Matrices

This companion follows the chapter's own build-up, section by section. It opens with the definition of a matrix and its order — the vocabulary of rows, columns and the $(i,j)$th entry that every later idea is stated in terms of. It then catalogues the special types of matrices (row, column, square, diagonal, scalar, identity, zero) that appear constantly as building blocks, followed by the precise meaning of equality of matrices. Next come the operations: addition and scalar multiplication (straightforward, entrywise, and law-abiding), and then matrix multiplication — the operation with the most exam traps, because it is only defined under a strict dimension-compatibility condition and does not commute. The chapter then turns to the transpose of a matrix and the two special square-matrix families it produces, symmetric and skew-symmetric matrices, proving that every square matrix is uniquely a sum of one of each. It closes with invertible matrices: what it means for a square matrix to have an inverse, why that inverse (when it exists) is unique, and how inverses of a product reverse and swap, $(AB)^{-1} = B^{-1}A^{-1}$. Work through the sections in this order — each is the toolkit the next section quietly assumes, and together they are the exact prerequisite Chapter 4 (Determinants) needs to actually compute an inverse using the adjoint formula.

What is a matrix, and what is its order

A matrix is an ordered rectangular array of numbers or functions, displayed inside brackets. Formally, an $m \times n$ matrix is written

where: $m$ is the number of rows (horizontal lines of entries) and $n$ is the number of columns (vertical lines of entries); the pair $m \times n$ (read "$m$ by $n$") is called the order of the matrix. The symbol $a_{ij}$ denotes the entry lying in the $i$th row and $j$th column — the $(i,j)$th element of $A$ — with the row index always written first. The $i$th row consists of the entries $a_{i1}, a_{i2}, \ldots, a_{in}$, and the $j$th column consists of $a_{1j}, a_{2j}, \ldots, a_{mj}$. A matrix of order $m \times n$ has exactly $mn$ elements, since there are $m$ rows and each contributes $n$ entries; consequently, if a matrix is known to have, say, $8$ elements, its possible orders are exactly the factor pairs of $8$: $1 \times 8$, $8 \times 1$, $2 \times 4$, $4 \times 2$ (NCERT's own Example 2). This chapter, following NCERT convention, considers only matrices whose entries are real numbers or real-valued functions. Matrices are not confined to tables of raw data — they can represent geometric information too. A point $P(x, y)$ in the plane can be written as a column matrix $\begin{bmatrix} x \\ y \end{bmatrix}$ or a row matrix $[x\ \ y]$, and the vertices of a closed figure can be stacked into a single matrix. For the quadrilateral $ABCD$ with $A(1,0)$, $B(3,2)$, $C(1,3)$, $D(-1,2)$, NCERT records the vertex data as the $2 \times 4$ matrix $X = \begin{bmatrix} 1 & 3 & 1 & -1 \\ 0 & 2 & 3 & 2 \end{bmatrix}$ (columns are vertices, rows are $x$- and $y$-coordinates) or equally as the $4 \times 2$ matrix $Y$ obtained by transposing it (rows are vertices). This dual reading — matrices as data tables and matrices as coordinate lists — recurs throughout coordinate geometry and computer graphics.

Types of matrices

A handful of special shapes recur so often that each has its own name. A column matrix has only one column, order $m \times 1$; a row matrix has only one row, order $1 \times n$. A square matrix has $m = n$ — equal numbers of rows and columns — and is then simply called a square matrix of order $n$; its entries $a_{11}, a_{22}, \ldots, a_{nn}$ (same row and column index) form its diagonal. Within square matrices, four nested families matter most: a diagonal matrix has every off-diagonal entry zero ($a_{ij} = 0$ whenever $i \neq j$, with no restriction on the diagonal entries themselves); a scalar matrix is a diagonal matrix whose diagonal entries are all equal to some constant $k$; an identity matrix, denoted $I_n$ or simply $I$, is the scalar matrix with $k = 1$ — every diagonal entry $1$, every off-diagonal entry $0$. Finally, a zero (or null) matrix, denoted $O$, has every entry equal to zero, of any order. These families nest inside one another: every identity matrix is a scalar matrix (the case $k=1$), every scalar matrix is a diagonal matrix, and every diagonal matrix is a square matrix — but the reverse inclusions fail (a scalar matrix is an identity matrix only when $k = 1$; not every diagonal matrix has equal diagonal entries). The table collects the definitions with a concrete $3\times 3$ example of each.

