Determinants - Class 12 Engineering Maths - Extra Questions

$$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$$
$$=2(-1)-4(-5)=-2+20=18$$

Given matrix:

$$\Delta =\left| \begin{matrix} 1 & { \omega  }^{ 3 } & { \omega  }^{ 2 } \\ { \omega  }^{ 3 } & 1 & \omega  \\ { \omega  }^{ 2 } & \omega  & 1 \end{matrix} \right| $$

$$D=\left| \begin{matrix} 1 & 1 & -2 \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{matrix} \right| \\ D=1(2+3)-1(-1-6)-2(-1+4)\\ D=5+7-6\\ D=6$$

If the points are collinear

$$\begin{array}{l}\left| A \right| = 0\\6\left( { - 2 - 10} \right) - x\left( {4 + 20} \right) + \left( {10 - 10} \right)\\ = 0\\so,\\ - 72 =   2 4x\\x =  - 3\end{array}$$

$$\triangle =\begin{vmatrix} 3 & 5 & 7 \\ 2 & -3 & 1 \\ 1 & 1 & 2 \end{vmatrix}$$

$$\begin{vmatrix} ax & by & cz \\ x^{ 2 } & y^{ 2 } & z^{ 2 } \\ 1 & 1 & 1 \end{vmatrix}$$
Multiplying $${ R }_{ 3 }$$ by $$xyz$$
$$=\cfrac { 1 }{ xyz } \begin{vmatrix} ax & by & cz \\ x^{ 2 } & y^{ 2 } & z^{ 2 } \\ xyz & xyz & xyz \end{vmatrix}$$
Taking $$x,y$$ and $$z$$ common from $${ C }_{ 1 },{ C }_{ 2 }$$ and $${ C }_{ 3 }$$ respectively, we get
$$=\cfrac { xyz }{ xyz } \begin{vmatrix} a & b & c \\ x & y & z \\ yz & xz & xy \end{vmatrix}=\begin{vmatrix} a & b & c \\ x & y & z \\ yz & zx & xy \end{vmatrix}$$

$$D=\begin{vmatrix} n! & (n+1)! & (n+2)! \\ (n+1)! & (n+2)! & (n+3)! \\ (n+2)! & (n+3)! & (n+4)! \end{vmatrix}$$
Taking $$\left( n! \right) ,\left( n+1 \right) !$$ and $$\left( n+2 \right) !$$ common from $${ C }_{ 1 },{ C }_{ 2 }$$ and $${ C }_{ 3 }$$, we get
$$D=\left( n! \right) \left( n+1 \right) !\left( n+2 \right) !\begin{vmatrix} 1 & 1 & 1 \\ n+1 & \left( n+2 \right)  & \left( n+3 \right)  \\ \left( n+2 \right) \left( n+1 \right)  & \left( n+3 \right) \left( n+2 \right)  & \left( n+4 \right) \left( n+3 \right)  \end{vmatrix}$$
Applying $${ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 },{ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 2 }$$
$$D=\left( n! \right) \left( n+1 \right) !\left( n+2 \right) !\begin{vmatrix} 1 & 0 & 0 \\ n+1 & 1 & 2 \\ \left( n+2 \right) \left( n+1 \right)  & 2\left( n+2 \right)  & 2\left( n+5 \right)  \end{vmatrix}\\ =\left( n! \right) \left( n+1 \right) !\left( n+2 \right) !\left( 2n+2 \right) $$
Hence
$$\left[ \frac { D }{ (n!)^{ 3 } } -4 \right] =\left[ \frac { \left( n! \right) \left( n+1 \right) !\left( n+2 \right) !\left( 2n+2 \right)  }{ { \left( n! \right)  }^{ 3 } } -4 \right] =\left[ \frac { \left( n+1 \right) \left( n+1 \right) \left( n+2 \right) \left( 2n+2 \right)  }{ 1 } -4 \right] \\ =\left( 2\left( { n }^{ 4 }+5{ n }^{ 3 }+9{ n }^{ 2 }+7n+2 \right) -4 \right) =2\left( { n }^{ 4 }+5{ n }^{ 3 }+9{ n }^{ 2 }+7n \right) $$

