GATE EC · EE · IN

Linear algebra

Linear algebra is the study of matrices, the systems of linear equations they encode and the vector spaces they act on. Rank decides whether a system has no solution, exactly one or infinitely many, and eigenvalues and eigenvectors describe what a matrix does to directions in space. Typical problems in this topic are short computations (a determinant, a rank, the parameter value that makes a system consistent, an eigenvalue of $A^{- 1}$ or $A^{2} + 2 A$ ) that become quick once the standard properties below are known; the same eigenvalue results return in state-space analysis.

46 previous-year questions · 45 solved · 6 practice questions

1. Notation, matrix algebra, determinants and special matrices

Conventions used in these notes. $j$ is the imaginary unit ( $j^{2} = - 1$ ) and $\overset{z}{ˉ}$ is the complex conjugate of $z$ . Subscripts such as $i, j, k, p, q$ (as in $a_{ij}$ or $T_{ij}$ ) are always indices, never the imaginary unit. For a matrix $A$ : $A^{T}$ is the transpose, $\overset{ˉ}{A}$ the matrix of conjugates and $A^{*} = \overset{ˉ}{A}^{T}$ the conjugate transpose (other texts write $A^{θ}$ or $A^{H}$ ). $∣ A ∣$ and $det A$ both mean the determinant, $tr A$ is the trace (the sum of the diagonal entries), $I$ or $I_{n}$ is the $n \times n$ identity and $O$ is the zero matrix. Vectors are column vectors. Scalars are real unless a statement says complex.

An $m \times n$ matrix $A$ can multiply an $n \times p$ matrix $B$ , and $A B$ is $m \times p$ with $(A B)_{ik} = \sum_{j} a_{ij} b_{j k}$ . Multiplication is associative and distributive but not commutative: in general $A B \neq = B A$ , and $A B = O$ does not force $A = O$ or $B = O$ . A square matrix is non-singular (invertible) exactly when $∣ A ∣ \neq = 0$ ; its inverse is unique and $A A^{- 1} = A^{- 1} A = I$ . The cofactor of $a_{ij}$ is $C_{ij} = (- 1)^{i + j} M_{ij}$ , where the minor $M_{ij}$ is the determinant left after deleting row $i$ and column $j$ . The adjoint (adjugate) $adj A$ is the transpose of the matrix of cofactors.

REVERSAL LAWS

(A B)^{T} = B^{T} A^{T}, (A B)^{- 1} = B^{- 1} A^{- 1}, (A^{T})^{- 1} = (A^{- 1})^{T}, (A B)^{*} = B^{*} A^{*}

The order reverses for any number of factors, e.g. $(A B C)^{- 1} = C^{- 1} B^{- 1} A^{- 1}$ . The transpose and conjugate-transpose laws hold whenever $A B$ is defined, square or not; the inverse laws need every factor square and non-singular.

MORE INVERSE RULES

(A^{- 1})^{- 1} = A, (k A)^{- 1} = \frac{1}{k} A^{- 1}, (A^{m})^{- 1} = (A^{- 1})^{m}, (A^{*})^{- 1} = (A^{- 1})^{*}

$A$ non-singular, $k \neq = 0$ a scalar and $m$ a positive integer. A diagonal matrix with non-zero diagonal entries $d_{1}, \dots, d_{n}$ has inverse $diag (1/ d_{1}, \dots, 1/ d_{n})$ , and the inverse of a non-singular upper (lower) triangular matrix is again upper (lower) triangular.

INVERSE THROUGH THE ADJOINT

A (adj A) = (adj A) A = ∣ A ∣ I, A^{- 1} = \frac{adj A}{∣ A ∣}

The first identity holds for every square matrix; the inverse formula needs $∣ A ∣ \neq = 0$ .

INVERSE OF A 2 × 2 MATRIX

[a c b d]^{- 1} = \frac{1}{a d - b c} [d - c - b a]

Swap the diagonal entries, change the sign of the off-diagonal entries, divide by the determinant. Valid only when $a d - b c \neq = 0$ .

Inverse by row reduction (Gauss–Jordan)

Write the $n \times 2 n$ matrix $[A ∣ I]$ .
Apply elementary row operations only (never column operations) until the left block becomes $I$ . The right block is then $A^{- 1}$ .
If a row of zeros appears in the left block, $∣ A ∣ = 0$ and $A$ has no inverse.

TRACE PROPERTIES

tr (A + B) = tr A + tr B, tr (k A) = k tr A, tr (A B) = tr (B A), tr A^{T} = tr A

$tr (A B) = tr (B A)$ holds whenever both products are defined, even for non-square $A$ and $B$ . The trace is unchanged by cyclic shifts, $tr (A B C) = tr (B C A) = tr (C A B)$ , but not by other reorderings: in general $tr (A B C) \neq = tr (A C B)$ . In general $tr (A B) \neq = tr A tr B$ .

SYMMETRIC PLUS SKEW-SYMMETRIC SPLIT

A = \frac{1}{2} (A + A^{T}) + \frac{1}{2} (A - A^{T})

Every square matrix is, in exactly one way, a symmetric part plus a skew-symmetric part. Likewise $A = \frac{1}{2} (A + A^{*}) + \frac{1}{2} (A - A^{*})$ splits a complex matrix into Hermitian and skew-Hermitian parts.

Properties of determinants (for an n × n matrix)

$∣ A^{T} ∣ = ∣ A ∣$ , so every row property below holds for columns too.
Interchanging two rows changes the sign of the determinant.
Multiplying one row by $k$ multiplies the determinant by $k$ . Hence $∣ k A ∣ = k^{n} ∣ A ∣$ , not $k ∣ A ∣$ .
Two equal or proportional rows, or a row of zeros, give $∣ A ∣ = 0$ .
Adding a multiple of one row to another row leaves the determinant unchanged. This is the operation to use when simplifying by hand.
If one row is a sum, the determinant splits: $a_{1} + b_{1} c_{1} a_{2} + b_{2} c_{2} = a_{1} c_{1} a_{2} c_{2} + b_{1} c_{1} b_{2} c_{2}$ . In general $∣ A + B ∣ \neq = ∣ A ∣ + ∣ B ∣$ .
Triangular (upper or lower) or diagonal matrix: the determinant is the product of the diagonal entries. Block-triangular: $det [A O B D] = det A \cdot det D$ for square blocks $A$ and $D$ .
A skew-symmetric matrix of odd order has determinant $0$ , because $∣ A ∣ = ∣ A^{T} ∣ = ∣ - A ∣ = (- 1)^{n} ∣ A ∣$ .
The determinant equals the product of the eigenvalues (Section 5).

