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+17 votes
Which one of the following does NOT equal

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

(A) \begin{vmatrix} 1& x(x+1) & x+1\\ 1& y(y+1) & y+1\\ 1& z(z+1) & z+1 \end{vmatrix} (B) \begin{vmatrix} 1& x+1 & x^{2}+1\\ 1& y+1 & y^{2}+1\\ 1& z+1 & z^{2}+1 \end{vmatrix} (C) \begin{vmatrix} 0& x-y & x^{2}-y^{2}\\ 0& y-z & y^{2}-z^{2}\\ 1& z & z^{2} \end{vmatrix} (D) \begin{vmatrix} 2& x+y & x^{2}+y^{2}\\ 2& y+z & y^{2}+z^{2}\\ 1& z & z^{2} \end{vmatrix}
asked in Linear Algebra by Veteran (342k points)
edited by | 1.2k views
This type of question I'm Go with options


and easily get option (A)

3 Answers

+19 votes
Best answer

Answer is A.

Operations are:

$C_{3} \leftarrow C_{3} + C_{2}$

$C_{2} \leftarrow C_{2} + C_{1}$

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

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

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

answered by Active (4.2k points)
edited by
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.

+8 votes
$\begin{vmatrix} 1 &x &x^{2} \\ 1 & y& y^{2}\\ 1& z& z^{2} \end{vmatrix}$

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

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

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

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

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

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


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


Hence,Option(A) is the correct choice.
answered by Boss (39.6k points)
edited by
+3 votes
the correct ans is option (A)
answered by (325 points)

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