JEE Advanced (IIT Entrance) · Mathematics (JEE & NDA)

Matrices and Determinants

Algebra and types of matrices, evaluation of determinants, adjoint and inverse, and solving linear systems.

Five concepts: matrix algebra, determinant rules, adjoint and inverse, linear systems, and the eigenvalue checks that close a JEE Main matrices question.

JEE Advanced (IIT Entrance)
Medium level
5 concepts
5 practice questions

1Matrix algebra and types

Matrices add entrywise and multiply by the row–column rule: the product A_{m\times n}B_{n\times p} is defined only when the inner dimensions match. Transpose reverses the product order, (AB)^T = B^T A^T. A square matrix is symmetric when A^T = A and skew-symmetric when A^T = -A.

Matrix addition and multiplication are entrywise / row-by-column algebra on arrays — nothing is placed in the plane.

Building a product

Check dimensionsWrite A as m\times n and B as n\times p. If the shared n disagrees, the product is undefined — stop.
Row into columnEntry (i,j) of AB is the dot product of row i of A with column j of B; the result is m\times p.
Transpose lastWhen the question asks for (AB)^T, compute B^T A^T rather than reversing a product you have not formed.

Read off the type from what transpose does — the criterion, not a memorised name.

Square-matrix types
Type	Defining relation	Immediate consequence
Symmetric	A^T = A	All a_{ij} = a_{ji}
Skew-symmetric	A^T = -A	Diagonal entries are 0 (over \mathbb{R})
Neither	A^T \neq \pm A	Still multiplies and adds as usual

If A and B are square of the same order, which identity is always true?

(AB)^T = A^T B^T
(AB)^T = B^T A^T
(AB)^T = AB

Transpose reverses the product: (AB)^T = B^T A^T. The order A^T B^T is the trap that keeps the factors in the original sequence.

2Determinant properties

For square matrices of the same order, |AB| = |A||B| and |A^T| = |A|. A scalar factors as |kA| = k^n|A| for an n\times n matrix. Swapping two rows (or columns) changes the sign of the determinant; a repeated row makes the determinant zero.

Row operations and the multiplicative property are algebraic rules on an array; the properties table carries them.

Using the scalar rule

Name nRead the order of A before touching k. For a 3\times3 matrix, every scalar factor contributes a cube.
Pull scalarsRewrite |kA| as k^n|A|. If several scalars sit outside nested expressions, apply the rule once per factor.
Row checksBefore expanding, glance for a repeated row or an obvious swap — either finishes the evaluation without cofactors.

Each row is a stored identity; the scalar row is the one that costs marks when n is ignored.

Determinant identities
Identity	Statement	Watch for
Product	\|AB\| = \|A\|\|B\|	Same order, both square
Transpose	\|A^T\| = \|A\|	Sign does not flip
Scalar	\|kA\| = k^n\|A\|	Exponent is the order n
Row swap	sign flips	One swap \Rightarrow one minus
Repeated row	\|A\| = 0	Two identical rows or columns

If A is 3\times3 and |A| = 5, what is |2A|?

|2A| = 2^3|A| = 8\times5 = 40. The distractor 10 pulls the scalar out to the first power; 80 is 16\times5, as if n were 4.

3Adjoint and inverse

The adjoint satisfies A\,\mathrm{adj}(A) = |A|I_n, so the inverse is A^{-1} = \dfrac{1}{|A|}\mathrm{adj}(A). The inverse exists only when the determinant is nonzero. Also |\mathrm{adj}(A)| = |A|^{n-1} and (AB)^{-1} = B^{-1}A^{-1}. For a 3\times3 matrix those adjoint exponents become |\mathrm{adj}(A)| = |A|^2 and |\mathrm{adj}(\mathrm{adj}(A))| = |A|^4.

Figure. With n=3 and |A|=5, |3A|=3^3\times5=135. Then |adj(3A)|=135^{2}=18225, and the outer factor of 2 contributes 2^3, so |2\,adj(3A)|=145800.

Evaluating a scaled adjoint

Absorb inner scalarsIf the matrix inside the adjoint is kA, replace it by B = kA and compute |B| = k^n|A| first.
Adjoint powerUse |\mathrm{adj}(B)| = |B|^{n-1}. For n = 3 that is a square.
Outer scalarA factor sitting outside the adjoint is another |cM| = c^n|M| on an n\times n matrix — apply n again, not 1.

Determinant of a scaled adjoint

A is a 3\times3 matrix with |A| = 5. Evaluate |2\,\mathrm{adj}(3A)|.

