Which one of the following determinants does NOT equal the determinant given…
2013
Which one of the following determinants does NOT equal the determinant given below?
1 | x | x2 |
1 | y | y2 |
1 | z | z2 |
- A.
1
x(x+1)
x+1
1
y(y+1)
y+1
1
z(z+1)
z+1
- B.
1
x+1
x2+1
1
y+1
y2+1
1
z+1
z2+1
- C.
0
x−y
x2−y2
0
y−z
y2−z2
1
z
z2
- D.
2
x+y
x2+y2
2
y+z
y2+z2
1
z
z2
Attempted by 5 students.
Show answer & explanation
Correct answer: A
Concept
A 3×3 determinant is unchanged when a multiple of one column (or row) is added to another column (or row), and it changes sign whenever two columns (or two rows) are interchanged. The given array is the Vandermonde determinant, whose value is (x−y)(y−z)(z−x). To test an alternative array, reduce its columns or rows back to the standard form 1, variable, variable-squared using only these operations and watch for any interchange that flips the sign.
Application
Reduce each candidate array using column and row operations:
Array with second column v+1 and third column v2+1: subtract column 1 from column 2 and from column 3. This gives columns 1, v, v2 with no interchange.
Array with rows 0, v−w, v2−w2 and bottom row 1, z, z2: its first two rows are the row-differences (row1−row2) and (row2−row3) of the standard array, formed only by adding multiples of rows, so the rows stay in standard order.
Array with rows 2, v+w, v2+w2 and bottom row 1, z, z2: the first two rows are row-sums of consecutive standard rows; expanding by linearity and removing the duplicated bottom row again leaves the standard order with no interchange.
Array with second column v(v+1)=v2+v and third column v+1: subtract column 3 from column 2 to get v2−1, then subtract column 1 from column 3 to get v, then add column 1 back to column 2 to get v2. The columns are now 1, v2, v, which is the standard array with its last two columns interchanged.
Cross-check
Three of the arrays reduce to the standard order 1, v, v2 with no interchange, so they keep the value (x−y)(y−z)(z−x). The remaining array reduces to 1, v2, v, a single column interchange away from standard, so its value is −(x−y)(y−z)(z−x), the negative of the given determinant. A numeric check confirms it: for (x, y, z) = (2, 3, 5) the given determinant is 6, while that array evaluates to −6.
Therefore the array built from the columns 1, v(v+1), v+1 is the one that does not equal the given determinant, since it differs by a sign.