DETERMINANTS OF PRODUCTS, POWERS AND MULTIPLES

∣ A B ∣ = ∣ A ∣ ∣ B ∣, ∣ A^{m} ∣ = ∣ A ∣^{m}, ∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}, ∣ k A ∣ = k^{n} ∣ A ∣

$A$ and $B$ are $n \times n$ and $k$ is a scalar. $∣ A^{m} ∣ = ∣ A ∣^{m}$ holds for every positive integer $m$ (for every integer $m$ when $A$ is non-singular), and $∣ A^{- 1} ∣ = 1/∣ A ∣$ needs $∣ A ∣ \neq = 0$ . $∣ A B ∣ = ∣ B A ∣$ even when $A B \neq = B A$ .

ADJOINT IDENTITIES

∣ adj A ∣ = ∣ A ∣^{n - 1}, adj (adj A) = ∣ A ∣^{n - 2} A, adj (A B) = (adj B) (adj A), adj (k A) = k^{n - 1} adj A

For $n \times n$ matrices with $n \geq 2$ ; all four hold for singular matrices too. The first follows from taking determinants of $A adj A = ∣ A ∣ I$ , which gives $∣ A ∣ ∣ adj A ∣ = ∣ A ∣^{n}$ . Consequently $∣ adj (adj A) ∣ = ∣ A ∣^{(n - 1)^{2}}$ . Also $adj (A^{T}) = (adj A)^{T}$ , and for non-singular $A$ , $adj (A^{- 1}) = (adj A)^{- 1} = A /∣ A ∣$ .

Special matrices

Matrix	Defining property	Consequences worth remembering
Symmetric	$A^{T} = A$	For real $A$ : eigenvalues real, eigenvectors of distinct eigenvalues orthogonal, always diagonalisable
Skew-symmetric	$A^{T} = - A$	Diagonal entries $0$ ; $∣ A ∣ = 0$ for odd order; for real $A$ , eigenvalues $0$ or purely imaginary
Orthogonal (real)	$A^{T} A = A A^{T} = I$	$A^{- 1} = A^{T}$ ; $∣ A ∣ = \pm 1$ ; columns (and rows) orthonormal; every eigenvalue has $∣ λ ∣ = 1$
Hermitian	$A^{*} = A$	Diagonal entries real; eigenvalues real; a real Hermitian matrix is symmetric
Skew-Hermitian	$A^{*} = - A$	Diagonal entries $0$ or purely imaginary; eigenvalues $0$ or purely imaginary
Unitary	$A^{} A = A A^{} = I$	$A^{- 1} = A^{*}$ ; $∣ det A ∣ = 1$ ; every eigenvalue has $∣ λ ∣ = 1$
Idempotent	$A^{2} = A$	Eigenvalues $0$ or $1$ ; $rank A = tr A$ ; $I - A$ is also idempotent
Nilpotent	$A^{k} = O$ for some positive integer $k$ (the least such $k$ is the index)	All eigenvalues $0$ ; $∣ A ∣ = 0$ and $tr A = 0$ ; $I - A$ is invertible
Involutory	$A^{2} = I$	$A^{- 1} = A$ ; eigenvalues $\pm 1$ ; $∣ A ∣ = \pm 1$
Triangular or diagonal	$a_{ij} = 0$ for $i > j$ (upper) or for $i < j$ (lower)	Determinant and eigenvalues read straight off the diagonal

PRODUCTS DO NOT COMMUTE

$(A + B)^{2} = A^{2} + A B + B A + B^{2}$ , which equals $A^{2} + 2 A B + B^{2}$ only when $A B = B A$ . Likewise $(A B)^{T} = B^{T} A^{T}$ , not $A^{T} B^{T}$ . The product of two symmetric matrices is symmetric exactly when they commute, because $(A B)^{T} = B^{T} A^{T} = B A$ .

SCALAR MULTIPLES INSIDE A DETERMINANT

For a $3 \times 3$ matrix with $∣ A ∣ = 4$ : $∣2 A ∣ = 2^{3} \times 4 = 32$ , $∣ - A ∣ = (- 1)^{3} \times 4 = - 4$ , $∣ adj A ∣ = 4^{2} = 16$ and $∣2 A^{- 1} ∣ = 2^{3} /4 = 2$ . Writing $∣2 A ∣ = 8$ is the usual slip.

2. Vector spaces, linear independence, basis and dimension

A vector space $V$ over the real numbers (or over the complex numbers) is a set in which vectors can be added and multiplied by scalars with the results staying in $V$ , and in which the usual rules hold: there is a zero vector, every vector has an additive inverse, $1 v = v$ , and the associative, commutative and distributive laws apply. $R^{n}$ , the set of all $m \times n$ matrices and the set of polynomials of degree at most $n$ are all vector spaces. A subspace is a non-empty subset $W$ that is itself a vector space under the same operations.

Subspace test

A subset $W$ of $V$ is a subspace if and only if all three hold:

The zero vector of $V$ lies in $W$ . If it does not, stop: $W$ is not a subspace.
$u, v \in W \Rightarrow u + v \in W$ (closed under addition).
$u \in W$ and $c$ any scalar $\Rightarrow c u \in W$ (closed under scalar multiplication). The last two together are equivalent to $a u + b v \in W$ for all scalars $a, b$ and all $u, v \in W$ .

Typical verdicts: the solution set of a homogeneous system $A X = 0$ is a subspace (the null space of $A$ ); the solution set of $A X = B$ with $B \neq = 0$ is not, since it misses the zero vector. In $R^{3}$ the subspaces are ${0}$ , lines through the origin, planes through the origin and $R^{3}$ itself. The intersection of two subspaces is always a subspace; their union is a subspace only when one of them contains the other.

The span of $v_{1}, \dots, v_{k}$ is the set of all linear combinations $c_{1} v_{1} + \dots + c_{k} v_{k}$ ; it is always a subspace. The vectors are linearly independent if $c_{1} v_{1} + c_{2} v_{2} + \dots + c_{k} v_{k} = 0$ forces $c_{1} = c_{2} = \dots = c_{k} = 0$ , and linearly dependent otherwise, in which case at least one of them is a combination of the others. A basis of $V$ is a linearly independent set that spans $V$ . Every basis of a finite-dimensional space $V$ has the same number of vectors, the dimension $dim V$ .