B = 3A: |B| = 3^3|A|27\times5 = 135
|\mathrm{adj}(B)| = |B|^{3-1} = 135^218225
|2\,\mathrm{adj}(B)| = 2^3\times18225145800

Pro tip. Apply |kM| = k^n|M| once for the 3A and again for the 2 outside — both use n = 3. For a 3\times3 matrix remember the adjoint exponents n-1 = 2 and (n-1)^2 = 4.

If A is 3\times3 and |A| = 2, what is |\mathrm{adj}(A)|?

|\mathrm{adj}(A)| = |A|^{n-1} = 2^{2} = 4. The distractor 2 forgets the power; 8 is |A|^3 or |A|\cdot n by mistake.

4Linear systems and Cramer

For AX = B with A square, a unique solution exists only when the coefficient determinant is nonzero; Cramer's rule then gives each unknown as x_i = \dfrac{|A_i|}{|A|}, where A_i replaces column i of A by B. If |A| = 0, the system has infinitely many solutions when (\mathrm{adj}\,A)B = O and no solution otherwise.

Figure. The coefficient determinant is |A|=-k. Unique solution needs |A|\neq0, so uniqueness fails exactly at k=0.

Testing uniqueness

Form |A|Write the coefficient determinant before expanding anything else — uniqueness is decided by whether it is zero.
ExpandUse a row or column with the parameter (or the most zeros). Record the simplified polynomial or linear expression in the parameter.
Read the conditionUnique solution needs |A| \neq 0. The roots of |A| = 0 are exactly the parameter values that lose uniqueness; consistency then needs the adjoint test.

Consistency of a linear system

For what value of k does x+y+z = 1,\ 2x+y+3z = 2,\ x+2y+kz = 3 fail to have a unique solution? Give the determinant condition.

Coefficient matrix determinant\begin{vmatrix}1&1&1\\2&1&3\\1&2&k\end{vmatrix}
Expand along first row: 1(k-6)-1(2k-3)+1(4-1)k-6-2k+3+3 = -k
Unique solution needs |A| \neq 0-k \neq 0, so fails at k = 0

Pro tip. Compute the coefficient determinant first: it is zero exactly when the system is not uniquely solvable. Only then spend time on (\mathrm{adj}\,A)B to separate infinite solutions from none.

For AX = B with A square, which statement is correct?

|A| = 0 always means no solution
|A| \neq 0 guarantees a unique solution
Cramer's rule still gives a unique x_i when |A| = 0

|A| \neq 0 is exactly the unique-solution case. When |A| = 0 the system may have infinitely many solutions or none, according to whether (\mathrm{adj}\,A)B = O; Cramer's quotients are undefined.

5Eigenvalues from the characteristic equation

Eigenvalues of a square matrix A are the roots of the characteristic equation |A - \lambda I| = 0. Their sum equals the trace and their product equals the determinant: \mathrm{tr}(A) = \sum\lambda_i and |A| = \prod\lambda_i. Those two checks catch an algebraic slip before it costs the whole question.

Figure. Eigenvalues 3 and -1 of a 2\times2 matrix must satisfy \mathrm{tr}(A)=3+(-1)=2 and |A|=3\times(-1)=-3. Those two checks catch a slip in the characteristic roots.

Reading eigenvalues

Form A - \lambda ISubtract \lambda from each diagonal entry; off-diagonal entries stay as in A.
Set determinant zeroExpand |A - \lambda I| = 0 to a polynomial in \lambda. The roots are the eigenvalues.
Check sum and productConfirm \sum\lambda_i = \mathrm{tr}(A) and \prod\lambda_i = |A| before moving on — a fast filter on the roots you just claimed.

A 2\times2 matrix has eigenvalues 3 and -1. Which pair must hold?

\mathrm{tr}(A) = 2 and |A| = -3
\mathrm{tr}(A) = 2 and |A| = 3
\mathrm{tr}(A) = -2 and |A| = -3

Sum 3 + (-1) = 2 = \mathrm{tr}(A); product 3\times(-1) = -3 = |A|. Option B flips the sign of the product; option C flips the sign of the sum.

