Matrix

$\Delta =\left|\begin{array}{ccc}\hfill -k\hfill & \hfill 3\hfill & \hfill -14\hfill \\ \hfill -15\hfill & \hfill 4\hfill & \hfill -k\hfill \\ \hfill -4\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right|$

$=\left(-k\right)\left(12-k\right)+3\left(4k+45\right)-14\left(-15+16\right)$

$\Delta =0\Rightarrow k=\pm 11$             

 For k = -11,

->11x + 3y – 14z = 25

-4x + y + 3z = 4

$\left\{\begin{array}{l}-11x+3y-14z=25\\ -15x+4y-11z=3\\ -4x+y+3z=4\end{array}\right\}\to$ No solution for k = $\pm$ 11

Let A = $\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill d\hfill & \hfill e\hfill & \hfill f\hfill \\ \hfill 9\hfill & \hfill h\hfill & \hfill i\hfill \end{array}\right]$  

Now A^TA

trace will be    $a^2+b^2+c^2+d^2+e^2+f^2+9^2+h^2=6$

total ways = 

Kindly go through the solution

Use characteristic equation = 0

$\left|adj\left(adj\left(A\right)\right)\right|={\left|A\right|}^{2^2}={\left|A\right|}^4$

$\therefore {\left|A\right|}^4=\left|\begin{array}{ccc}\hfill 14\hfill & \hfill 28\hfill & \hfill -14\hfill \\ \hfill -14\hfill & \hfill 14\hfill & \hfill 28\hfill \\ \hfill 25\hfill & \hfill -14\hfill & \hfill 14\hfill \end{array}\right|$

$={\left(14\right)}^3\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right|$

$={\left(14\right)}^3\left(3-2\left(-5\right)-1\left(-1\right)\right)$

${\left|A\right|}^4={\left(14\right)}^4\Rightarrow \left|A\right|=14$

x + 2y + z = 2

$\alpha x+3y-z=\alpha$

$-\alpha x+y+2z=-\alpha$            

$\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill \alpha \hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill -\alpha \hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=1\left(6+1\right)-2\left(2\alpha -\alpha \right)+1\left(\alpha +3\alpha \right)=7+2a$            

$\alpha =-\frac{7}{2}$                

$A=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]$

$EA=\left[\begin{array}{cc}\hfill a\hfill & \hfill c\hfill \\ \hfill b\hfill & \hfill d\hfill \end{array}\right]\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill -a+c\hfill & \hfill 2a-c\hfill \\ \hfill -b+d\hfill & \hfill 2b-d\hfill \end{array}\right]$

For a = c For $\begin{array}{cc}\hfill -a+c=0\hfill & \hfill 2a-c=1\hfill \end{array}]\to a=1,c=1E=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$  

d = b + 1, d = 1, b = 0

$\begin{array}{cc}\hfill b+d=1\hfill & \hfill 2b-d=-1\hfill \end{array}]\to b=0,d=1R_1\to R_1\to R_2\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

For  $\begin{array}{cc}\hfill -a+c=1\hfill & \hfill 2a-c=1\hfill \end{array}]\to a=0,c=1$

$For\begin{array}{cc}\hfill -a+c=-1\hfill & \hfill 2a-c=2\hfill \end{array}]\to a=1,c=0$

$\begin{array}{cc}\hfill -b+d=-2\hfill & \hfill 2b-d=7\hfill \end{array}]\to b=5,d=3\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 5\hfill & \hfill 3\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

R₂ -> 5R₁ + 3R₂

For $For\begin{array}{cc}\hfill -a+c=-1\hfill & \hfill 2a-c=2\hfill \end{array}]\to a=1,c=1$

$\begin{array}{cc}\hfill -b+d=-1\hfill & \hfill 2b-d=3\hfill \end{array}]\to b=2,d=1$

(A) ® R₁ ® R₁ + R₂

_{(B) ® R2 ® R2 + 2R1 $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$}

(C) ® R₂ ® 3R₂ + 5R₁

R₁ =  $\left\{\left(a,1\right)\in N*N:\left|a-b\right|\le 13\right\}$

 $\left(a,a\right)\in R_{1}as\left|a-a\right|\le 13\left(Reflexive\right)$

$\left(a,b\right)\&\left(b,a\right)\in R_{1}as\left|a-b\right|=\left|b-a\right|\to \left(symmetric\right).$

But it is not necessary that if (a,b) & (b, c)  $\in R,then\left(a,c\right)\in R$

Eg   $-\left(21,10\right)\in R\&\left(10,1\right)\in Rbut\left(21,1\right)\notin R_1$

R₂ =   $\left\{\left(a,b\right)\in N*N:\left|a-b\right|\ne 13\therefore R_2\to Notequivalencesolutoin.\right\}$

$\left(a,b\right)\&\left(b,a\right)\in R_{2}as\left|a-b\right|=\left|b-a\right|$

But it is not necessary that if (a, b) & (b, c)  $\in R_2$  then (a, c) also $\in R_2$ .

Eg – (21, 1)   $\in R_2\&\left(1,8\right)\in R_{2}but\left(21,8\right)\notin R_2$

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