We know that in a skew-symmetric matrix, diagonal elements are always 0.
So, its trace is 0.

Given, $A=\left[ \begin{matrix} 1 & -5 & 7 \ 0 & 7 & 9 \ 11 & 8 & 9 \end{matrix} \right] $
$tr(A)=sum\ of\ diagonal\ matrix$
$tr(A)=1+7+9=17$

By definition of trace of a scalar matrix of order n, 
$tr(A)=a _{11}+ a _{22}+a _{33}+.....+a _{nn}$

$=k+k+....k $ (upto n times)

$tr(A)=nk$

By definition of trace of a matrix of order n, 
$tr(A)=a _{11}+ a _{22}+a _{33}+.....+a _{nn}$
$\displaystyle =: \sum _{ i=j }  { a } _{ ij } =: \sum _{ i } a _{ ij }$
Hence, option 'B' is correct.

We know that for a skew-symmetric matrix the sum of diagonal elements is zero.
$a _{ij}=0    \forall i=j$
So,$ tr (A)=0$

$A=\begin{bmatrix} 1 & -5 & 7 \ 0 & 7 & 9 \ 11 & 8 & 9 \end{bmatrix}$
$trace(A)=$sum of diagonal elements$=1+7+9=17$

Ans: A

Given $A=[a _{i,j}]$ and  $a _{i,j}=i(i+j)$ 

$A+2B=\begin{bmatrix} 1 & 2 & 0 \ 6 & -3 & 3 \ -5 & 3 & 1 \end{bmatrix}$         .....(i)

$2A-B=\begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix}$         .....(ii)

$\Rightarrow 4A-2B=\begin{bmatrix} 4 & -2 & 10 \ 4 & -2 & 12 \ 0 & 2 & 4 \end{bmatrix}$      ....(iii)

Adding (i) and (iii), we get
$5A=\begin{bmatrix} 5 & 0 & 10 \ 10 & -5 & 15 \ -5 & 5 & 5 \end{bmatrix}$

$\Rightarrow A=\begin{bmatrix} 1 & 0 & 2 \ 2 & -1 & 3 \ -1 & 1 & 1 \end{bmatrix}$
So, $tr(A)=1$

Now, by eq(ii),
$B=\begin{bmatrix} 2 & 0 & 4 \ 4 & -2 & 6 \ -2 & 2 & 2 \end{bmatrix}-\begin{bmatrix} 2 & -1 & 5 \ 2 & -1 & 6 \ 0 & 1 & 2 \end{bmatrix}$

$\Rightarrow B=\begin{bmatrix} 0 & 1 & -1 \ 2 & -1 & 0 \ -2 & 1 & 0 \end{bmatrix}$
So, $tr(B)=-1$
Now, $tr(A)-tr(B)=1+1=2$

Given $A = \begin{bmatrix}2 & 3 & 4\ 5 & -3 & 8\ 9 & 2 & 16\end{bmatrix}$

$tr(A)=2-3+16=15$

Given $tr(A)=4$
$\Rightarrow a _{11}+a _{22}=4$

$a _{ ij }={ { 0,1,2,3,4}  }$

So, diagonal entries of A can be 0 and 4 , 4 and 0, 1 and 3, 3 and 1, 2 and 2,
Hence, 5 matrices are possible

Given,  $A = \begin{bmatrix}1 & -5 & 7\ 0 & 7 & 9\ 11 & 8 & 9\end{bmatrix}$ .

Given $A = [a _{ij}]$ is a scalar matrix of order $n\times n$ such that $a _{ii} = k$ for all $i$.

As $A$ is a skew symmetric matrix
                   $A' = -A$
$\Rightarrow  a _{ii}=0 : \forall : i \Rightarrow trace : (A) =0$
Also             $|A|=|A'|=|-A|=(-1)^3|A|$
$\Rightarrow  2|A|=0\Rightarrow |A|=0$
Hence, option B.

If $A = 0,tr(A)=0$.

$tr(A)=1+\omega^n+\omega^{2n}$
if $n=3k$ i.e Mutiple of $3$.
$\Rightarrow tr(A)=1+\omega^{3k}+\omega^{6k}=1+1+1=3$
if $n\neq 3k$ i.e not a multiple of $3$.
then $n=3k+1$ or $n=3k+2$
$\Rightarrow tr(A)=1+\omega^{3k+1}+\omega^{2(3k+1)}=1+\omega+\omega^2=0$
Hence, options B and C.