INNER PRODUCT, LENGTH AND ORTHOGONALITY

u \cdot v = u^{T} v = i = 1 \sum n u_{i} v_{i}, ∥ u ∥ = u^{T} u, u ⊥ v ⟺ u^{T} v = 0

For real vectors in $R^{n}$ . For complex vectors use $u^{*} v$ and $∥ u ∥ = u^{*} u$ . A set of vectors is orthonormal if they are mutually orthogonal and each has length $1$ . In an orthonormal basis $e_{1}, \dots, e_{n}$ the coordinates of any $v$ are simply $c_{i} = e_{i}^{T} v$ .

Tests for linear independence

Write the vectors as the rows (or the columns) of a matrix. They are independent if and only if its rank equals the number of vectors.
$n$ vectors in $R^{n}$ : independent if and only if the $n \times n$ determinant they form is non-zero.
More than $n$ vectors in an $n$ -dimensional space are always dependent.
Any set that contains the zero vector is dependent.
Two vectors are dependent if and only if one is a scalar multiple of the other.
Non-zero, mutually orthogonal vectors are independent, and eigenvectors belonging to distinct eigenvalues are independent.
In an $n$ -dimensional space, any $n$ independent vectors form a basis, and any $n$ vectors that span the space form a basis.

GRAM–SCHMIDT ORTHOGONALISATION

u_{1} = v_{1}, u_{k} = v_{k} - i = 1 \sum k - 1 \frac{u _{i}^{T} v _{k}}{u _{i}^{T} u _{i}} u_{i}, e_{k} = \frac{u _{k}}{∥ u _{k} ∥}

Turns independent real vectors $v_{1}, \dots, v_{m}$ into mutually orthogonal vectors $u_{1}, \dots, u_{m}$ , and orthonormal vectors $e_{1}, \dots, e_{m}$ , spanning the same subspace. Each step subtracts from $v_{k}$ its projections on the vectors already built. Getting $u_{k} = 0$ at some step means $v_{k}$ was a combination of the earlier vectors.

DIMENSION OF A SUM OF SUBSPACES

dim (U + W) = dim U + dim W - dim (U \cap W)

$U$ and $W$ are subspaces of a finite-dimensional space and $U + W$ is the set of all sums $u + w$ . Two distinct planes through the origin in $R^{3}$ meet in a line: $3 = 2 + 2 - 1$ .

Dimensions of common spaces

Space	Dimension	Basis or remark
$R^{n}$	$n$	standard basis $e_{1}, \dots, e_{n}$
All real $m \times n$ matrices	$mn$	matrices with a single entry $1$ and the rest $0$
$n \times n$ symmetric matrices	$\frac{n ( n + 1 )}{2}$	choose the diagonal and the entries above it
$n \times n$ skew-symmetric matrices	$\frac{n ( n - 1 )}{2}$	diagonal forced to $0$ ; choose the entries above it
$n \times n$ upper (or lower) triangular matrices	$\frac{n ( n + 1 )}{2}$	same count as the symmetric matrices
$n \times n$ diagonal matrices	$n$	one free entry per diagonal position
$n \times n$ matrices with trace $0$	$n^{2} - 1$	one linear condition on $n^{2}$ entries
Polynomials of degree $\leq n$	$n + 1$	$1, x, x^{2}, \dots, x^{n}$
Null space of an $m \times n$ matrix $A$	$n - rank A$	rank-nullity theorem (Section 3)
$C^{n}$	$n$ over $C$ ; $2 n$ over $R$	the field of scalars matters

CHECK THE ZERO VECTOR FIRST

${(x, y, z) : x + 2 y - z = 3}$ fails at once because $(0, 0, 0)$ is not in it. A condition that is not linear, such as $x y = 0$ , usually fails closure instead: $(1, 0, 0)$ and $(0, 1, 0)$ both satisfy $x y = 0$ , but their sum $(1, 1, 0)$ does not.

3. Rank of a matrix and the rank-nullity theorem

The rank of an $m \times n$ matrix $A$ has four equivalent descriptions: the maximum number of linearly independent rows; the maximum number of linearly independent columns; the order of the largest square submatrix with a non-zero determinant (the largest non-vanishing minor); and the number of non-zero rows in any row echelon form of $A$ . Row rank always equals column rank, so $rank A = rank A^{T}$ . Elementary row or column operations do not change the rank. The zero matrix has rank $0$ ; every other matrix has rank at least $1$ .

Finding the rank by row reduction

Use elementary row operations (swap two rows, multiply a row by a non-zero constant, add a multiple of one row to another) to reach row echelon form: zero rows at the bottom, and the first non-zero entry (pivot) of each row to the right of the pivot in the row above.
Count the non-zero rows. That count is the rank.
Shortcut for a square matrix: if $∣ A ∣ \neq = 0$ the rank is $n$ and no reduction is needed. If $∣ A ∣ = 0$ , look for a non-zero minor of order $n - 1$ , and so on down.
A matrix whose rows are all multiples of one non-zero row has rank $1$ . Every rank-one matrix can be written as $u v^{T}$ with non-zero column vectors $u$ and $v$ .

BOUNDS ON RANK

0 \leq rank A \leq min (m, n), rank A^{T} = rank A

For an $m \times n$ matrix. A square $n \times n$ matrix has rank $n$ exactly when $∣ A ∣ \neq = 0$ .

RANK OF A PRODUCT (SYLVESTER'S INEQUALITY ON THE LEFT)

rank A + rank B - n \leq rank (A B) \leq min (rank A, rank B)

$A$ is $m \times n$ and $B$ is $n \times p$ , so $n$ is the shared inner dimension. Multiplying by a non-singular matrix does not change the rank: $rank (P A) = rank (A Q) = rank A$ for invertible $P$ and $Q$ .

RANK OF A SUM

∣ rank A - rank B ∣ \leq rank (A + B) \leq rank A + rank B

$A$ and $B$ of the same size.

RANK-NULLITY THEOREM

rank A + nullity A = n

$n$ is the number of columns of $A$ (the number of unknowns in $A X = 0$ ), and $nullity A$ is the dimension of the solution space of $A X = 0$ .

The four subspaces of a matrix. For a real $m \times n$ matrix $A$ of rank $r$ : the column space (the set of all products $A X$ ) is a subspace of $R^{m}$ of dimension $r$ ; the row space is a subspace of $R^{n}$ of dimension $r$ ; the null space (all $X$ with $A X = 0$ ) is a subspace of $R^{n}$ of dimension $n - r$ ; and the null space of $A^{T}$ is a subspace of $R^{m}$ of dimension $m - r$ . $A X = B$ is consistent exactly when $B$ lies in the column space of $A$ . In the language of linear maps, $T (X) = A X$ from $R^{n}$ to $R^{m}$ has range equal to the column space and kernel equal to the null space, so rank-nullity reads $dim (range) + dim (kernel) = n$ . $T$ is one-to-one exactly when $r = n$ , and onto exactly when $r = m$ .

