Determinants Test - 28

Put x=0
$$\begin{vmatrix}0 &2  &0 \\  6&  0&0 \\  0& 0 &6 \end{vmatrix}=0=e$$

Diff. the determinant w.r.t. x
$$\begin{vmatrix}1 &0  &1 \\  x^{2}& x &6 \\  x& x &6\end{vmatrix}+\begin{vmatrix}x &2  &x \\  2x&1  &0 \\  x& x &6 \end{vmatrix}+\begin{vmatrix}x &2  &x \\  x^{2}&  x&6 \\  1& 1 &0 \end{vmatrix}$$

Put  $$x=0$$
$$0+0+12=12=d$$

Diff. again w.r.t. x
$$\begin{vmatrix}0 &0  &0 \\  x^{2}&  x&6 \\  x& x &6 \end{vmatrix}+\begin{vmatrix}1 &0  &1 \\  2x& 1 &0 \\  x& x &6\end{vmatrix}+\begin{vmatrix}1 &0  &1 \\  x^{2}& x &6 \\  1& 1 &0\end{vmatrix}+\begin{vmatrix}1 &0  &1 \\  2x& 1 &0 \\  x& x &6\end{vmatrix}+\begin{vmatrix}x &2  &x \\  2& 0 &0 \\  x& x &6\end{vmatrix}+\begin{vmatrix}x &2  &x \\  2x& 1 &0 \\  1& 1 &0\end{vmatrix}+\begin{vmatrix}1 &0  &1 \\  x^{2}& x &6 \\  1& 1 &0\end{vmatrix}+\begin{vmatrix}x &2  &x \\  2x& 1 &0 \\  1& 1 &0\end{vmatrix}+\begin{vmatrix}x &2  &x \\  x^{2}& x &6 \\  0& 0 &0\end{vmatrix}$$

Put  $$ x=0$$
$$C = -12\  \  Similarly\ \  a=1\ \,   b=-1   $$
$$5-4-36+24=-11$$

$$f(x)=\begin{vmatrix}
\sin\, \, x &1 &0 \\
1& 2\sin\, \, x& 1\\
0& 1 & 2\sin\, \, x
\end{vmatrix}$$
Expanding along first column we get,
$$f(x)=\sin x(4\sin^2x-1)-1(2\sin x-0)=4\sin^3x-3\sin x$$
Clearly $$f(-x)=f(x)\Rightarrow$$ f is an odd function
Hence $$\displaystyle \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}} f\left ( x \right )=0$$

$$A=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$$

$${ A }^{ 2 }=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}$$

$${ A }^{ 2 }-4A+3I=\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}+\begin{bmatrix} -8 & -4 \\ -4 & -8 \end{bmatrix}+\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

$$\Rightarrow |{ A }^{ 2 }-4A+3I|=0$$

So,$$\sin  \theta =0$$
Hence, the general solution is $$\theta =n\pi $$

$$\displaystyle \begin{vmatrix} 1 & -1 & 0 \\ 0 & 4 & 2 \\ 3 & -4 & 6 \end{vmatrix}$$

$$1+i[i(1-i)-(1+i)^{2}]-(1-i)\left [ (1-i)^{2}-i(1+i) \right ]+1\left [ (1-i)(1+i)-i^{2} \right ]$$

$$=(1+i)\left [ 1-l^{2}-2i \right ]-(1-i)\left [ 1+l^{2}-2i-i-l^{2} \right ]+\left [ 1-l^{2}-l^{2} \right ]$$

$$=(1+i)[-2i^{2}-2i]-(1-i)(1-3i)+[1-2i^{2}]$$

$$=(1+i)[+2-2i]-(1-i) (1-3i)-1$$

$$2[1-l^2]-(1-i)(1-3i)-1$$

$$4-(1-3i-i+3i^2)-1$$

$$3-(1-4i-3)$$

$$3-(-2-4i)$$

$$=5+4i$$

so (c) complex number

Taking  $$a=b=c=1$$ in the given matrix, we get
$$\begin{pmatrix}
-2 & 2 & 2\\
2 & -2 & 2\\
2 & 2 & -2
\end{pmatrix}$$

$$\begin{vmatrix} 1 & 2 \\ 6 & 4 \end{vmatrix}+\begin{vmatrix} \frac { 1 }{ 2 }  & 2 \\ 2 & 4 \end{vmatrix}+\begin{vmatrix} \frac { 1 }{ 4 }  & 2 \\ \frac { 2 }{ 3 }  & 4 \end{vmatrix}+.......$$

$$\displaystyle =4-12+2-4+1-\frac { 4 }{ 3 } +.....$$

$$\displaystyle=(4+2+1+...)-(12+4+\frac { 4 }{ 3 }+....)$$

$$\displaystyle=\frac{4}{1-\frac{1}{2}}- \frac{12}{1-\frac{1}{3}}$$

$$=8-18=-10$$

$$ \displaystyle f(x)=\sec x \cos^2x- \cos x \times \cos^2x.$$
$$ \displaystyle =\cos x- \cos^3x.$$
$$ \displaystyle \int_{0}^{\pi/2}\left(cosx-cos^{3}x \right)dx=\int_{0}^{\pi/2} \cos x \sin^{2}xdx$$
Substitute $$ \sin x=t$$
$$ \Rightarrow \cos x dx=dt$$
The integral becomes 
$$ \displaystyle \int_{0}^{1}t^{2}dt$$
$$ \displaystyle =\left[ \frac{t^{3}}{3} \right]_{0}^{1}=1/3$$

Solving $$\displaystyle \begin{vmatrix} x & 2 & 3 \\ 2 & 3 & x \\ 3 & x & 2 \end{vmatrix}=\begin{vmatrix} 0 & 5 & x \\ 5 & x & 0 \\ x & 0 & 5 \end{vmatrix}$$

 Let $$common\ diff\ = m$$
$$R_1\rightarrow R_1-R_2$$
$$\Delta(x)=\begin{vmatrix}
a-b & b-c & a-c+1\\
x+b & x+c & x-1\\
x+c & x+d & x-b+d
\end{vmatrix}$$
$$R_3\rightarrow R_3-R_2$$
$$\Delta (x)=\begin{vmatrix}
a-b & b-c & a-c+1\\
x+b & x+c & x-1\\
c-b & d-c & d-b+1
\end{vmatrix}$$
$$=\begin{vmatrix}
m & m & -2m+1\\
x+b & x+c & x-1\\
m & m & 2m+1
\end{vmatrix}$$