $$A{A^T} = I$$
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}} 0&{2b}&c\\ a&b&{ - c}\\ a&{ - b}&c \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&a&a\\ {2b}&b&{ - b}\\ c&{ - c}&c \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\]
Equating the corresponding elements, we get,
$$4{b^2} + {c^2} = 1\,\,\,\,.....\left( 1 \right)$$
$$\eqalign{ & 2{b^2} - {c^2} = 0\,\,\,\,.....\left( 2 \right) \cr & {a^2} + {b^2} + {c^2} = 1\,\,\,\,.....\left( 3 \right) \cr} $$
On solving (1), (2) and (3), we get,
$$a = \pm \frac{1}{{\sqrt 2 }};b = \pm \frac{1}{{\sqrt 6 }};c = \pm \frac{1}{{\sqrt 3 }}$$