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Which one of the following does NOT equal $$\begin{vmatrix} 1 & x & x^{2}\\ 1& y & y^{2}\\ 1 & z & z^{2} \end{vmatrix} \quad ?$$

1. $\begin{vmatrix} 1& x(x+1)& x+1\\ 1& y(y+1) & y+1\\ 1& z(z+1) & z+1 \end{vmatrix}$
2. $\begin{vmatrix} 1& x+1 & x^{2}+1\\ 1& y+1 & y^{2}+1\\ 1& z+1 & z^{2}+1 \end{vmatrix}$
3. $\begin{vmatrix} 0& x-y & x^{2}-y^{2}\\ 0 & y-z & y^{2}-z^{2}\\ 1 & z & z^{2} \end{vmatrix}$
4. $\begin{vmatrix} 2& x+y & x^{2}+y^{2}\\ 2 & y+z & y^{2}+z^{2}\\ 1 & z & z^{2} \end{vmatrix}$

edited
Yes, try to substitute different $-$different values and eliminate the options.
What is a,b and c here?
see above comment $a,b,c$ is nothing but $x,y,z$

Operations are:

$C_{3} \leftarrow C_{3} + C_{2}$

$C_{2} \leftarrow C_{2} + C_{1}$

Swap $C_{2}\, \& \,C_{3}$

The Swapping operations make the determinant as $(-1)*|A|$

whereas the other options have their determinant as $|A|.$

why i can not use exchange command here?
Bcz ,determinant property-- if any  two rows (or any two columns)of a determinant are interchanged then the value of determinant is multiplied by "-1".

This is special type of matrix called vandermonde matrix, it is special as it has some important applications.

even exchanges won't change it tho right.
$\begin{vmatrix} 1 &x &x^{2} \\ 1 & y& y^{2}\\ 1& z& z^{2} \end{vmatrix}$

$(B):$    $C2\rightarrow C2+C1 ,\ C3\rightarrow C3+C1$

$\begin{vmatrix} 1 &x+1 &x^{2}+1 \\ 1 & y+1& y^{2}+1\\ 1& z+1& z^{2}+1 \end{vmatrix}$

$(C):$   $R1\rightarrow R1-R2 ,\ R2\rightarrow R3-R2$

$\begin{vmatrix} 0 &x-y &x^{2}-y^{2} \\ 0 & y-z& y^{2}-z^{2}\\ 1& z& z^{2} \end{vmatrix}$

$(D):$   $R1\rightarrow R1+R2 ,\ R2\rightarrow R3+R2$

$\begin{vmatrix} 2 &x+y &x^{2}+y^{2} \\ 2 & y+z& y^{2}+z^{2}\\ 1& z& z^{2} \end{vmatrix}$

We can't get option(A) from given Determinant.

Hence,Option(A) is the correct choice.

### 1 comment

Applying row operations mentioned in D, does not yield the last determinant. How you got the last one?
the correct ans is option (A)
by

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