Matrices

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XII, Chapter 3.

Can one multiplication produce every bill at once?

Ravi buys 2 pens and 3 notebooks. Sana buys 4 pens and 1 notebook. A pen costs ₹10 and a notebook ₹40. Instead of working out each bill separately, write the purchases as a table and the prices as a column, and multiply:

$$\begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 10 \\ 40 \end{bmatrix} = \begin{bmatrix} 2\times 10 + 3 \times 40 \\ 4 \times 10 + 1 \times 40 \end{bmatrix} = \begin{bmatrix} 140 \\ 80 \end{bmatrix}$$

Ravi pays ₹140 and Sana ₹80. That is the entire idea of a matrix: a rectangular table of numbers that you can add, scale and multiply as a single object. The rest of the chapter is the rulebook for doing this safely, and the places where it differs from ordinary number arithmetic.

What a matrix is

A matrix is an ordered rectangular array of numbers (or functions) arranged in rows and columns. A matrix with $m$ rows and $n$ columns has order $m \times n$ and contains $mn$ elements. Write it as $A = [a_{ij}]{m \times n}$, where $a{ij}$ is the element in the $i$-th row and $j$-th column. Row number comes first, always.

Worked example: building a matrix from a rule. Write the $2 \times 3$ matrix whose elements are $a_{ij} = 2i - j$.

Common wrong step. Reading $a_{ij}$ with $j$ as the row. That produces a $3 \times 2$ matrix with the entries in the wrong places.

Counting matrices. If every entry can be chosen independently from $k$ allowed values, an $m \times n$ matrix has $mn$ entries and so $k^{mn}$ possibilities. For a $3 \times 3$ matrix with entries $0$ or $1$ that is $2^9 = 512$.

Types of matrices

The types are nested: every identity matrix is a scalar matrix, every scalar matrix is a diagonal matrix, and every diagonal matrix is a square matrix. The reverse never holds in general. A scalar matrix is the identity only when $k = 1$.

Two matrices are equal only when they have the same order and every pair of matching entries is equal. For example, $\begin{bmatrix} x + y & 2 \\ 3 & x - y \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ 3 & 1 \end{bmatrix}$ gives $x + y = 5$ and $x - y = 1$, so $x = 3$ and $y = 2$.

Adding and scaling

Matrices add entry by entry, and only when they have the same order: $A + B = [a_{ij} + b_{ij}]$. Scaling multiplies every entry: $kA = [k\,a_{ij}]$, and the difference is $A - B = A + (-1)B$. Addition behaves like ordinary addition: it is commutative, associative, has the zero matrix as its identity and $-A$ as its inverse, and scalars distribute over it: $k(A + B) = kA + kB$ and $(k + l)A = kA + lA$.

Multiplication: row by column

To find the entry in row $i$, column $k$ of $AB$, take row $i$ of $A$ and column $k$ of $B$, multiply matching entries and add. This is exactly what the bill calculation above did.

$$c_{ik} = \sum_{j} a_{ij}\, b_{jk}$$

That rule only works if the row and the column have the same length. So $AB$ is defined only when the number of columns of $A$ equals the number of rows of $B$. If $A$ is $m \times n$ and $B$ is $n \times p$, then $AB$ is $m \times p$. Write the orders side by side: the two inner numbers must match, and the two outer numbers give the order of the answer.

Worked example: a $2 \times 2$ times a $2 \times 3$. Find $AB$ for $A = \begin{bmatrix} 6 & 9 \\ 2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 6 & 0 \\ 7 & 9 & 8 \end{bmatrix}$.

Common wrong step. Multiplying matching entries one by one, like addition. That is not matrix multiplication, and here it could not even be done because the orders differ.

Pause and predict: $A$ is $2 \times 3$ and $B$ is $3 \times 2$. What are the orders of $AB$ and $BA$?

Check your answer: $AB$ is $2 \times 2$ and $BA$ is $3 \times 3$. Both exist, but they cannot possibly be equal, because they do not even have the same order.

Where matrix multiplication breaks the rules of numbers

Three familiar facts from ordinary arithmetic fail. Each is a regular CUET question.

What does hold: multiplication is associative, $(AB)C = A(BC)$, and distributive, $A(B + C) = AB + AC$ and $(A + B)C = AC + BC$, and $AI = IA = A$ for a square $A$ of the same order as $I$. Diagonal matrices of the same order also commute.

Worked example: $AB \ne BA$, and $AB = O$. Show that $AB \ne BA$ for $A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. Then find $AB$ for $P = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix}$ and $Q = \begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix}$.

Common wrong step. Concluding from $PQ = O$ that $P = O$ or $Q = O$. That argument works for numbers only. Notice also that $QP = \begin{bmatrix} 0 & 7 \\ 0 & 0 \end{bmatrix}$, which is not $O$, so the order of the product matters here too.

