Q.Find minors and cofactors of the elements a11, a21 in the determinant Δ=a11a21a31a12a22a32a13a23a33.
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Determinant Evaluation Using Identities
Expanding a 4×4 or 5×5 determinant term by term is painful and error-prone. The smarter route is to transform the determinant into an easy form using properties (the "identities") that change its value in a known, controlled way — then read the answer off a triangular matrix.
The geometric intuition
A determinant measures the signed "volume" of the box spanned by the rows in n-dimensional space. Sliding one row parallel to another doesn't change that volume; swapping two rows flips its sign; scaling a row scales the volume. The algebraic identities are just these facts translated into rules.
The three row (or column) operations
- Swap two rows: det→−det (sign flips).
- Scale a row by k: det→kdet (the factor comes out).
- Add a multiple of one row to a different row (Ri→Ri+λRj, i=j): det unchanged.
The identical rules hold for columns. There is also row-wise linearity: if a row is a sum Ri=Ri′+Ri′′, the determinant splits into the sum of two determinants with all other rows fixed.
Row-wise linearity is not det(A+B)=detA+detB — that is false. The splitting works one row at a time.
The strategy
- Use operation 3 to create zeros in a row or column (value unchanged).
- Factor out common factors with operation 2.
- Swap rows if needed to reach upper-triangular form (track the sign change).
- The determinant is then the product of the diagonal entries.
Worked example
det1472583610.
Apply R2→R2−4R1 and R3→R3−7R1 (no change), then R3→R3−2R2: …
Concept: Minor and Cofactor Definition
The minor Mij of element aij is the determinant obtained by deleting the i-th row and j-th column. The cofactor Cij is (−1)i+jMij.
Step 1: Minor of a11
Delete row 1 and column 1:
M11=a22a32a23a33=a22a33−a23a32
Step 2: Cofactor of a11
C11=(−1)1+1M11=M11.
Step 3: Minor of a21
Delete row 2 and column 1: …
The minor of an element is the determinant left after deleting its row and column; the cofactor is (−1)i+j times the minor. For a11, minor M11=a22a32a23a33 and cofactor A11=M11. For a21, minor M21=a12a32a13a33 and cofactor A21=−M21.
The idea is simple: every element in a determinant has a "shadow" — the smaller determinant you get by removing its row and column. That shadow is the minor. But the sign matters when you use it to expand the determinant, so we attach a sign based on the element's position: (−1)row+column. That signed minor is the cofactor.
Let’s walk through it for the two elements asked.
- For a11 (first row, first column): Delete row 1 and column 1. What remains is the 2×2 determinant formed by the other rows (2 and 3) and other columns (2 and 3):
M11=a22a32a23a33
The cofactor A11 is (−1)1+1M11=(+1)⋅M11=M11.
So A11=a22a32a23a33.
- For a21 (second row, first column): Delete row 2 and column 1. The remaining rows are 1 and 3, remaining columns are 2 and 3:
M21=a12a32a13a33
The cofactor A21=(−1)2+1M21=(−1)3M21=−M21.
So A21=−a12a32a13a33. …
Method: Finding Minors and Cofactors in a General (Symbolic) Determinant
This method finds the minor and cofactor of a named element when the determinant's entries are symbols (aij) rather than numbers — exactly the same deletion rule applies, just carried out algebraically.
Steps
Step 1: Identify the element's exact position
Read off the row i and column j from the subscript of the element itself — e.g. a21 sits in row 2, column 1.
Step 2: Delete that row and column symbolically
Cross out every symbol in row i and every symbol in column j from the full 3×3 array, keeping the remaining symbols in their original relative positions.
Step 3: Write the resulting 2×2 symbolic minor
Mij=apqasqaprasr=apqasr−aprasq
where p,s are the two surviving row indices and q,r are the two surviving column indices, in their original order.