RANK OF THE GRAM MATRIX

rank (A^{T} A) = rank (A A^{T}) = rank A (A real)

For complex $A$ the same holds with $A^{*}$ in place of $A^{T}$ . So the $n \times n$ matrix $A^{T} A$ is invertible exactly when the $n$ columns of $A$ are independent.

RANK OF THE ADJOINT

rank (adj A) = ⎩ ⎨ ⎧ n, 1, 0, rank A = n rank A = n - 1 rank A \leq n - 2

For an $n \times n$ matrix with $n \geq 2$ . It follows from $A adj A = ∣ A ∣ I$ and the fact that every cofactor is a minor of order $n - 1$ .

RANK IS NOT THE COUNT OF NON-ZERO EIGENVALUES

For a diagonalisable matrix (in particular any real symmetric matrix) the rank equals the number of non-zero eigenvalues counted with multiplicity. For other matrices the rank can be larger, never smaller: $[0010]$ has rank $1$ but both eigenvalues are $0$ . (In general $n - rank A$ is the geometric multiplicity of the eigenvalue $0$ , which cannot exceed its algebraic multiplicity; see Section 5.) What always holds is that $A$ is singular exactly when $0$ is an eigenvalue.

4. Systems of linear equations: existence and uniqueness

Write $m$ equations in $n$ unknowns as $A X = B$ , where $A$ is the $m \times n$ coefficient matrix and $[A ∣ B]$ is the augmented matrix ( $A$ with $B$ appended as an extra column). The system is consistent if it has at least one solution. Reduce $[A ∣ B]$ to row echelon form: the rank of its first $n$ columns is $rank A$ , and the rank of the whole is $rank [A ∣ B]$ , which is either $rank A$ or $rank A + 1$ .

CONSISTENCY CONDITION (ROUCHÉ–CAPELLI THEOREM)

A X = B is consistent ⟺ rank A = rank [A ∣ B]

When consistent, the solution is unique if this common rank equals the number of unknowns $n$ , and has $n - r$ free parameters if the common rank is $r < n$ .

Rank test for AX = B in n unknowns

Ranks	Nature of the solution
$rank A < rank [A ∣ B]$	Inconsistent: no solution
$rank A = rank [A ∣ B] = n$	Consistent: unique solution
$rank A = rank [A ∣ B] = r < n$	Consistent: infinitely many solutions, with $n - r$ free parameters

Homogeneous systems $A X = 0$ are always consistent, because the trivial solution $X = 0$ satisfies them. Non-trivial solutions exist if and only if $rank A < n$ , and the solutions then form a subspace of dimension $n - rank A$ . For a square system this means $∣ A ∣ = 0$ . With fewer equations than unknowns ( $m < n$ ), $rank A \leq m < n$ , so non-trivial solutions always exist.

NON-TRIVIAL SOLUTIONS OF A SQUARE HOMOGENEOUS SYSTEM

A X = 0 has a non-trivial solution ⟺ ∣ A ∣ = 0 ⟺ 0 is an eigenvalue of A

$A$ is $n \times n$ . The number of linearly independent solutions is $n - rank A$ .

Square systems. If $∣ A ∣ \neq = 0$ , the system $A X = B$ has exactly one solution, $X = A^{- 1} B$ , for every $B$ . If $∣ A ∣ = 0$ , it has either no solution or infinitely many, never exactly one; the rank test decides which. More generally, a linear system has $0$ , $1$ or infinitely many solutions: if $X_{1} \neq = X_{2}$ are two solutions, then $X_{1} + t (X_{2} - X_{1})$ is a solution for every scalar $t$ .

CRAMER'S RULE

x_{i} = \frac{Δ _{i}}{Δ}, Δ = ∣ A ∣ \neq = 0

$Δ_{i}$ is the determinant of $A$ with its $i$ -th column replaced by $B$ . Applies only to a square system with $Δ \neq = 0$ . If $Δ = 0$ and some $Δ_{i} \neq = 0$ , the system is inconsistent.

WORKED EXAMPLE · CONSISTENCY WITH TWO PARAMETERS

For the system $x + 2 y + z = 4$ , $2 x + y + 3 z = 5$ , $3 x + 3 y + λ z = μ$ , find the values of $λ$ and $μ$ for which it has (i) a unique solution, (ii) no solution, (iii) infinitely many solutions, and write the solution in case (iii).

Solution

Augmented matrix: $123213 13 λ 45 μ$ . $R_{2} \to R_{2} - 2 R_{1}$ and $R_{3} \to R_{3} - 3 R_{1}$ give the rows $[0 - 3 1 - 3]$ and $[0 - 3 λ - 3 μ - 12]$ .
$R_{3} \to R_{3} - R_{2}$ gives the echelon form $100 2 - 3 0 11 λ - 4 4 - 3 μ - 9$ .
(i) If $λ \neq = 4$ : $rank A = rank [A ∣ B] = 3 = n$ , so there is a unique solution for every $μ$ . (Check: $∣ A ∣ = - 3 (λ - 4)$ , non-zero exactly when $λ \neq = 4$ .)
(ii) If $λ = 4$ and $μ \neq = 9$ : the last row reads $0 = μ - 9 \neq = 0$ , so $rank A = 2 < 3 = rank [A ∣ B]$ and there is no solution.
(iii) If $λ = 4$ and $μ = 9$ : both ranks are $2 < 3$ , so there are infinitely many solutions with $3 - 2 = 1$ free parameter. Put $y = t$ . The second row, $- 3 y + z = - 3$ , gives $z = 3 t - 3$ , and the first gives $x = 4 - 2 t - z = 7 - 5 t$ .
Check in the third equation: $3 (7 - 5 t) + 3 t + 4 (3 t - 3) = 21 - 12 = 9$ for every $t$ .

Answer: Unique solution for $λ \neq = 4$ (any $μ$ ); no solution for $λ = 4$ , $μ \neq = 9$ ; infinitely many for $λ = 4$ , $μ = 9$ , namely $(x, y, z) = (7 - 5 t, t, 3 t - 3)$ with $t$ real.

CRAMER'S RULE CANNOT CERTIFY INFINITELY MANY SOLUTIONS

When $Δ = 0$ and every $Δ_{i} = 0$ , the system may have infinitely many solutions or none: $x + y + z = 1$ , $x + y + z = 2$ , $x + y + z = 3$ has $Δ = Δ_{1} = Δ_{2} = Δ_{3} = 0$ (each determinant has two equal columns) and no solution. Settle the $Δ = 0$ case with ranks.

