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.