Notes

Matrix algebra: Matrices add entrywise and multiply by the row-column rule, with A_{m\times n}B_{n\times p} defined only when inner dimensions match. Transpose satisfies (AB)^T=B^TA^T; a matrix is symmetric if A^T=A and skew-symmetric if A^T=-A.
Determinant properties: |AB|=|A||B|, |A^T|=|A| and |kA|=k^n|A| for an n\times n matrix. Swapping two rows changes the sign; a repeated row makes the determinant zero.
Adjoint and inverse: A\,\text{adj}(A)=|A|I_n, so A^{-1}=\dfrac{1}{|A|}\text{adj}(A) exists iff |A|\ne0. Also |\text{adj}(A)|=|A|^{n-1} and (AB)^{-1}=B^{-1}A^{-1}.
Systems of linear equations: For AX=B, a unique solution exists when |A|\ne0 (Cramer's rule x_i=\dfrac{|A_i|}{|A|}). If |A|=0, the system has infinitely many solutions when (\text{adj }A)B=O and no solution otherwise.
Eigenvalues: The characteristic equation |A-\lambda I|=0 gives eigenvalues whose sum equals the trace and whose product equals the determinant, \text{tr}(A)=\sum\lambda_i, |A|=\prod\lambda_i.
Cayley-Hamilton (every square matrix satisfies its own characteristic equation) lets you reduce high powers A^n to a linear combination of lower powers and I.
For a system AX=B: unique solution iff \det A\ne0; if \det A=0 examine (\mathrm{adj}A)B to distinguish 'no solution' from 'infinitely many'.
Rank via row-reduction determines consistency; a homogeneous system AX=0 has non-trivial solutions iff \det A=0.
Useful determinant facts: \det(\mathrm{adj}A)=(\det A)^{n-1} and \mathrm{adj}(\mathrm{adj}A)=(\det A)^{n-2}A for an n\times n matrix.

Formulas

Inverse: A^{-1}=\dfrac{1}{|A|}\text{adj}(A),\quad A\,\text{adj}(A)=|A|I_n
Scalar / product: |kA|=k^n|A|,\quad |AB|=|A||B|
Adjoint: |\text{adj}(A)|=|A|^{n-1},\quad (AB)^{-1}=B^{-1}A^{-1}
Cramer's rule: x_i=\dfrac{|A_i|}{|A|}
Characteristic: |A-\lambda I|=0
Eigenvalues: \text{tr}(A)=\sum\lambda_i,\quad |A|=\prod\lambda_i
A^{-1}=\dfrac{1}{\det A}\,\mathrm{adj}(A)
\det(\mathrm{adj}A) = (\det A)^{n-1}
Cayley-Hamilton: A satisfies \det(A-\lambda I)=0

Exam traps & shortcuts

For a 3\times3 matrix, |\text{adj}(A)|=|A|^2 and |\text{adj}(\text{adj}(A))|=|A|^4; remember the exponents n-1 and (n-1)^2.
When applying |kA|=k^n|A|, watch the dimension n: a scalar pulled out of a 3\times3 matrix comes out cubed, not to the first power.
The determinant and trace give a fast check on eigenvalues: their product must equal |A| and their sum the trace.
To find A^{100}, compute the characteristic equation, reduce A^2 (or A^3) to lower terms, then telescope.
For a 3\times3 system with a parameter, set \det A=0 to find the critical parameter, then test consistency separately.

Reference tables

The identities that recur on every JEE Main matrices paper, collected in one place.

Matrices and determinants — formula sheet
Name	Formula
Inverse	A^{-1} = \dfrac{1}{\|A\|}\mathrm{adj}(A), with A\,\mathrm{adj}(A) = \|A\|I_n
Scalar / product	\|kA\| = k^n\|A\|, \quad \|AB\| = \|A\|\|B\|
Adjoint size	\|\mathrm{adj}(A)\| = \|A\|^{n-1}; for n = 3, \|\mathrm{adj}(A)\| = \|A\|^2
Inverse of product	(AB)^{-1} = B^{-1}A^{-1}
Cramer	x_i = \dfrac{\|A_i\|}{\|A\|} when \|A\| \neq 0
Characteristic	\|A - \lambda I\| = 0
Eigenvalue checks	\mathrm{tr}(A) = \sum\lambda_i, \quad \|A\| = \prod\lambda_i

Recap

Read only this the night before.

Product order: (AB)^T = B^T A^T and (AB)^{-1} = B^{-1}A^{-1} — both reverse the factors.
Scalar power: |kA| = k^n|A|. On a 3\times3 matrix the scalar comes out cubed, twice if it appears twice.
Inverse gate: A^{-1} exists iff |A| \neq 0, and A\,\mathrm{adj}(A) = |A|I_n builds it.
Systems: Unique solution \Leftrightarrow |A| \neq 0. At |A| = 0, use (\mathrm{adj}\,A)B to separate infinite solutions from none.
Eigen check: Roots of |A - \lambda I| = 0 must sum to the trace and multiply to |A|.

Practise Matrices and Determinants

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