Equality of matrices

Two matrices $A = [a_{ij}]$ and $B = [b_{ij}]$ are declared equal, written $A = B$, exactly when two conditions both hold:

where: condition (i) is checked first and is purely about shape — two matrices of different orders are never equal, no matter what their entries are — and condition (ii) is entrywise, entry by entry. The practical consequence is that a single matrix equation between two matrices of order $m \times n$ silently packages $mn$ separate ordinary equations, one per entry. This is the standard device for extracting several unknowns from one matrix statement: NCERT's own Example 4 sets $\begin{bmatrix} x+3 & 2y-7 \\ z+4 & a-1 \end{bmatrix} = \begin{bmatrix} 0 & 3y-2 \\ 6 & -3 \end{bmatrix}$ (with two further conditions on $b, c$) and reads off $x+3=0$, $2y-7=3y-2$, $z+4=6$, $a-1=-3$ as four ordinary equations, solved independently to give $x=-3,\ y=-5,\ z=2,\ a=-2$.

Addition and scalar multiplication of matrices

If $A = [a_{ij}]$ and $B = [b_{ij}]$ are matrices of the same order $m \times n$, their sum is defined entrywise:

where: addition of matrices, like addition of numbers, is only meaningful when the two objects have compatible shape — if $A$ and $B$ have different orders, $A + B$ is simply not defined; there is no notion of "padding with zeros." Addition inherits the algebraic properties of ordinary numbers directly, entry by entry: it is commutative ($A+B=B+A$), associative ($(A+B)+C=A+(B+C)$), has an additive identity, the zero matrix $O$ of the same order (so $A+O=O+A=A$), and every $A$ has an additive inverse $-A = (-1)A$ with $A+(-A) = O$. The difference of two same-order matrices is then simply $A - B = A + (-1)B$, subtracting entrywise. Scalar multiplication scales every entry by the same number $k$ (called a scalar):

where: $k$ can be any real number (or, in the general definition, any function-valued scalar consistent with the matrix's entries). Two laws link scalar multiplication to addition: $k(A+B) = kA + kB$ (a scalar distributes over a matrix sum) and $(k+l)A = kA + lA$ (a sum of scalars distributes over a single matrix) — both proved by expanding the entrywise definitions and using the distributive law for ordinary numbers, exactly as NCERT does. These two operations combine routinely in "find $X$" problems: given $2A + 3X = 5B$, the same additive-identity and additive-inverse manipulations used for ordinary linear equations isolate $X = \tfrac13(5B - 2A)$, entry by entry.

Multiplication of matrices

Matrix multiplication is the operation that most departs from ordinary number algebra, and it is defined only under a strict compatibility condition. If $A = [a_{ij}]$ is an $m \times n$ matrix and $B = [b_{jk}]$ is an $n \times p$ matrix — note that the number of columns of $A$ must equal the number of rows of $B$ — then the product $AB$ is the $m \times p$ matrix $C = [c_{ik}]$ defined by

where: to get the entry in row $i$, column $k$ of the product, take the $i$th row of $A$, the $k$th column of $B$, multiply them entry by entry, and add — the familiar "row dot column" rule. If the column count of $A$ does not match the row count of $B$, the product $AB$ is undefined, full stop; there is no matrix "you almost get." This single dimension rule has sharp consequences that exam questions probe constantly. If $A$ is $m \times n$ and $B$ is $k \times l$, both $AB$ and $BA$ are defined only when $n = k$ and $l = m$; in particular, if $AB$ is defined it does not follow that $BA$ is defined at all (NCERT's Example 12: $A$ is $2\times 2$, $B$ is $2\times 3$, so $AB$ exists as a $2\times 3$ matrix, but $BA$ is not defined since $B$'s $3$ columns do not match $A$'s $2$ rows). Even when both products exist and share the same order, matrix multiplication is not commutative: $AB \neq BA$ in general — NCERT's Example 13 and Example 14 both exhibit $2\times 2$ matrices with $AB \neq BA$. A further surprise is that the product of two non-zero matrices can be the zero matrix: NCERT's Example 15 shows $\begin{bmatrix}0&-1\\0&2\end{bmatrix}\begin{bmatrix}3&5\\0&0\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}$ with neither factor equal to $O$ — the familiar real-number fact "$ab=0 \Rightarrow a=0$ or $b=0$" simply fails for matrices, so matrices possess zero divisors. Despite this loss of commutativity, multiplication keeps three structural laws (each stated by NCERT without proof, and provable by expanding the summation definition entry by entry): the associative law $(AB)C = A(BC)$ whenever both sides are defined; the distributive laws $A(B+C) = AB+AC$ and $(A+B)C = AC+BC$ whenever both sides are defined; and the existence of a multiplicative identity — for every square matrix $A$ of order $n$, the identity matrix $I_n$ satisfies $IA = AI = A$.