${ T } _{ r }(A)={ \alpha  }^{ 2 }+{ \beta  }^{ 2 }+{ \gamma  }^{ 2 }$

Given, $B=diag(1    2    5)$
$\Rightarrow B^{2}=diag(1     4    25)$

Now, $AB^{2}=\begin{bmatrix}1 &0  &0 \ 0 &24  &0 \ 0 &0  &50 \end{bmatrix}$

$\Rightarrow tr(AB^{2})=1+24+50=75$

$A=\left| \begin{matrix} 2 & 1 \ 4 & 1 \end{matrix} \right| ,\quad B=\left| \begin{matrix} 3 & 4 \ 2 & 3 \end{matrix} \right| ,\quad C=\left| \begin{matrix} 3 & -4 \ -2 & 3 \end{matrix} \right| \ AB\quad =\quad \left| \begin{matrix} 2 & 1 \ 4 & 1 \end{matrix} \right| \left| \begin{matrix} 3 & 4 \ 2 & 3 \end{matrix} \right| \ \qquad =\quad \left| \begin{matrix} 6+2 & 8+3 \ 12+2 & 16+3 \end{matrix} \right| \ \quad \quad \quad =\quad \left| \begin{matrix} 8 & 11 \ 14 & 19 \end{matrix} \right| \ ABC\quad =\quad \left| \begin{matrix} 8 & 11 \ 14 & 19 \end{matrix} \right| \left| \begin{matrix} 3 & -4 \ -2 & 3 \end{matrix} \right| \ \quad \quad \quad \quad \quad =\quad \left| \begin{matrix} 24-22 & -32+33 \ 42-38 & -56+57 \end{matrix} \right| \ \qquad \quad \quad =\quad \left| \begin{matrix} 2 & 1 \ 4 & 1 \end{matrix} \right| \ { (BC) }^{ 2 }\quad =\quad \left| \begin{matrix} 3 & 4 \ 2 & 3 \end{matrix} \right| \left| \begin{matrix} 3 & -4 \ -2 & 3 \end{matrix} \right| \ \qquad \quad \quad =\quad { \left| \begin{matrix} 1 & 0 \ 0 & 1 \end{matrix} \right|  }^{ 2 }$

Here, $A=\begin{bmatrix} { \omega  }^{ 2 } & { \omega  }^{ 3 } & ... & { \omega  }^{ 11 } \ { \omega  }^{ 3 } & { \omega  }^{ 4 } & ... & { \omega  }^{ 12 } \ ... & ... & ... & ... \ { \omega  }^{ 11 } & { \omega  }^{ 12 } &  & { \omega  }^{ 20 } \end{bmatrix}$

$tr(A)={ \omega  }^{ 2 }+{ \omega  }^{ 4 }+{ \omega  }^{ 6 }+.....+{ \omega  }^{ 20 }$

$=\displaystyle \frac { { \omega  }^{ 2 }(1-{ \omega  }^{ 20 }) }{ 1-{ \omega  }^{ 2 } } $

 $A=\left[\begin{matrix}2&0&7\0&1&0\1&-2&1\end{matrix}\right]$ and $B=\left[\begin{matrix}-x&14x&7x\0&1&0\x&-4x&-2x\end{matrix}\right]$
$AB=\left[\begin{matrix}2&0&7\0&1&0\1&-2&1\end{matrix}\right]\left[\begin{matrix}-x&14x&7x\0&1&0\x&-4x&-2x\end{matrix}\right]=\begin{bmatrix} 5x & 14x & 0 \ 0 & 1 & 0 \ 0 & 10x-2 & 5x \end{bmatrix}$
but,$AB=(AB)^{-1}\Rightarrow (AB)^2=I$
$\Rightarrow (AB)^2=\begin{bmatrix} 5x & 14x & 0 \ 0 & 1 & 0 \ 0 & 10x-2 & 5x \end{bmatrix}\begin{bmatrix} 5x & 14x & 0 \ 0 & 1 & 0 \ 0 & 10x-2 & 5x \end{bmatrix}=\begin{bmatrix} 25x^{ 2 } & 70x^{ 2 }+14x & 0 \ 0 & 1 & 0 \ 0 & (5x+1)(10x-2) & 25x^{ 2 } \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow x=\displaystyle\frac{-1}{5}$
$Tr.\left(AB+(AB)^2+(AB)^3+...+(AB)^{100}\right)=Tr.\left(AB+I+(AB)+...+I\right)$
$tr(AB)=10x+1=-1$ and $tr(I)=3$
$\therefore Tr.\left(AB+I+(AB)+...+I\right)=(-1+3-1+3......+3)=50(3-1)=100$
Hence, option C.

