AP EAPCET (Engineering) · 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.
- AP EAPCET (Engineering)
- 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.
| 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.
| 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|?
- 10
- 40
- 80
|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)|?
- 2
- 4
- 8
|\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.
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
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.
Reference tables
The identities that recur on every JEE Main matrices paper, collected in one place.
| 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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