COMPARE THE RANK WITH THE NUMBER OF UNKNOWNS

Uniqueness needs the common rank to equal the number of unknowns $n$ , not the number of equations. Four equations in three unknowns can still have a unique solution (both ranks $3$ ), and a homogeneous system with more unknowns than equations always has non-trivial solutions.

5. Eigenvalues and eigenvectors

A scalar $λ$ is an eigenvalue of the $n \times n$ matrix $A$ if $A X = λ X$ for some non-zero vector $X$ , called an eigenvector for $λ$ . Rewritten as $(A - λ I) X = 0$ , a non-zero solution exists exactly when $∣ A - λ I ∣ = 0$ , the characteristic equation, which is a polynomial equation of degree $n$ . It has $n$ roots counted with multiplicity, possibly complex even when $A$ is real. The eigenvectors for $λ$ are the non-zero solutions of $(A - λ I) X = 0$ ; together with the zero vector they form the eigenspace of $λ$ .

Convention. An eigenvector is fixed only up to a non-zero scalar multiple: if $X$ is an eigenvector, so is $c X$ for every $c \neq = 0$ , so $(1, - 1)^{T}$ and $(- 2, 2)^{T}$ are the same answer. The zero vector is never an eigenvector, but $λ = 0$ can be an eigenvalue, and it is one exactly when $A$ is singular.

CHARACTERISTIC EQUATION OF A 2 × 2 MATRIX

λ^{2} - (tr A) λ + ∣ A ∣ = 0

For $A = [a c b d]$ : $λ^{2} - (a + d) λ + (a d - b c) = 0$ , so $λ = \frac{( a + d ) \pm ( a - d ) ^{2} + 4 b c}{2}$ . The eigenvalues are real whenever $b c \geq 0$ , in particular whenever $A$ is real symmetric.

CHARACTERISTIC EQUATION OF A 3 × 3 MATRIX

λ^{3} - (tr A) λ^{2} + (C_{11} + C_{22} + C_{33}) λ - ∣ A ∣ = 0

This is $∣ A - λ I ∣ = 0$ multiplied through by $- 1$ . $C_{11} + C_{22} + C_{33}$ is the sum of the three principal $2 \times 2$ minors (the cofactors of the diagonal entries); it also equals $λ_{1} λ_{2} + λ_{2} λ_{3} + λ_{3} λ_{1}$ .

SUM AND PRODUCT OF EIGENVALUES

λ_{1} + λ_{2} + \dots + λ_{n} = tr A, λ_{1} λ_{2} \dots λ_{n} = ∣ A ∣

Each eigenvalue is counted as often as its algebraic multiplicity, complex ones included. More generally $tr (A^{m}) = λ_{1}^{m} + λ_{2}^{m} + \dots + λ_{n}^{m}$ for every positive integer $m$ ; for instance $tr (A^{2})$ is the sum of the squares of the eigenvalues. A quick check on any eigenvalue calculation.

Eigenvalues of matrices built from A (eigenvalue λ, eigenvector X)

Matrix	Eigenvalue	Eigenvector	Condition
$k A$	$k λ$	$X$	any scalar $k$
$A + k I$	$λ + k$	$X$	any scalar $k$
$A^{m}$	$λ^{m}$	$X$	$m$ a positive integer
$A^{- 1}$	$\frac{1}{λ}$	$X$	$A$ non-singular (so $λ \neq = 0$ )
$(A - k I)^{- 1}$	$\frac{1}{λ - k}$	$X$	$k$ not an eigenvalue of $A$
$adj A$	$\frac{∣ A ∣}{λ}$	$X$	$A$ non-singular
$p (A) = a_{0} I + a_{1} A + \dots + a_{m} A^{m}$	$p (λ)$	$X$	any polynomial $p$
$A^{T}$	$λ$	in general different	same characteristic polynomial as $A$
$\overset{ˉ}{A}$	$\overset{ˉ}{λ}$	$\overset{ˉ}{X}$	complex $A$
$A^{*}$	$\overset{ˉ}{λ}$	in general different	complex $A$
$P^{- 1} A P$	$λ$	$P^{- 1} X$	$P$ non-singular (similar matrices)
$A B$ and $B A$	the same eigenvalues for both, with the same multiplicities	in general different	$A$ and $B$ both $n \times n$ ( $A B$ and $B A$ have the same characteristic polynomial)

Eigenvalues of special matrices

Matrix	Eigenvalues
Triangular or diagonal	the diagonal entries
Block triangular $[A_{1} O B A_{2}]$ , with $A_{1}$ and $A_{2}$ square	the eigenvalues of $A_{1}$ together with those of $A_{2}$
Real symmetric, or Hermitian	all real; eigenvectors of distinct eigenvalues are orthogonal
Real skew-symmetric, or skew-Hermitian	$0$ or purely imaginary ( $\pm j β$ )
Orthogonal, or unitary	modulus $1$ , $∣ λ ∣ = 1$ ; the real ones are $\pm 1$
Idempotent ( $A^{2} = A$ )	$0$ or $1$
Nilpotent	all $0$
Involutory ( $A^{2} = I$ )	$+ 1$ or $- 1$
Rank one, $A = u v^{T}$ ( $n \times n$ , $u$ and $v$ non-zero)	$v^{T} u = tr A$ once, and $0$ repeated $n - 1$ times
All entries equal to $1$ ( $n \times n$ )	$n$ once, and $0$ repeated $n - 1$ times
Every row sums to the same number $s$	$s$ is an eigenvalue, with eigenvector $(1, 1, \dots, 1)^{T}$ ; if instead every column sums to $s$ , $s$ is still an eigenvalue (apply the row rule to $A^{T}$ ), but that eigenvector need not work
Any real matrix	non-real eigenvalues occur in conjugate pairs $α \pm j β$ ; a real matrix of odd order has at least one real eigenvalue

Algebraic and geometric multiplicity. The algebraic multiplicity (AM) of $λ$ is its multiplicity as a root of the characteristic equation. The geometric multiplicity (GM) is the number of linearly independent eigenvectors for $λ$ , that is, the dimension of its eigenspace.

GEOMETRIC MULTIPLICITY

GM (λ) = n - rank (A - λ I), 1 \leq GM (λ) \leq AM (λ)

An eigenvalue with AM $= 1$ (a simple root) always has GM $= 1$ . A repeated eigenvalue can have any GM from $1$ up to its AM.