Step 4: Attach the sign for the cofactor
Cij=(−1)i+jMij …
Common Mistakes
Mistake 1: Forgetting the negative sign on the cofactor of a21
Why it's wrong: since 2+1=3 is odd, (−1)2+1=−1, so A21=−M21 — writing A21=M21 (without the negative) is a very common slip, especially since a11's cofactor (an even position) does NOT need the sign flip, and it's easy to apply that same "no flip" habit to every element. Correct approach: check the parity of i+j for every single element individually — never assume the pattern from a neighboring element.
Mistake 2: Deleting the wrong row/column pair symbolically …
Showing the 12 most recent of 59 on this concept.
- CBSE 2026Set 65/1/11 markMCQQ.If Δ1=100020003 and Δ2=010200006, then (A) Δ1=2Δ2 (B) Δ2=−2Δ1 (C) Δ1=Δ2 (D) Δ2=−Δ1
›Reveal solutionSolution
The first determinant is diagonal; the second requires one row interchange to reach diagonal form. Each interchange flips the sign; evaluating both determinants gives Δ2=−2Δ1. The answer is (B).
Why determinants change under row operations
A determinant measures the signed volume of the parallelepiped spanned by the row vectors. When you swap two rows, you reflect the figure across a hyperplane—the volume stays the same in magnitude but the orientation reverses, flipping the sign.
The diagonal determinant is the easiest to compute: the product of the diagonal entries. The second determinant looks scrambled, but a single row swap will bring it into a form we recognize.
Step-by-step evaluation
1. Compute Δ1 directly.
The matrix is diagonal:
Δ1=100020003=1⋅2⋅3=6.
2. Recognize the structure of Δ2.
Δ2=010200006.
The first two rows are out of order compared to a diagonal form. Swap rows 1 and 2 to bring the 1 into the top-left position.
3. Apply the row-interchange property.
Swapping rows 1 and 2:
Δ2=−100020006.
The negative sign comes from the single interchange.
4. Evaluate the new diagonal determinant.
100020006=1⋅2⋅6=12.
So Δ2=−12. …
- CBSE 2024Set 65/1/11 markMCQQ.x+1x2+x+1x−1x2−x+1 is equal to : (A) 2x3 (B) 2 (C) 0 (D) 2x3−2
›Reveal solutionSolution
Expand the 2×2 determinant as ad−bc; the cube-sum and cube-difference collapse to a constant. The value is 2, option (B).
For a 2×2 determinant, acbd=ad−bc.
Δ=x+1x2+x+1x−1x2−x+1=(x+1)(x2−x+1)−(x−1)(x2+x+1)
Use the standard factorisations a3+b3=(a+b)(a2−ab+b2) and a3−b3=(a−b)(a2+ab+b2) with a=x, b=1: …
- CBSE 2026Set A1 markMCQQ.233663121026112637=(a) 1(b) −1(c) 0(d) 2
›Reveal solutionSolution
The determinant equals 0 because one column is the sum of the other two.
Inspect the columns of
233663121026112637.
Check: 12+11=23, 10+26=36, 26+37=63.
…
- CBSE 2026Set A1 markMCQQ.cos15∘sin75∘sin15∘cos75∘=(a) 1(b) 0(c) −1(d) 21
›Reveal solutionSolution
The determinant equals cos90∘=0.
Expand:
cos15∘sin75∘sin15∘cos75∘=cos15∘cos75∘−sin15∘sin75∘.
…
- CBSE 2026Set A1 markMCQQ.a+ib−c+idc+ida−ib=(a) a2+b2+c2+d2(b) a2−b2−c2−d2(c) a2−b2+c2+d2(d) a2+b2+c2−d2
›Reveal solutionSolution
Expand the 2×2 determinant and simplify the complex products.
a+ib−c+idc+ida−ib=(a+ib)(a−ib)−(c+id)(−c+id).
