Q.Evaluate xyx+yyx+yxx+yxy.
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Determinant Properties Cyclic
Among the several determinant properties in the NCERT Class 12 Determinants chapter, one specific result deals with what happens when the rows (or columns) of a determinant are shuffled in a cyclic order, rather than simply interchanged two at a time. This is subtly different from the familiar "swap two rows flips the sign" rule, and it is a genuinely useful shortcut on its own.
What "cyclic order" means
For a 3×3 determinant with rows R1,R2,R3, a cyclic rearrangement sends
R1→R2,R2→R3,R3→R1,
so the new determinant has its rows in the order R2,R3,R1. This moves every row at once, in the same rotational direction — it is not the same as interchanging just one pair of rows.
The property
Cyclically permuting the rows (or, equivalently, the columns) of a 3×3 determinant does not change its value.
Why it's true
A single row interchange flips the sign of a determinant. Sending R1 to the bottom while R2 and R3 shift up can be built from two successive interchanges: first swap R1↔R2 (sign flips once), then swap the row now in the second slot with R3 (sign flips again). Two sign flips cancel:
(−1)×(−1)=1.
So a full cyclic shift of three rows is an even permutation and leaves the determinant unchanged.
This is specific to an odd count of rows. In general, cyclically permuting n rows carries a sign of (−1)n−1: for n=3 that's (−1)2=+1 (unchanged), but for n=4 rows a full cyclic shift does flip the sign, since (−1)3=−1.
A numeric check
Δ=1472583610=−3.
Cyclically shift the rows to R2,R3,R1:
Δ′=4715826103.
Expanding Δ′ along its first row: 4(8⋅3−10⋅2)−5(7⋅3−10⋅1)+6(7⋅2−8⋅1)=4(4)−5(11)+6(6)=16−55+36=−3 — the same value as Δ, confirming the cyclic shift left the determinant unchanged even though every row moved.
Don't confuse this with a single row swap, which does flip the sign. The cyclic-invariance property only holds when all rows (or all columns) rotate together in one direction — shifting just one pair changes the value in the ordinary way.
Where this shows up: "cyclic determinants"
This property is the reason a whole family of board-exam "prove that" questions work, where the entries themselves are arranged cyclically, e.g. …
Concept: Determinant Properties Cyclic — use row/column operations to exploit symmetry and simplify.
Step 1: Apply C1→C1+C2+C3. Each entry in C1 becomes x+y+(x+y)=2(x+y):
2(x+y)2(x+y)2(x+y)yx+yxx+yxy
Step 2: Factor 2(x+y) from C1:
2(x+y)111yx+yxx+yxy
Step 3: Perform R2→R2−R1 and R3→R3−R1: …
Adding all columns to the first factors out 2(x+y); row reduction then gives the value −2(x3+y3).
We evaluate
Δ=xyx+yyx+yxx+yxy.
1. Column operation C1→C1+C2+C3. Every row sum is x+y+(x+y)=2(x+y), so
Δ=2(x+y)111yx+yxx+yxy.
2. Row operations R2→R2−R1 and R3→R3−R1:
Δ=2(x+y)100yxx−yx+y−y−x.
3. Expand along the first column: …
Method: Column-Sum Simplification for Cyclic/Symmetric Determinants
This method solves determinants where each row (or column) sums to the same expression — a strong hint to use a column/row operation to factor that sum out before expanding.
Steps
Step 1: Check whether every row (or column) has the same sum
Add the entries across each row (or down each column). If the sum is identical for every row/column, that repeated value is a signal to combine columns (or rows) into one, since the corresponding operation leaves the determinant's value unchanged while revealing a common factor.
Step 2: Apply the combining operation C1→C1+C2+C3 (or the row analogue)
By operation 3 of the determinant row/column rules, adding one column's multiple to another doesn't change the determinant's value. Replacing C1 by C1+C2+C3 turns every entry of the first column into the same repeated sum S:
Δ=SSS∗∗∗∗∗∗
Step 3: Factor S out of the column
Since every entry of C1 is now S, factor it out using operation 2 (scaling a row/column pulls out a common multiple):
Δ=S111∗∗∗∗∗∗
Step 4: Clear the remaining 1's with row operations, then expand …
Common Mistakes
Mistake 1: Forgetting the sign when subtracting rows after factoring out 2(x+y)
Why it's wrong: after C1→C1+C2+C3 and factoring, students often apply R2→R2−R1 but subtract entry-by-entry inconsistently, producing a row that doesn't actually have a zero in the first position. Correct approach: subtract the entire row 1 from row 2 (and row 3) term by term, one column at a time, and double-check the first column really becomes 0 before moving on.
Mistake 2: Missing that all three rows share the same sum, and expanding the original 3×3 directly …
- AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQQ.If α,β,γ are the roots of the equation x3+bx+c=0, then αβγβγαγαβ= (A) −b3 (B) b3−3c (C) b2−3c (D) 0
›Reveal solutionSolution
The determinant equals 0 because its rows are cyclic permutations of the same three numbers, making the rows linearly dependent when the sum of the roots is zero — a condition given by the cubic x3+bx+c=0. The correct option is (D).
We are given that α,β,γ are the roots of x3+bx+c=0. Notice the cubic has no x2 term, so the sum of the roots is zero:
α+β+γ=0.
This fact is the key that unlocks the determinant.
The determinant is
Δ=αβγβγαγαβ.
Each row is a cyclic shift of (α,β,γ). When the sum of the three numbers is zero, the rows become linearly dependent. Let’s see why.
-
Observe the row sums.
Add all three columns of the first row: α+β+γ=0. The same holds for every row. So each row sums to zero.
-
Construct a linear dependence.
If we add all three rows together, we get the row vector
(α+β+γ,β+γ+α,γ+α+β)=(0,0,0).
That means the sum of the three rows is the zero row. Hence the rows are linearly dependent.
- Consequence for the determinant. …
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- AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQQ.Let Z1,Z2,Z3 be three non zero complex numbers such that a=∣Z1∣,b=∣Z2∣,c=∣Z3∣. if the determinant abcbcacab=0, then (A) ∣Z1∣=∣Z2∣=∣Z3∣=abc (B) ∣Z1∣+∣Z2∣+∣Z3∣=0 (C) ∣Z1∣+∣Z2∣+∣Z3∣=abc (D) ∣Z1−Z2∣=∣Z2−Z3∣
›Reveal solutionSolution
The determinant vanishes only when ∣Z1∣=∣Z2∣=∣Z3∣.
Concept and Intuition
A cyclic (circulant) determinant with entries a,b,c factors neatly using the identity a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca). Since a,b,c are moduli of non-zero complex numbers, they are strictly positive.
Step-by-Step Solution
- Expand: abcbcacab=3abc−(a3+b3+c3).
- Set equal to zero: a3+b3+c3=3abc, i.e. (a+b+c)(a2+b2+c2−ab−bc−ca)=0.
- Because a,b,c>0, a+b+c=0, so a2+b2+c2−ab−bc−ca=0.
- This is 21[(a−b)2+(b−c)2+(c−a)2]=0, forcing a=b=c.
- Hence ∣Z1∣=∣Z2∣=∣Z3∣.
Common Mistakes
- Concluding a+b+c=0, which is impossible for positive moduli. …
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