Consider first part 

$$\Delta =\left| \begin{matrix} 1 & \omega  & { \omega  }^{ 2 } \\ \omega  & { \omega  }^{ 2 } & 1 \\ { \omega  }^{ 2 } & 1 & \omega  \end{matrix} \right| \\ { R }_{ 1 }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }\\ \Rightarrow \Delta =\left| \begin{matrix} 1+\omega +{ \omega  }^{ 2 } & 1+\omega +{ \omega  }^{ 2 } & 1+\omega +{ \omega  }^{ 2 } \\ \omega  & { \omega  }^{ 2 } & 1 \\ { \omega  }^{ 2 } & 1 & \omega  \end{matrix} \right| \\ \Rightarrow \Delta =\left| \begin{matrix} 0 & 0 & 0 \\ \omega  & { \omega  }^{ 2 } & 1 \\ { \omega  }^{ 2 } & 1 & \omega  \end{matrix} \right| $$

Apply $${C}_{1}+{C}_{2}+{C}_{3}$$ and take out $${ \left( a+b \right) }^{ 2 }$$ common from $${C}_{1}$$
$$\triangle ={ \left( a+b \right) }^{ 2 }\begin{vmatrix} 1 & { a }^{ 2 } & { b }^{ 2 } \\ 1 & { b }^{ 2 } & 2ab \\ 1 & 2ab & { a }^{
2 } \end{vmatrix}$$
Apply $${R}_{2}-{R}_{1}$$; $${R}_{3}-{R}_{1}$$ to make two zeros.
$$\triangle ={ \left( a+b \right) }^{ 2 }\begin{vmatrix} 1 & { a }^{ 2 } & { b }^{ 2 } \\ 0 & { b }^{ 2 }-{ a }^{ 2 } & b\left( 2a-b
\right) \\ 0 & a\left( 2b-a \right) & { a }^{ 2 }-{ b }^{ 2 } \end{vmatrix}$$
$$={ \left( a+b \right) }^{ 2 }\begin{vmatrix} { b }^{ 2 }-{ a }^{ 2 } & b\left( 2a-b \right) \\ a\left( 2b-a \right) & { a }^{ 2
}-{ b }^{ 2 } \end{vmatrix}$$
$$={ \left( a+b \right) }^{ 2 }\left[ -{ \left( { a }^{ 2 }-b^{ 2 } \right) }^{ 2 }-ab\left( 2b-a \right) \left( 2a-b \right)
\right]$$
$$={ -\left( a+b \right) }^{ 2 }\left[ { \left( { a }^{ 2 }-b^{ 2 } \right) }^{ 2 }+4{ a }^{ 2 }{ b }^{ 2 }-2ab\left( { a }^{ 2
}+b^{ 2 } \right) +{ a }^{ 2 }{ b }^{ 2 } \right]$$
$$={ -\left( a+b \right) }^{ 2 }\left[ { \left( { a }^{ 2 }-b^{ 2 } \right) }^{ 2 }-2ab\left( { a }^{ 2 }+b^{ 2 } \right) +{ a }^{ 2
}{ b }^{ 2 } \right]$$
$$={ -\left( a+b \right) }^{ 2 }{ \left[ { a }^{ 2 }+b^{ 2 }-ab \right] }^{ 2 }=-{ \left( { a }3+{ b }^{ 3 } \right) }^{ 2 }$$

Multiply $${R}_{1}$$, $${R}_{2}$$ and $${R}_{3}$$ by $$c$$, $$a$$
and $$b$$ respectively and hence divide $$\triangle$$ by $$abc$$
$$\therefore \Delta =\cfrac { 1 }{ abc } \begin{vmatrix} { a }^{ 2
}+{ b }^{ 2 } & { c }^{ 2 } & { c }^{ 2 } \\ { a }^{ 2 } & { b }^{ 2
}+{ {
c }^{ 2 } } & { a }^{ 2 } \\ { b }^{ 2 } & { b }^{ 2 } & c^{ 2 }+{
a^{ 2 } } \end{vmatrix}$$
Now apply $${ R }_{ 1 }-\left( { R }_{ 2 }+{ R }_{ 3 } \right)$$
$$\therefore \Delta =\dfrac { 1 }{ abc } \begin{vmatrix} 0 & { -2b }^{
2 } & { -2a }^{ 2 } \\ { a }^{ 2 } & { b }^{ 2 }+{ { c }^{ 2 } } & {
a }^{ 2 } \\ { b }^{ 2 } & { b }
^{ 2 } & c^{ 2 }+{ a^{ 2 } } \end{vmatrix}$$
$$=-\cfrac { 2 }{ abc } \begin{vmatrix} 0 & { b }^{ 2 } & { a }^{ 2
} \\ { a }^{ 2 } & { b }^{ 2 }+{ { c }^{ 2 } } & { a }^{ 2 } \\ {
b }^{ 2 } & { b }^{ 2 } & c^{ 2 }+{ a^{ 2 } } \end{vmatrix}$$
Apply $${R}_{2}-{R}_{1}$$ and $${R}_{3}-{R}_{1}$$
$$=-\cfrac { 2 }{ abc } \begin{vmatrix} 0 & { b }^{ 2 } & { a }^{ 2
} \\ { a }^{ 2 } & { { c }^{ 2 } } & 0 \\ { b }^{ 2 } & { b }^{ 2
} & { c^{ 2 } } \end{vmatrix}$$ Expand.
$$=-\cfrac { 2 }{ abc } \left[ 0-b^{ 2 }\left( { a }^{ 2 }{ c }^{ 2 }-0 \right) +{ a }^{ 2 }\left( 0-{ b }^{ 2 }{ c }^{ 2 } \right)
\right]$$
$$=-\cfrac { 2 }{ abc } \left( -2{ a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 } \right) =4abc$$

