Question 3 : On Matrix Det

3.                                                                          |1    x    x^2|

    Which one of the following does NOT equal           |1    y    y^2|

                                                                             |1    z    z^2|

     | 1    x(x+1)     x+1|

A. | 1    y(y+1)     y+1|
     | 1    z(z+1)      z+1|

     | 1    x+1     x^2+1|
B. | 1    y+1     y^2+1|
     | 1    z+1     z^2+1|

     | 0    x-y     x^2-y^2|
C. | 0    y-z     y^2-z^2|
     | 1    z               z^2|

     | 2    x+y     x^2 + y^2|
D. | 2    y+z     y^2 + z^2|
     | 1    z                   z^2|

Answer : (A)

Explanation :

Detailed Method
TBD (Using row/column manipulations).


Substitution Method :
Lets assume x = 1, y = 2, z = 3.

Original Matrix Det
|1    x    x^2|    |  1  1  1  |

|1    y    y^2| = |  1  2  4  | = 1(18-12)-1(9-4)+1(3-2) = 6-5+1 = 2

|1    z    z^2|     |  1  3  9 |

Option A matrix Det
| 1    x(x+1)     x+1|       | 1    2   2 |
| 1    y(y+1)     y+1|  =   | 1    6   3 | = -2
| 1    z(z+1)      z+1|       | 1  12  4 |

The det of this matrix is not equal to det of original matrix, so answer is A.

But, anyway, lets calculate the det of other matrices for safety.
     | 1    x+1     x^2+1|        | 1   2   2  |
B. | 1    y+1     y^2+1| =     | 1   3   5  | = 2
     | 1    z+1     z^2+1|        | 1   4   10|

     | 0    x-y     x^2-y^2|      |  0   -1   -3  |
C. | 0    y-z     y^2-z^2| =    |  0   -1   -5 | = 2
     | 1    z               z^2|       |  1    3     9 |

     | 2    x+y     x^2 + y^2|      | 2   3   5  |
D. | 2    y+z     y^2 + z^2| =   | 2   5   13| = 2
     | 1    z                   z^2|      | 1   3    9 |

you can clearly see Det of other matrices is 2, which is equal to det of the original matrix in the question.