Given, $A=\left[\begin{matrix}1 & \displaystyle\frac{3}{2}\1 & 2\end{matrix}\right], B = \left[\begin{matrix}4 & -3\-2 & 2\end{matrix}\right] $

So, $AB=\left[ \begin{matrix} 1 & 0 \ 0 & 1 \end{matrix} \right] $
$\Rightarrow AB=I$

Now, $(AB)^{r} C _r =I \left[\begin{matrix}r.3^r & 2^r\0 & (r-1)3^r\end{matrix}\right]$

$\Rightarrow (AB)^{r} C _r=\left[\begin{matrix}r.3^r & 2^r\0 & (r-1)3^r\end{matrix}\right]$

$\sum _{ r=1 }^{ 50 }{ tr.\left( (AB)^{ r }C _{ r } \right)  } =\sum _{ r=1 }^{ 50 }{ tr\left( C _{ r } \right)  } =\sum _{ r=1 }^{ 50 }{ (2r } -1)3^{ r }$

It is an arithmetico-geometric series
$S _{50}=1.3+3.3^{2}+5.3^{3}+......+99.3^{50}$
$3S _{50}=        3^{2}+3.3^{3}+......+97.3^{50}+99.3^{51}$

$\Rightarrow -2S=3+2.3^{2}+2.{3}^{3}+.....+2.3^{50}-99.3^{51}$

$\Rightarrow -2S=3-99.3^{ 51 }+2(3^{ 2 }+{ 3 }^{ 3 }+.....+3^{ 50 })$

$\Rightarrow -2S=3-99.3^{ 51 }+3^{ 51 }-3^{ 2 }$

$\Rightarrow S=3(1+49.3^{ 50 })$

Given $A=\left[\begin{matrix}3x^2\1\6x\end{matrix}\right], B=[a,b,c]$ 

Here, $AB=\left[ \begin{matrix} 3x^{ 2 } \ 1 \ 6x \end{matrix} \right] [a,b,c]$

$\Rightarrow AB=\left[ \begin{matrix} 3ax^{ 2 } & 3bx^{ 2 } & 3cx^{ 2 } \ a & b & c \ 6ax & 6bx & 6cx \end{matrix} \right] $

$tr(AB)=3ax^{2}+b+6cx$

Now, given $C=\left[\begin{matrix}(x+2)^2&5x^2&2x\5x^2&2x&(x+2)^2\2x&(x+2)^2&5x^2\end{matrix}\right]$

$tr(C)=(x+2)^2+2x+5x^{2}$
$\Rightarrow tr(C)=6x^{2}+6x+4$

Since, we have $tr(AB)=tr(C)$
$\Rightarrow 3ax^{2}+b+6cx=6x^{2}+6x+4$

On comparing, we get $a=2,b=4,c=1$
So, $a+b+c=7$

$f(x,y) = x^2 + y^2 - 2xy, \space (x,y \in R)$ and
$\quad A = \begin{bmatrix}f(x _1,y _1) & f(x _1,y _2) & f(x _1,y _3) \ f(x _2,y _1) & f(x _2,y _2) & f(x _2,y _3) \ f(x _3,y _1) & f(x _3,y _2) & f(x _3,y _3) \end{bmatrix}$ such that trace $(A) = 0$
$\Rightarrow A=\begin{bmatrix} (x _{ 1 }-y _{ 1 })^{ 2 } & \quad (x _{ 1 }-y _{ 2 })^{ 2 } & \quad (x _{ 1 }-y _{ 3 })^{ 2 } \ (x _{ 2 }-y _{ 1 })^{ 2 } & \quad (x _{ 2 }-y _{ 2 })^{ 2 } & \quad (x _{ 2 }-y _{ 3 })^{ 2 } \ (x _{ 3 }-y _{ 1 })^{ 2 } & \quad (x _{ 3 }-y _{ 2 })^{ 2 } & \quad (x _{ 3 }-y _{ 3 })^{ 2 } \end{bmatrix}$
Trace$(A)=0$
$\Rightarrow (x _1-y _1)^2+(x _2-y _2)^2+(x _3-y _3)^2=0$
$\Rightarrow x _1=y _1; x _2=y _2; x _3=y _3$
$A=\begin{bmatrix} 0 & \quad (y _{ 1 }-y _{ 2 })^{ 2 } & \quad (y _{ 1 }-y _{ 3 })^{ 2 } \ (y _{ 2 }-y _{ 1 })^{ 2 } & 0 & \quad (y _{ 2 }-y _{ 3 })^{ 2 } \ (y _{ 3 }-y _{ 1 })^{ 2 } & \quad (y _{ 3 }-y _{ 2 })^{ 2 } & 0 \end{bmatrix}$
$|A|=2(y _1-y _2)^2(y _2-y _3)^2(y _3-y _1)^2\geq 0$
$\therefore |A|\geq 0$
Hence, option A.