WORKED EXAMPLE · EIGENVALUES, MULTIPLICITY AND A POLYNOMIAL IN A

Let $A = 311131113$ . Find the eigenvalues of $A$ , an eigenvector for the largest one, the geometric multiplicity of the repeated eigenvalue, and the trace and determinant of $B = A^{2} - 4 A + I$ .

Solution

Write $A = 2 I + J$ , where $J$ is the $3 \times 3$ matrix of all ones. $J$ has rank $1$ , so its eigenvalues are $tr J = 3$ and $0, 0$ .
Adding $2 I$ adds $2$ to each eigenvalue: $A$ has eigenvalues $5, 2, 2$ . Check: $5 + 2 + 2 = 9 = tr A$ , and $5 \times 2 \times 2 = 20 = ∣ A ∣$ (direct expansion: $3 (9 - 1) - 1 (3 - 1) + 1 (1 - 3) = 20$ ).
For $λ = 5$ : $A - 5 I$ has rows $(- 2, 1, 1)$ , $(1, - 2, 1)$ and $(1, 1, - 2)$ . Each row sums to zero, so $X = (1, 1, 1)^{T}$ (or any non-zero multiple) is an eigenvector.
For $λ = 2$ : $A - 2 I = J$ has rank $1$ , so GM $= 3 - 1 = 2$ , equal to its AM. The eigenspace is the plane $x_{1} + x_{2} + x_{3} = 0$ , spanned for instance by $(1, - 1, 0)^{T}$ and $(1, 0, - 1)^{T}$ , and $A$ is diagonalisable.
With $p (λ) = λ^{2} - 4 λ + 1$ , $B = p (A)$ has eigenvalues $p (5) = 6$ and $p (2) = - 3$ (twice).
$tr B = 6 - 3 - 3 = 0$ and $∣ B ∣ = 6 \times (- 3) \times (- 3) = 54$ .

Answer: Eigenvalues $5, 2, 2$ ; eigenvector $(1, 1, 1)^{T}$ for $λ = 5$ ; GM of $λ = 2$ is $2$ ; $tr B = 0$ and $∣ B ∣ = 54$ .

DO NOT ROW-REDUCE BEFORE FINDING EIGENVALUES

Row operations preserve rank but change eigenvalues. $[1221]$ has eigenvalues $3$ and $- 1$ ; after $R_{2} \to R_{2} - 2 R_{1}$ it becomes $[10 2 - 3]$ , with eigenvalues $1$ and $- 3$ . Solve $∣ A - λ I ∣ = 0$ on the original matrix.

SUMS AND PRODUCTS OF MATRICES

In general the eigenvalues of $A + B$ are not sums, and those of $A B$ are not products, of the eigenvalues of $A$ and $B$ . The rules in the table work because $k A$ , $A^{m}$ , $A^{- 1}$ and $p (A)$ share the eigenvector $X$ of $A$ .

A MATRIX EQUATION LIMITS THE POSSIBLE EIGENVALUES

If $p (A) = O$ for a polynomial $p$ , every eigenvalue satisfies $p (λ) = 0$ (apply both sides to an eigenvector). From $A^{2} - 5 A + 6 I = O$ , each eigenvalue is $2$ or $3$ . The converse fails: which of those roots actually occur has to be decided separately.

ROTATION MATRICES

The orthogonal matrix $[cos θ sin θ - sin θ cos θ]$ has trace $2 cos θ$ and determinant $1$ , so its eigenvalues are $cos θ \pm j sin θ$ , of modulus $1$ . They are real only for $θ = 0$ or $π$ (in $[0, 2 π)$ ).

6. Diagonalisation and the Cayley–Hamilton theorem

An $n \times n$ matrix $A$ is diagonalisable if $P^{- 1} A P = D$ for some non-singular $P$ and diagonal $D$ . This happens exactly when $A$ has $n$ linearly independent eigenvectors: the columns of $P$ are those eigenvectors, and the diagonal of $D$ lists the matching eigenvalues in the same order. Here $P$ and $D$ may be complex: a real matrix with non-real eigenvalues, such as a rotation, can be diagonalised only with complex $P$ and $D$ . Similar matrices ( $B = P^{- 1} A P$ ) have the same characteristic polynomial, hence the same eigenvalues with the same multiplicities, the same trace and the same determinant; they also have the same rank, because multiplying by a non-singular matrix preserves rank. The converse fails: $[2012]$ and $2 I$ share the characteristic polynomial $(λ - 2)^{2}$ but are not similar, since $P^{- 1} (2 I) P = 2 I$ for every $P$ .

When is a matrix diagonalisable?

Necessary and sufficient: $GM (λ) = AM (λ)$ for every eigenvalue; equivalently, the GMs add up to $n$ .
Sufficient: $n$ distinct eigenvalues. This is not necessary: $I$ has a single repeated eigenvalue and is already diagonal.
Sufficient: $A$ real symmetric. The eigenvectors can then be chosen orthonormal, $P$ is orthogonal and $P^{T} A P = D$ (orthogonal diagonalisation). A Hermitian matrix is likewise unitarily diagonalisable, $U^{*} A U = D$ .
Fails whenever some GM $<$ AM. The standard case is $[a 0 1 a]$ : $λ = a$ has AM $= 2$ , but $rank (A - a I) = 1$ , so GM $= 1$ .
A non-zero nilpotent matrix is never diagonalisable: its only eigenvalue is $0$ , so $D = O$ would force $A = P D P^{- 1} = O$ .

POWERS AND FUNCTIONS THROUGH DIAGONALISATION

A = P D P^{- 1} \Rightarrow A^{k} = P D^{k} P^{- 1}, f (A) = P f (D) P^{- 1}

$f (D) = diag (f (λ_{1}), \dots, f (λ_{n}))$ acts entry by entry on the diagonal. $f$ is a polynomial, or a convergent power series such as the exponential: $e^{A t} = P diag (e^{λ_{1} t}, \dots, e^{λ_{n} t}) P^{- 1}$ . For a real symmetric $A$ , take orthonormal eigenvectors as the columns of $P$ (normalise each one and, inside the eigenspace of a repeated eigenvalue, also make them orthogonal); then $P^{- 1} = P^{T}$ . Normalising alone is not enough when an eigenvalue repeats: $(1, - 1, 0)^{T}$ and $(1, 0, - 1)^{T}$ in the Section 5 example are not orthogonal.