A will be 3 X 3 matrix.
$$\therefore \ \ \ \ A = \begin{bmatrix}
a^2 & ab & -a^2 \\[0.3em]
ba & b^2 & ba \\[0.3em]
-a^2 & -ab & a^2
\end{bmatrix}$$
Clearly |A| = 0 because $$R_1$$ and $$R_3$$ are identical. Therefore A is singular and hence its inverse does not exists.

Now,

Equation of line through $$(3,2)$$ & $$(-1,3)$$ is 

$$\left| \begin{matrix} 3  & 1 \\ 5 & 2 \end{matrix} \right|=6-5=1$$

$$\begin{vmatrix} 2i & -3i \\ { i }^{ 3 } & -2i^{ 8 } \end{vmatrix}\\ =2i\left( -2i^{ 8 } \right) -{ i }^{ 3 }\left( -3i \right) \\ =-4i+3{ i }^{ 4 }\quad \left( \because { i }^{ 8 }={ \left( { i }^{ 4 } \right)  }^{ 2 }=1 \right) \\ =3-4i\quad \left( { i }^{ 4 }=1 \right) $$

$$\Delta =\left| \begin{matrix} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{matrix} \right| \\ { C }_{ 1 }+{ C }_{ 2 }+{ C }_{ 3 }\\ \Delta =\left| \begin{matrix} x+9 & 3 & 5 \\ x+9 & x+2 & 5 \\ x+9 & 3 & x+4 \end{matrix} \right| \\ =(x+9)\left| \begin{matrix} 1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4 \end{matrix} \right| \\ { R }_{ 2 }={ R }_{ 2 }-{ R }_{ 1 }\quad \quad \quad { R }_{ 3 }={ R }_{ 3 }-{ R }_{ 1 }\\ \Delta =(x+9)\left| \begin{matrix} 1 & 3 & 5 \\ 0 & x-1 & 0 \\ 0 & 0 & x-1 \end{matrix} \right| $$

$$A=\begin{bmatrix}1+x&7\\3-x&8\end{bmatrix}$$

$$\displaystyle \sum_{r=1}^nD_r=\begin{vmatrix}
\sum_{r=1}^n2^{r-1} & \sum_{r=1}^n2(3^{r-1}) & \sum_{r=1}^n4(5^{r-1})\\
x & y & z\\
2^n-1 & 3^n-1 & 5^n-1
\end{vmatrix}$$
$$=\begin{vmatrix}
1+2+...2^{n-1} & 2\left \{1+3....3^{n-1}\right \} & 4(1+5+....5^{n-1})\\
x & y & z\\
2^n-1 & 3^n-1 & 5^n-1
\end{vmatrix}$$

$$=\begin{vmatrix}
\frac {2^n-1}{2-1} & \frac {2(3^n-1)}{3-1} & \frac {4(5^n-1)}{5-1}\\
x & y & z\\
2^n-1 & 3^n-1 & 5^n-1
\end{vmatrix}=\begin{vmatrix} { 2 }^{ n-1 } & { 3 }^{ n-1 } & { 5 }^{ n-1 } \\ x & y & z \\ { 2 }^{ n-1 } & { 3 }^{ n-1 }

& { 5 }^{ n-1 } \end{vmatrix}\quad=0$$
Since $$R_1 = R_3$$

Let $$A = \begin{bmatrix} a & b \\ c & 3 - a \end{bmatrix}$$