First term: (a+ib)(a−ib)=a2−(ib)2=a2+b2. …
- CBSE 2026Set ANNUAL1 markMCQQ.Value of x2−x+1x+1x−1x+1 will be(a) x2−x+2(b) x3+x2−2(c) x3−x2+2(d) x3+x2+4
›Reveal solutionSolution
Expand the 2×2 determinant using acbd=ad−bc.
Δ=(x2−x+1)(x+1)−(x−1)(x+1)
(x2−x+1)(x+1)=x3+1 (the middle terms cancel).
(x−1)(x+1)=x2−1.
…
- CBSE 2026Set ANNUAL1 markQ.The value of determinant Δ=1−14231400 is __________.
›Reveal solutionSolution
Expand the 3×3 determinant along the first row.
Δ=1−14231400
Expanding along row 1: …
- CBSE 2026Set ANNUAL1 markQ.Find the value of determinant Δ=0−sinαcosαsinα0−sinβ−cosαsinβ0.
›Reveal solutionSolution
The matrix is skew-symmetric (each aij=−aji) and every odd-order skew-symmetric matrix has determinant 0.
Check: a12=sinα=−a21, a13=−cosα=−a31, a23=sinβ=−a32, and all diagonal entries are 0 — so the matrix is skew-symmetric.
…
- CBSE 2026Set ANNUAL1 markMCQQ.cos30∘sin30∘sin30∘cos30∘=(a) 21(b) 23(c) 0(d) None of these
›Reveal solutionSolution
This determinant has the form cos2θ−sin2θ=cos2θ.
cos30∘sin30∘sin30∘cos30∘=cos30∘⋅cos30∘−sin30∘⋅sin30∘=cos230∘−sin230∘
…
- CBSE 2026Set ANNUAL1 markQ.Evaluate the determinant \Delta = \begin{vmatrix}1 & 2 & 4\ -1 & 3 & 0\ 4 & 1 & 0\end{vmatrix}.
›Reveal solutionSolution
Expand the 3×3 determinant along the first row (or any row/column) using cofactors.
Working: Expanding along Row 1:
Δ=1−14231400 …
- CBSE 2025Set 65/4/11 markMCQQ.If M and N are square matrices of order 3 such that det(M)=m and MN=mI, then det(N) is equal to : (A) −1 (B) 1 (C) −m2 (D) m2
›Reveal solutionSolution
The key idea is that MN=mI implies N=mM−1, so det(N)=m3det(M−1)=m3⋅m1=m2. The correct option is (D).
The problem gives us two square matrices M and N of order 3, with det(M)=m and MN=mI, where I is the 3×3 identity matrix. We need det(N).
The central concept here is the relationship between matrix multiplication and determinants. When two matrices multiply to give a scalar times the identity, that scalar is intimately connected to the determinant of the first matrix. The equation MN=mI is not just a product — it tells us that N is essentially a scaled inverse of M.
Why? Because if MN=mI, then multiplying both sides on the left by M−1 (assuming M is invertible) gives N=mM−1. But we must first check: is M invertible? Yes — since det(M)=m=0 (the problem doesn't state m=0 explicitly, but if m=0, then MN=0, which would make N singular and the answer ambiguous; in standard exam contexts, m is taken as a non-zero scalar, often a real number, and the options suggest m=0). So M−1 exists.
Now, the determinant of a scalar multiple of a matrix: for an n×n matrix A, det(kA)=kndet(A). Here n=3, so det(mM−1)=m3det(M−1).
And we know det(M−1)=det(M)1=m1.
Putting it together:
- From MN=mI, take determinant on both sides: det(MN)=det(mI).
- det(MN)=det(M)⋅det(N)=m⋅det(N). …
- CBSE 2025Set E1 markMCQQ.212564111527101037=(a) 1190(b) 841(c) 0(d) 1
›Reveal solutionSolution
A column that is the sum of the other two makes the determinant zero.
Examine the columns of
212564111527101037.
Check C2+C3 against C1:
11+10=21,15+10=25,27+37=64.
…
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