Beyond the textbook. For the exam: $AB = O$ does not force $A = O$ or $B = O$, and you may be asked for an example. In reality: the failure is not random. If $A$ has an inverse, then $AB = O$ gives $A^{-1}AB = A^{-1}O$, so $B = O$. The troublemakers are exactly the matrices with no inverse, which are the ones with determinant zero in the next chapter. Why it matters: it is why you may never "cancel" a matrix from both sides of an equation unless it is invertible.

Beyond the textbook. For the exam: $AB \ne BA$ is just a property to remember. In reality: it is the reason computer graphics works the way it does. A rotation, a stretch and a reflection of an image are each a matrix, and doing one after another is multiplying them. "Rotate then stretch" gives a different picture from "stretch then rotate" precisely because the two products differ.

Transpose

The transpose $A'$ (also written $A^T$) swaps rows and columns: if $A = [a_{ij}]{m \times n}$, then $A' = [a{ji}]_{n \times m}$. Four rules cover everything:

$$(A')' = A, \qquad (kA)' = kA', \qquad (A + B)' = A' + B', \qquad (AB)' = B'A'$$

The last one reverses the order, like taking off your socks and shoes: whatever went on last comes off first. The orders prove it must be reversed. If $A$ is $m \times n$ and $B$ is $n \times p$, then $(AB)'$ is $p \times m$. Now $B'$ is $p \times n$ and $A'$ is $n \times m$, so $B'A'$ is $p \times m$, a perfect match. But $A'B'$ would multiply an $n \times m$ matrix by a $p \times n$ matrix, which does not even exist unless $m = p$.

Symmetric and skew symmetric matrices

A square matrix is symmetric if $A' = A$, meaning $a_{ji} = a_{ij}$, and skew symmetric if $A' = -A$, meaning $a_{ji} = -a_{ij}$.

Why the diagonal of a skew symmetric matrix is zero. Put $j = i$ in $a_{ji} = -a_{ij}$ to get $a_{ii} = -a_{ii}$, so $2a_{ii} = 0$ and $a_{ii} = 0$.

Splitting any square matrix. For every square $A$, the matrix $A + A'$ is symmetric and $A - A'$ is skew symmetric. The first holds because $(A + A')' = A' + (A')' = A' + A$. The second holds because $(A - A')' = A' - A = -(A - A')$. Adding the halves gives back $A$:

$$A = \dfrac{1}{2}(A + A') + \dfrac{1}{2}(A - A')$$

Worked example: symmetric plus skew symmetric. Express $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ as the sum of a symmetric and a skew symmetric matrix.

Common wrong step. Taking $A + A'$ itself as the symmetric part. It is symmetric, but the two parts then add to $2A$ instead of $A$. The factor $\dfrac{1}{2}$ is essential.

Pause and predict: if $A$ is skew symmetric and the entry $a_{13} = 7$, what is $a_{31}$?

Check your answer: $-7$, since $a_{31} = -a_{13}$. The diagonal entries $a_{11}$, $a_{22}$ and $a_{33}$ must be $0$.

Invertible matrices

A square matrix $A$ of order $n$ is invertible if there is a square matrix $B$ of the same order with

$$AB = BA = I_n$$

and then $B$ is written $A^{-1}$. Four facts matter. Only square matrices can be invertible, so a rectangular matrix never has an inverse. The inverse, if it exists, is unique. If $B = A^{-1}$ then $A = B^{-1}$. And for invertible $A$ and $B$ of the same order,

$$(AB)^{-1} = B^{-1}A^{-1}$$

Why the order reverses. Check that $B^{-1}A^{-1}$ undoes $AB$: $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = I$, and $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = I$. By uniqueness, it is the inverse. The same sock-and-shoe reversal as the transpose.

Worked example: testing an inverse. Is $B = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}$ the inverse of $A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$?

Common wrong step. Inverting entry by entry, giving $\begin{bmatrix} \frac{1}{2} & 1 \\ \frac{1}{5} & \frac{1}{3} \end{bmatrix}$. The inverse of a matrix is not the reciprocal of each entry.

Pause and predict: $A$, $B$ and $C$ are invertible matrices of the same order. What is $(ABC)^{-1}$?

Check your answer: $C^{-1}B^{-1}A^{-1}$. The reversal law extends to any number of factors: the last one in is the first one out.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Inverse Trigonometric Functions closes the functions half of the course.
• Next: Determinants gives the number that decides whether a square matrix has an inverse, and the formula for finding it.
• Same topic, earlier: Relations and Functions introduces invertible functions, the idea that $A^{-1}$ extends to matrices.
• Same idea, elsewhere: Vector Algebra uses the same sum of products of matching components in a dot product that row by column multiplication is built from.