Cayley–Hamilton theorem. Every square matrix satisfies its own characteristic equation. If the characteristic polynomial is $p (λ) = det (λ I - A) = λ^{n} + c_{n - 1} λ^{n - 1} + \dots + c_{1} λ + c_{0}$ , then $p (A) = O$ , with the constant term read as $c_{0} I$ . The theorem reduces any polynomial in $A$ to one of degree at most $n - 1$ , and gives $A^{- 1}$ without cofactors.

CAYLEY–HAMILTON THEOREM

A^{n} + c_{n - 1} A^{n - 1} + \dots + c_{1} A + c_{0} I = O, c_{n - 1} = - tr A, c_{0} = (- 1)^{n} ∣ A ∣

Using $∣ A - λ I ∣$ instead of $det (λ I - A)$ only multiplies the polynomial by $(- 1)^{n}$ , which does not change the conclusion.

INVERSE FROM CAYLEY–HAMILTON

A^{- 1} = - \frac{1}{c _{0}} (A^{n - 1} + c_{n - 1} A^{n - 2} + \dots + c_{1} I)

Multiply the Cayley–Hamilton identity by $A^{- 1}$ . Needs $c_{0} \neq = 0$ , that is, $∣ A ∣ \neq = 0$ .

THE 2 × 2 CASE

A^{2} - (tr A) A + ∣ A ∣ I = O, A^{- 1} = \frac{( tr A ) I - A}{∣ A ∣}

For any $2 \times 2$ matrix; the inverse needs $∣ A ∣ \neq = 0$ .

HIGHER POWERS BY THE REMAINDER METHOD

λ^{m} = q (λ) p (λ) + r (λ), de g r \leq n - 1 \Rightarrow A^{m} = r (A)

Here $p$ is the characteristic polynomial (degree $n$ ), and the result follows because $p (A) = O$ . With distinct eigenvalues, find the $n$ coefficients of $r$ from $r (λ_{i}) = λ_{i}^{m}$ . For an eigenvalue of algebraic multiplicity $k$ , also match the first $k - 1$ derivatives; for a double root this is $r^{'} (λ_{i}) = m λ_{i}^{m - 1}$ .

The minimal polynomial of $A$ is the monic polynomial $m (λ)$ of least degree with $m (A) = O$ . It divides the characteristic polynomial and has every eigenvalue as a root; $A$ is diagonalisable (over the complex numbers) exactly when $m (λ)$ has no repeated root. For $[2012]$ it is $(λ - 2)^{2}$ , and for $2 I$ it is $λ - 2$ .

WORKED EXAMPLE · INVERSE AND A POWER BY CAYLEY–HAMILTON

For $A = [4213]$ , use the Cayley–Hamilton theorem to find $A^{- 1}$ and $A^{4}$ .

Solution

$tr A = 7$ and $∣ A ∣ = 12 - 2 = 10$ , so the characteristic equation is $λ^{2} - 7 λ + 10 = 0$ , with roots $λ = 5$ and $λ = 2$ .
Cayley–Hamilton: $A^{2} - 7 A + 10 I = O$ . Multiplying by $A^{- 1}$ gives $A - 7 I + 10 A^{- 1} = O$ , so $A^{- 1} = \frac{7 I - A}{10} = \frac{1}{10} [3 - 2 - 1 4]$ .
Check: $A [3 - 2 - 1 4] = [100010]$ .
For $A^{4}$ , write $λ^{4} = q (λ) (λ^{2} - 7 λ + 10) + aλ + b$ . At $λ = 5$ : $625 = 5 a + b$ . At $λ = 2$ : $16 = 2 a + b$ . Subtracting, $3 a = 609$ , so $a = 203$ and $b = 16 - 406 = - 390$ .
Hence $A^{4} = 203 A - 390 I = [812 - 390 406 203 609 - 390] = [422406203219]$ .
Check: $tr A^{4} = 641 = 5^{4} + 2^{4}$ , and $∣ A^{4} ∣ = 422 \times 219 - 203 \times 406 = 10000 = 1 0^{4}$ .

Answer: $A^{- 1} = \frac{1}{10} [3 - 2 - 1 4]$ and $A^{4} = [422406203219]$ .

A REPEATED EIGENVALUE IS NOT A VERDICT

A repeated eigenvalue does not by itself make a matrix non-diagonalisable; compute $GM = n - rank (A - λ I)$ . In the worked example of Section 5, $λ = 2$ is repeated and $A$ is still diagonalisable. By contrast, $[2012]$ has $λ = 2$ with AM $= 2$ and GM $= 1$ , and is not.

CAYLEY–HAMILTON IS A MATRIX IDENTITY

The constant term becomes $c_{0} I$ , not the scalar $c_{0}$ . And the theorem cannot be 'proved' by putting $λ = A$ into $∣ A - λ I ∣$ : that gives the scalar $∣ A - A ∣ = 0$ , which is a different statement.

7. Quadratic forms and tensors (in brief)

A quadratic form in $x_{1}, \dots, x_{n}$ is $Q = X^{T} A X = \sum_{i} \sum_{j} a_{ij} x_{i} x_{j}$ with $A$ real symmetric. The coefficient of $x_{i}^{2}$ is $a_{ii}$ , and the coefficient of a cross term $x_{i} x_{j}$ ( $i \neq = j$ ) is $2 a_{ij}$ , so each off-diagonal entry is half the cross-term coefficient. For example, $Q = 2 x_{1}^{2} + 6 x_{1} x_{2} + 5 x_{2}^{2}$ has $A = [2335]$ . The orthogonal change of variables $X = P Y$ , with $P$ diagonalising $A$ , turns $Q$ into a sum of squares.

CANONICAL (SUM-OF-SQUARES) FORM

Q = X^{T} A X = λ_{1} y_{1}^{2} + λ_{2} y_{2}^{2} + \dots + λ_{n} y_{n}^{2}, X = P Y, P^{T} A P = D

$P$ is orthogonal, with orthonormal eigenvectors of $A$ as its columns. The rank of $Q$ is the number of non-zero $λ_{i}$ , and its index is the number of positive $λ_{i}$ .

Definiteness of a quadratic form

$Q = X^{T} A X$ with $A$ real symmetric; $D_{k}$ is the leading principal minor of order $k$ (the determinant of the top-left $k \times k$ block of $A$ ).

