You already know what an arithmetic progression (AP) is — a sequence where each term differs from the previous one by a fixed number called the common difference d. For example, 2,5,8,11,… is an AP with d=3.
Now imagine you take three terms from an AP — any three, not necessarily consecutive — and arrange them in a 3×3 determinant like this:
apxbqycrz
If the numbers in each row (or each column) form an arithmetic progression, something remarkable happens: the determinant is always zero. That is the core idea.
The Intuition
Why should that be true? Think about what a determinant measures. Geometrically, a 3×3 determinant gives the volume of a parallelepiped formed by three vectors. If those vectors are "linearly dependent" — meaning one can be written as a combination of the others — the volume collapses to zero.
When numbers in a row (or column) are in AP, that row is of the form:
a,a+d,a+2d
This is a linear function of the column index. The same linear pattern repeats across rows (or columns). That repetition creates a dependency: the second row is a linear combination of the first and third, or something similar. The determinant detects this dependency and returns zero.
Note
The determinant being zero does not mean the AP is "trivial" or the numbers are equal. It means the rows (or columns) are not independent — they are related by the AP structure.
The Precise Statement
If the elements of each row (or each column) of a 3×3 determinant are in arithmetic progression, then the value of the determinant is zero.
Let's write it clearly. Suppose we have a determinant:
Δ=a1b1c1a2b2c2a3b3c3
Case 1 — Rows in AP:
If a1,a2,a3 are in AP, and b1,b2,b3 are in AP, and c1,c2,c3 are in AP, then Δ=0.
Case 2 — Columns in AP:
If a1,b1,c1 are in AP, and a2,b2,c2 are in AP, and a3,b3,c3 are in AP, then Δ=0.
Watch out
The condition must hold for every row (or every column) simultaneously. If only one row is an AP and the others are not, the determinant is not necessarily zero.
A Quick Proof (for the curious)
Take the row-AP case. Let the first row be a,a+d1,a+2d1, the second row b,b+d2,b+2d2, and the third row c,c+d3,c+2d3. Write the determinant:
Δ=abca+d1b+d2c+d3a+2d1b+2d2c+2d3
Now perform column operations: C2→C2−C1 and C3→C3−C1. This gives:
Δ=abcd1d2d32d12d22d3
Factor 2 from the third column:
Δ=2abcd1d2d3d1d2d3
Now the second and third columns are identical. A determinant with two equal columns is zero. Hence Δ=0. …
Determinant identities are proved cleanly by column/row operations (which do not change the value) until the determinant collapses to an obvious product. Both OR parts are proved below. …
Column operations C2→C2−C1,C3→C3−C1 reduce it to x3.
C1→C1−C2−C3 then R2→R2−R3 gives 4xyz.
Elementary operations that preserve a determinant's value: adding a multiple of one row (column) to another row (column). A common factor in a row/column can be taken outside.
(a) Prove x+y5x+4y10x+8yx4x8xx2x3x=x3
Apply C1→C1−C2−C3 on the first column:
Row 1: (x+y)−x−x=y−x... instead use the cleaner route below.
Apply C2→C2−C1 and C3→C3−C1:
x+y5x+4y10x+8y−y−x−4y−2x−8y−y−3x−4y−7x−8y.
Rather than track every term, evaluate the original directly by expansion along Row 1:
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2025Set 465/S/WXYZ/43 marks
Q.(a) Prove that x+y5x+4y10x+8yx4x8xx2x3x=x3
(OR)
(b) Prove that y+zzyzz+xxyxx+y=4xyz
›Reveal solutionSolution
Column operations C2→C2−C1,C3→C3−C1 reduce it to x3.
C1→C1−C2−C3 then R2→R2−R3 gives 4xyz.
Elementary operations that preserve a determinant's value: adding a multiple of one row (column) to another row (column). A common factor in a row/column can be taken outside.
(a) Prove x+y5x+4y10x+8yx4x8xx2x3x=x3
Apply C1→C1−C2−C3 on the first column:
Row 1: (x+y)−x−x=y−x... instead use the cleaner route below.
Apply C2→C2−C1 and C3→C3−C1:
x+y5x+4y10x+8y−y−x−4y−2x−8y−y−3x−4y−7x−8y.
Rather than track every term, evaluate the original directly by expansion along Row 1: