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Ask iit jee aieee pet cbse icse state board experts Expert Question: Determinant 2008
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Mr.IITIAN007 (2990)

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Dear Experts ,please solve this deteminant and bring the answer in simplified form.AND PLEASE SOLVE THIS USING THE PROPERTIES.
 
| x    y     z|
| y    x     y|
| z    y     x| 
 
(NOTE:These are not a broken lines.They are continuous lines for showing the determinant .) 

Ken
From: UNITED STATES, Green Bay, Wisconsin
    
shikharbaid (175)

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(x-z)(x^2+xy-2y^2)

<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>


<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
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nik_d_1 (177)

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hey it is (x-z)(x^2+xZ-2y^2)




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nik_d_1 (177)

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C1 -> C1-C3
TAKE OUT (X-Z) FROM R1

THEN r1-> R1 + R3
THEN EXPAND ALONG C1




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avinash.sharma (1189)

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X
Y
Z
Y
X
Y
 Z
Y
x
C1-C3 ->C1
|x-z     y            z |
|0         x            y |
|z-x     y            x |
 
= (x-z) |1            y            z  |
 |0            x            y  |
 |-1            y            x|
 
R1+R3->R1
 
= (x-z) | 0            2y            z+x           |
 | 0            x            y                 |
 |-1            y            x                 |
 
=(x-z) (-1) (2y2 ?xz -x2)
 
=(x-z) (x2 +xz ?2y2)
 
(NOTE:These are not a broken lines.They are continuous lines for showing the determinant .) 
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arunitha (16)

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I think the answer is (x-z)(x^2 +xz-2y^2)
(C1)prime=C1-C3
| x-z  y  z |
|  0    x  y |
| z-x  y   x|
 
Take (x-z) from C1
         | 1  y  z  |
(x-z)   | 0  x  y  |
         |-1   y  x  |
 
Let R1(prime)= R1+R3
 
        |0     2y   x+z  |
(x-z)  |0      x     y    |
        |-1      y      x   |
 
(x-z)[-1(2y^2-x(x+z))]
(x-z)[x^2 +xz -2y^2] 
 
 
(NOTE:These are not a broken lines.They are continuous lines for showing the determinant .) 
 
Please rate for my efforts if satisfactory or inform me if I am wrong.
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nik_d_1 (177)

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hey dats wat i hv been doin....,,
lol...




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Mr.IITIAN007 (2990)

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Hey guys and respected expert!!!!!!!!!!! You all are correct and thank you all to show me the correct way of solving this determinant using the properties.

Ken
From: UNITED STATES, Green Bay, Wisconsin
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