Question

Appropriately matching the information given in the three columns of the following table.

$\begin{array}{cccccc}& \text{Column 1}& & \text{Column 2}& & \text{Column 3}\\ \text{(I)}& \text{If}a,b,cϵR-\left\{0\right\}\text{such that}a\ne b\ne c& & & & \\ & \text{and}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\text{and}A=\left[\begin{array}{ccc}1+a& 1& 1\\ 1& 1+b& 1\\ 1& 1& 1+c\end{array}\right],\text{then}& \left(i\right)& \text{A is singular matrix}& \text{(P)}& |adjA|=|A{|}^2\\ \text{(II)}& \text{If}\alpha ,\beta ,\gamma ϵR,\text{and}& & & & \\ & A=\left[\begin{array}{ccc}1& \cos (\alpha -\beta )& \cos (\alpha -\gamma )\\ \cos (\beta -\alpha )& 1& \cos (\beta -\gamma )\\ \cos (\gamma -\alpha )& \cos (\gamma -\beta )& 1\end{array}\right],\text{then}& \left(ii\right)& \text{A is singular matrix}& \left(Q\right)& adj\left(adjA\right)=|A|A\\ \left(III\right)& \text{If}\omega \ne 1\text{be cube root of unity and}& \left(iii\right)& \text{A is non-singular matrix}& \left(R\right)& |A|\text{is equal to}\\ & A=\left[\begin{array}{ccc}1+2{\omega }^{100}+{\omega }^{200}& {\omega }^2& 1\\ 1& 1+{\omega }^{101}+2{\omega }^{202}& \omega \\ \omega & {\omega }^2& 2+{\omega }^{100}+2{\omega }^{200}\end{array}\right]& & & & \text{is equal to minimum value of}\\ & ,then& & & & {\cos }^{-1}(x-\frac{1}{x})\\ & & & & & +{\cos }^{-1}\left(\frac{y^2}{y+1}\right)\\ & & & & & +{\cos }^{-1}(z^2+z+1)\\ & & & & & \text{(where}x,y,z\text{are real numbers)}\\ \text{(IV)}& \text{If}a,b,cϵR-\left\{0\right\}\text{such that}a\ne b\ne c,\text{and}& & & & \\ & A=\left[\begin{array}{ccc}0& (a-b{)}^3& (a-c{)}^3\\ (b-a{)}^3& 0& (b-c{)}^3\\ (c-a{)}^3& (c-b{)}^3& 0\end{array}\right],\text{then}& \left(iv\right)& \text{Invertible}& \left(S\right)& |A^{-1}|=\frac{1}{|A|}\end{array}$

Which of the following is only correct combination?

• A. (I)(i)(R)
• B. (I)(ii)(R)
• C. (I)(iii)(P)
• D. (I)(ii)(S)

Answer 1

The correct option is C

(I)(iii)(P)

$\text{(I)}\Delta =abc\left|\begin{array}{ccc}\frac{1}{a}+1& \frac{1}{b}& \frac{1}{c}\\ \frac{1}{a}& \frac{1}{b}+1& \frac{1}{c}\\ \frac{1}{a}& \frac{1}{b}& \frac{1}{c}+1\end{array}\right|=(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1)abc\left|\begin{array}{ccc}1& \frac{1}{b}& \frac{1}{c}\\ 1& \frac{1}{b}+1& \frac{1}{c}\\ 1& \frac{1}{b}& \frac{1}{c}+1\end{array}\right|=abc\ne 0$

$\therefore$ Non-singular, so invertible

Also it is symmetric

(II) $\left[\begin{array}{ccc}1& cos(\alpha -\beta )& cos(\alpha -\gamma )\\ cos(\beta -\alpha )& 1& cos(\beta -\gamma )\\ cos(\gamma -\alpha )& cos(\gamma -\beta )& 1\end{array}\right]=\left|\begin{array}{ccc}cos\alpha & sin\alpha & 0\\ cos\beta & sin\beta & 0\\ cos\gamma & sin\gamma & 0\end{array}\right|\left|\begin{array}{ccc}cos\alpha & cos\beta & cos\gamma \\ sin\alpha & sin\beta & sin\gamma \\ 0& 0& 0\end{array}\right|$

(III) $\left[\begin{array}{ccc}1+2{\omega }^{100}+{\omega }^{200}& {\omega }^2& 1\\ 1& 1+w^{101}+2{\omega }^{202}& \omega \\ \omega & {\omega }^2& 2+{\omega }^{100}+2{\omega }^{200}\end{array}\right]=\left|\begin{array}{ccc}\omega & {\omega }^2& 1\\ 1& \omega & \omega \\ \omega & {\omega }^2& -\omega \end{array}\right|=\left|\begin{array}{ccc}0& 0& 1+\omega \\ 1& \omega & \omega \\ \omega & {\omega }^2& -\omega \end{array}\right|=(1+\omega )\times 0=0$

$\therefore$ Singular

(IV) This is a skew - symmetric matrix of odd order

$\Rightarrow$ singular.