Transpose of a matrix

The transpose of $A = [a_{ij}]_{m\times n}$, written $A'$ or $A^T$, is obtained by interchanging its rows and columns:

where: the first row of $A$ becomes the first column of $A'$, the second row becomes the second column, and so on; consequently an $m \times n$ matrix transposes to an $n \times m$ matrix, and transposing a square matrix keeps its order unchanged. Four properties, each verified directly from the definition (and the fourth proved rigorously below), govern how transpose interacts with the other operations: $(A')' = A$ (transposing twice returns the original), $(kA)' = kA'$ (scalar multiplication commutes with transpose), $(A+B)' = A' + B'$ (transpose distributes over addition), and — the one exam-favourite that reverses order — $(AB)' = B'A'$.

Symmetric and skew-symmetric matrices

Two named families of square matrices arise directly from the transpose. A square matrix $A$ is symmetric if $A' = A$, that is $a_{ij} = a_{ji}$ for every $i, j$ — the matrix is a mirror image of itself across its own diagonal. A square matrix $A$ is skew-symmetric if $A' = -A$, that is $a_{ji} = -a_{ij}$ for every $i, j$:

where: setting $i = j$ in the skew-symmetric condition gives $a_{ii} = -a_{ii}$, so $2a_{ii} = 0$ and every diagonal entry of a skew-symmetric matrix must be zero — a fast diagnostic that immediately rules out any skew-symmetric matrix with a non-zero diagonal entry. The two families are proved below to be more than a curiosity: every square matrix, symmetric or not, can be split uniquely into a symmetric part and a skew-symmetric part that add back up to the original.

Invertible matrices

A square matrix $A$ of order $m$ is called invertible if there exists another square matrix $B$, of the same order $m$, such that

where: $B$ is then called the inverse of $A$, written $B = A^{-1}$, and by symmetry of the definition $A$ is equally the inverse of $B$, so $A = B^{-1}$. Only square matrices can possibly be invertible: for the products $AB$ and $BA$ to both be defined and both equal to the same matrix $I$, $A$ and $B$ must be square matrices of matching order — a rectangular matrix never has an inverse. Two structural theorems, proved in full below, complete the picture: the inverse of a matrix, when it exists, is unique, and the inverse of a product reverses order, $(AB)^{-1} = B^{-1}A^{-1}$. Neither theorem tells you how to compute $A^{-1}$ in practice — that formula, using the determinant and adjoint, is the subject of the next chapter.

Board level - Constructing a matrix from a formula for its general entry

Construct the $3 \times 2$ matrix $A = [a_{ij}]$ whose entries are given by $a_{ij} = \dfrac{1}{2}\,|i - 3j|$, for $i = 1, 2, 3$ and $j = 1, 2$.

Step 1. A general $3\times 2$ matrix has the form $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix}$, so six entries must be computed, one for each pair $(i,j)$ with $i \in \{1,2,3\}$ and $j \in \{1,2\}$.

Step 2. Compute the first column ($j=1$): $a_{11} = \tfrac12|1-3(1)| = \tfrac12|{-2}| = 1$; $a_{21} = \tfrac12|2-3(1)| = \tfrac12|{-1}| = \tfrac12$; $a_{31} = \tfrac12|3-3(1)| = \tfrac12(0) = 0$.

Step 3. Compute the second column ($j=2$): $a_{12} = \tfrac12|1-3(2)| = \tfrac12|{-5}| = \tfrac52$; $a_{22} = \tfrac12|2-3(2)| = \tfrac12|{-4}| = 2$; $a_{32} = \tfrac12|3-3(2)| = \tfrac12|{-3}| = \tfrac32$.

Step 4. Assemble the six entries into the matrix in their correct positions. Answer: $A = \begin{bmatrix} 1 & \tfrac52 \\[2pt] \tfrac12 & 2 \\[2pt] 0 & \tfrac32 \end{bmatrix}$.

Board level - Solving for unknowns using equality of matrices

Find the values of $a, b, c, d$ from the matrix equation $\begin{bmatrix} 2a+b & a-2b \\ 5c-d & 4c+3d \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 11 & 24 \end{bmatrix}$.

Step 1. Both matrices are $2\times 2$, so equality of matrices applies: equate corresponding entries to get four ordinary linear equations: $2a+b=4$, $a-2b=-3$, $5c-d=11$, $4c+3d=24$.

Step 2. Solve the pair in $a,b$. From the first equation, $b = 4-2a$. Substitute into the second: $a - 2(4-2a) = -3 \Rightarrow a - 8 + 4a = -3 \Rightarrow 5a = 5 \Rightarrow a = 1$, and then $b = 4-2(1) = 2$.

Step 3. Solve the pair in $c,d$. From the third equation, $d = 5c - 11$. Substitute into the fourth: $4c + 3(5c-11) = 24 \Rightarrow 4c+15c-33=24 \Rightarrow 19c = 57 \Rightarrow c=3$, and then $d = 5(3)-11 = 4$.