Nature	Meaning	Eigenvalues of $A$	Minor test
Positive definite	$Q > 0$ for every $X \neq = 0$	all $> 0$	$D_{1} > 0, D_{2} > 0, \dots, D_{n} > 0$
Negative definite	$Q < 0$ for every $X \neq = 0$	all $< 0$	$D_{1} < 0, D_{2} > 0, D_{3} < 0, \dots$ (the sign of $D_{k}$ is that of $(- 1)^{k}$ )
Positive semi-definite (not definite)	$Q \geq 0$ for every $X$ , and $Q = 0$ for some $X \neq = 0$	all $\geq 0$ , at least one $= 0$	every principal minor $\geq 0$ and $∣ A ∣ = 0$ (the leading minors alone are not enough)
Negative semi-definite (not definite)	$Q \leq 0$ for every $X$ , and $Q = 0$ for some $X \neq = 0$	all $\leq 0$ , at least one $= 0$	every principal minor of order $k$ is $0$ or has the sign of $(- 1)^{k}$ , and $∣ A ∣ = 0$
Indefinite	$Q$ takes both signs	some $> 0$ and some $< 0$	none of the above

Consequences worth knowing. A positive definite matrix has every eigenvalue positive, so $∣ A ∣ > 0$ and $A$ is invertible; every diagonal entry is positive too ( $a_{ii} = Q (e_{i}) > 0$ ), but positive diagonal entries alone do not make a matrix positive definite. For any real $m \times n$ matrix $B$ , the matrix $B^{T} B$ is symmetric and positive semi-definite, because $X^{T} B^{T} B X = ∥ B X ∥^{2} \geq 0$ ; it is positive definite exactly when $rank B = n$ .

HALF THE CROSS TERM

In $Q = x_{1}^{2} + 4 x_{1} x_{2} + x_{2}^{2}$ the matrix is $[1221]$ ( $2$ off the diagonal, not $4$ ). Its eigenvalues are $3$ and $- 1$ , so $Q$ is indefinite although every coefficient is positive: at $(x_{1}, x_{2}) = (1, - 1)$ , $Q = 1 - 4 + 1 = - 2$ .

LEADING MINORS DO NOT SETTLE SEMI-DEFINITENESS

$A = [00 0 - 1]$ has leading minors $D_{1} = 0$ and $D_{2} = 0$ , both $\geq 0$ , yet $Q = - x_{2}^{2}$ is negative semi-definite. When a leading minor is zero, use the eigenvalues or all the principal minors.

Tensors (listed for IN). A tensor extends the idea of scalars, vectors and matrices to quantities with any number of indices. In three-dimensional Cartesian coordinates a tensor of order $r$ has $3^{r}$ components written with $r$ indices, each running over $1, 2, 3$ . (The order is also called the rank of the tensor, which has nothing to do with the rank of a matrix.) Order $0$ is a scalar ( $1$ component, the same in every frame), order $1$ a vector ( $3$ components $v_{i}$ ), order $2$ a quantity such as stress, strain or electrical conductivity ( $9$ components $T_{ij}$ ). What makes a set of components a tensor is the rule by which they change when the axes are rotated: one direction-cosine factor per index.

Index notation

Summation convention: an index repeated in a single term is summed over $1, 2, 3$ and is called a dummy index: $a_{i} b_{i} = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$ , and $T_{ii} = T_{11} + T_{22} + T_{33}$ (the trace).
An index that appears once in a term is a free index. It must appear in every term of the equation, and the equation holds for each of its values: $y_{i} = A_{ij} x_{j}$ is three equations, one for each $i$ . No index may appear more than twice in one term.
Kronecker delta: $δ_{ij} = 1$ if $i = j$ and $0$ otherwise. It replaces an index, $δ_{ij} v_{j} = v_{i}$ , and $δ_{ii} = 3$ in three dimensions.
Levi-Civita (permutation) symbol: $ε_{ij k} = + 1$ if $(i, j, k)$ is an even permutation of $(1, 2, 3)$ , $- 1$ if it is odd, and $0$ if any two indices are equal. Then $(a \times b)_{i} = ε_{ij k} a_{j} b_{k}$ and $a \cdot b = a_{i} b_{i}$ .
Contraction (setting two indices equal and summing) lowers the order by $2$ : $T_{ii}$ is a scalar and $A_{ij} x_{j}$ is a vector.
A second-order tensor splits into a symmetric part ( $T_{ij} = T_{j i}$ , $6$ independent components) and an antisymmetric part ( $T_{ij} = - T_{j i}$ , $3$ independent components, zero diagonal): $T_{ij} = \frac{1}{2} (T_{ij} + T_{j i}) + \frac{1}{2} (T_{ij} - T_{j i})$ .

TRANSFORMATION LAW FOR CARTESIAN TENSORS

v_{i}^{'} = a_{ij} v_{j}, T_{ij}^{'} = a_{i p} a_{j q} T_{pq}

Under a rotation of axes, with $a_{ij}$ the cosine of the angle between the new axis $x_{i}^{'}$ and the old axis $x_{j}$ . The matrix $[a_{ij}]$ is orthogonal, $a_{ik} a_{j k} = δ_{ij}$ . A third-order tensor picks up three factors: $T_{ij k}^{'} = a_{i p} a_{j q} a_{k r} T_{pq r}$ .

EPSILON-DELTA IDENTITY

ε_{ij k} ε_{imn} = δ_{j m} δ_{k n} - δ_{j n} δ_{k m}

Summed over $i$ . It turns a double cross product into dot products, e.g. $a \times (b \times c) = (a \cdot c) b - (a \cdot b) c$ .

PREVIOUS-YEAR QUESTIONS

46 questions from the EC, EE and IN papers

Try each question first, then open the answer. One shows the official answer only. We placed each question under a topic ourselves; the exam paper does not label topics.

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GATE 2026 EC Q.26

GATE 2026 · ECQ.26 · 1 mark · MSQ

Consider the matrix $M = 21 - 1 13 a 10 b$ .

Which of the following options is/ are TRUE if $det (M) \neq = 0$ ?

(A)
$a = - \frac{1}{2}$ and $b = - \frac{1}{2}$
(B)
$a = \frac{1}{2}$ and $b = \frac{1}{2}$
(C)
$a = - 3$ and $b = 0$
(D)
$a = \frac{1}{2}$ and $b = - 3$

Answer and solution

GATE 2026 EC Q.26: answer and solution

See the question in the paper

Linear algebra

1. Notation, matrix algebra, determinants and special matrices

2. Vector spaces, linear independence, basis and dimension

3. Rank of a matrix and the rank-nullity theorem

4. Systems of linear equations: existence and uniqueness

5. Eigenvalues and eigenvectors

6. Diagonalisation and the Cayley–Hamilton theorem

7. Quadratic forms and tensors (in brief)

46 questions from the EC, EE and IN papers

6 questions to test yourself