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.
Note
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.
Watch out
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.
abcbcacab.
Each row is a cyclic shift of the same three entries a,b,c, so the sum a+b+c is common to every row and every column. Applying C1→C1+C2+C3 pulls that common factor out immediately — a technique that specifically relies on recognising the cyclic structure of the determinant, not on the general row/column identities used for an arbitrary, non-cyclic determinant.
Tip
Spotting that a determinant's rows (or columns) are cyclic permutations of the same entries is the cue to try "add all columns into one" first — it almost always exposes a clean common factor before any further reduction.
Recognising the cyclic-permutation invariance of a determinant is a specific "properties of determinants" result in the CBSE Class 12 syllabus, distinct from the general row/column-operation rules, and cyclic-entry determinants like the a-b-c example above are a recurring board and JEE Main "prove that" question type. Searching "cyclic property of determinants class 12" or "prove using properties of determinants cyclic" points straight at this row/column-shift argument.
The key idea is that if one row (or column) of a determinant is a scalar multiple of another, the determinant is zero.
Step 1: Write the given determinant:
Δ=2−43−6
Step 2: Observe that the second row is (−2) times the first row:
(−2)×[2,3]=[−4,−6]
Step 3: Since the rows are linearly dependent (one is a multiple of the other), the determinant is zero.
✓Final answer
The value is 0.
The determinant of a 2×2 matrix [acbd] is ad−bc. For this matrix, 2(−6)−3(−4)=−12+12=0, so the value is 0.
The determinant is a single number that captures key properties of a matrix — whether it's invertible, how it scales area, and so on. For a 2×2 matrix, the formula is straightforward: multiply the top-left and bottom-right entries, then subtract the product of the top-right and bottom-left entries. This is the definition you need to apply here.
Let's work through it step by step.
Identify the entries.
The matrix is [2−43−6]. Label them as:
a=2, b=3, c=−4, d=−6.
Apply the determinant formula.
For any 2×2 matrix [acbd], the determinant is ad−bc.
So here:
det=(2)(−6)−(3)(−4).
Compute each product.
2×(−6)=−12.
3×(−4)=−12, but note the minus sign in the formula: we subtract bc, so it becomes −(−12)=+12.
Combine the results.
−12+12=0.
Watch out
A common mistake is forgetting the minus sign in ad−bc, or mishandling the negative signs in the products. Here, 3×(−4)=−12, and subtracting that gives +12, not −12. Always write the subtraction explicitly to avoid sign errors.
The determinant is zero. This tells us the rows (or columns) are linearly dependent — in fact, the second row is exactly −2 times the first row. A zero determinant means the matrix is singular (non-invertible), which is consistent with the rows being multiples of each other.
✓Final answer
The value of the determinant is 0.
Method: Recognising a Zero Determinant From Proportional Rows
A shortcut to try before computing ad−bc by brute force.
Steps
Step 1: Compare the two rows for a scalar relationship
Check whether row 2 is a constant multiple of row 1, i.e. (c,d)=λ(a,b) for some λ.
Step 2: If so, conclude the determinant is 0 immediately
Two proportional rows (or columns) always force the determinant to 0 — this can be stated without any multiplication.
Step 3: Confirm with the direct formula as a check
Evaluate ad−bc directly; it should come out exactly 0, matching the shortcut.
Common Mistakes
Mistake 1: Computing ad−bc by brute force and making an arithmetic error, instead of first checking for proportional rows
Why it's wrong: here row 2 is exactly −2 times row 1, which guarantees the determinant is 0 as an instant sanity check — computing 2(−6)−3(−4) directly without this check risks a sign slip that produces a wrong nonzero "answer." Correct approach: always scan for a proportional relationship between the rows before computing, and use it both as a shortcut and as a check on the direct calculation.
Mistake 2: Concluding the matrix is invertible despite the determinant being 0
Why it's wrong: a 0 determinant means the matrix is singular — students sometimes proceed to (incorrectly) look for an inverse anyway, or assume the "value" being asked for must be nonzero. Correct approach: recognise that a zero determinant is a valid, meaningful answer — it directly tells you the matrix has no inverse, which is consistent with the proportional-rows structure.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
Q.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.
A matrix with linearly dependent rows has determinant zero. Therefore
Δ=0.
Watch out
A common mistake is to try to expand the determinant fully and then use Vieta’s formulas. That works but is much longer. The cyclic structure plus the zero-sum condition gives the answer instantly.
Tip
Whenever you see a circulant-like matrix (each row is a cyclic shift) and the sum of the entries is zero, the determinant is zero because the vector (1,1,1) is in the nullspace of the transpose.
Thus, without any heavy algebra, we conclude:
✓Final answer
The correct option is (D).
ANSWER: D
AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
Q.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.
Forgetting that the sum-of-squares expression forces strict equality of all three.
✓Final answer
The correct option is (A) — ∣Z1∣=∣Z2∣=∣Z3∣.
ANSWER: A
Note
This solution was worked out by our team and cross-checked by a second independent solve. The official answer key for this question could not be confirmed, so please cross-verify with the official paper where possible.