Step 4. Check quickly: $2(1)+2=4$ ✓, $1-2(2)=-3$ ✓, $5(3)-4=11$ ✓, $4(3)+3(4)=24$ ✓, so all four original equations are satisfied. Answer: $a=1,\ b=2,\ c=3,\ d=4$.

Competitive level (JEE / NEET) - Verifying a matrix polynomial identity

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}$, show that $A^3 - 23A - 40I = O$.

Step 1. Compute $A^2 = A \cdot A$ entry by entry, each entry a row-of-$A$ dot column-of-$A$. Row 1 of $A^2$: $[1{\cdot}1{+}2{\cdot}3{+}3{\cdot}4,\ 1{\cdot}2{+}2{\cdot}({-}2){+}3{\cdot}2,\ 1{\cdot}3{+}2{\cdot}1{+}3{\cdot}1] = [19,\,4,\,8]$. Row 2: $[3{\cdot}1{-}2{\cdot}3{+}1{\cdot}4,\ 3{\cdot}2{-}2{\cdot}({-}2){+}1{\cdot}2,\ 3{\cdot}3{-}2{\cdot}1{+}1{\cdot}1]=[1,\,12,\,8]$. Row 3: $[4{\cdot}1{+}2{\cdot}3{+}1{\cdot}4,\ 4{\cdot}2{+}2{\cdot}({-}2){+}1{\cdot}2,\ 4{\cdot}3{+}2{\cdot}1{+}1{\cdot}1]=[14,\,6,\,15]$. So $A^2 = \begin{bmatrix}19&4&8\\1&12&8\\14&6&15\end{bmatrix}$.

Step 2. Compute $A^3 = A^2 \cdot A$ the same way. Row 1: $[19{\cdot}1{+}4{\cdot}3{+}8{\cdot}4,\ 19{\cdot}2{+}4{\cdot}({-}2){+}8{\cdot}2,\ 19{\cdot}3{+}4{\cdot}1{+}8{\cdot}1] = [63,\,46,\,69]$. Row 2: $[1{\cdot}1{+}12{\cdot}3{+}8{\cdot}4,\ 1{\cdot}2{+}12{\cdot}({-}2){+}8{\cdot}2,\ 1{\cdot}3{+}12{\cdot}1{+}8{\cdot}1] = [69,\,{-}6,\,23]$. Row 3: $[14{\cdot}1{+}6{\cdot}3{+}15{\cdot}4,\ 14{\cdot}2{+}6{\cdot}({-}2){+}15{\cdot}2,\ 14{\cdot}3{+}6{\cdot}1{+}15{\cdot}1] = [92,\,46,\,63]$. So $A^3 = \begin{bmatrix}63&46&69\\69&-6&23\\92&46&63\end{bmatrix}$.

Step 3. Form $23A = \begin{bmatrix}23&46&69\\69&-46&23\\92&46&23\end{bmatrix}$ and $40I = \begin{bmatrix}40&0&0\\0&40&0\\0&0&40\end{bmatrix}$, then subtract both from $A^3$ entrywise: $A^3-23A = \begin{bmatrix}63{-}23&46{-}46&69{-}69\\69{-}69&{-}6{-}({-}46)&23{-}23\\92{-}92&46{-}46&63{-}23\end{bmatrix} = \begin{bmatrix}40&0&0\\0&40&0\\0&0&40\end{bmatrix}$.

Step 4. Subtract $40I$: $(A^3-23A) - 40I = \begin{bmatrix}40{-}40&0&0\\0&40{-}40&0\\0&0&40{-}40\end{bmatrix} = \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix} = O$. Answer: $A^3 - 23A - 40I = O$, verified exactly. Timing note: about four minutes at exam speed. The entire calculation is arithmetic, not insight, so the real skill being tested is organised bookkeeping — compute $A^2$ completely and check it once before starting $A^3$, since any slip in $A^2$ silently corrupts every later entry. Writing $23A$ and $40I$ out in full before subtracting (rather than subtracting term-by-term in your head) is the fastest way to avoid a sign error in the final line.

Competitive level (JEE / NEET) - Finding an unknown matrix from a product-equality condition

Let $A = \begin{bmatrix}2&-1\\3&4\end{bmatrix}$, $B=\begin{bmatrix}5&2\\7&4\end{bmatrix}$, $C=\begin{bmatrix}2&5\\3&8\end{bmatrix}$. Find a matrix $D$ such that $CD - AB = O$.

Step 1. Since $CD - AB = O$ means $CD = AB$, and $C$ is a $2\times 2$ square matrix, $D$ must also be $2\times 2$ for $CD$ to make sense; write $D = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ with four unknowns to find.