An m × n matrix is a rectangular array of numbers with m rows and n columns. A typical m × n matrix A is written as

The matrix above may be represented compactly as A = [aij]m×n. Each entry aij is called the (i, j)th element of A. The element aij lies in the ith row and jth column of the matrix.

Let A and B be square matrices of order n, and let In denote the n × n identity (unit) matrix. The following formulas and properties involving the adjoint (or classical adjoint) and determinant are useful.
(a) A(adj A) = |A| I_n = (adj A)A
(b) |adj A| = |A|n-1
(Thus A · adj A is always a scalar matrix.)
(c) adj(adj A) = |A|^(n-2) A
(d) adj(A + B) does not have a simple general expansion analogous to determinants; treat on a case-by-case basis.
(e) adj(AB) = (adj B)(adj A)
(f) adj(Am) = (adj A)m, for integer m ≥ 1
(g) adj(kA) = kn-1adj A, for scalar k ∈ ℝ
(h) adj(I_n) = I_n
(i) adj(0) = 0
(j) If A is symmetric then adj A is also symmetric
(k) If A is diagonal then adj A is diagonal
(l) If A is triangular then adj A is triangular
(m) If A is singular then |adj A| = 0


The trace of a square matrix is the sum of the elements on its main diagonal. For an n × n matrix A = [aij],
The trace is denoted by tr(A).

The transpose of a matrix A is the matrix obtained by interchanging its rows and columns. The transpose is denoted by AT or A'. If A has order m × n then AT has order n × m.



The adjoint (classical adjoint) of an n × n matrix A, denoted by adj A or \(\operatorname{adj}A\), is the transpose of the cofactor matrix of A. Each entry of adj A is the cofactor of the corresponding entry of A, placed in transposed position.

An n × n matrix A has an inverse A-1 if and only if A is non-singular, i.e., \(|A|\ne 0\). When the inverse exists, it satisfies \(AA^{-1}=A^{-1}A=I_n\).
The inverse of A (when \(|A|\ne 0\)) can be written in terms of the adjoint:
\(A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A\).

The order of a matrix is given by the number of rows by number of columns, denoted m × n. For example:

The matrix shown above has order 2 × 3.
Sol:
Compute the element in the ith row and jth column by substituting i and j in the formula aij=2i+3j.
a11=2·1+3·1=5
a12=2·1+3·2=8
a13=2·1+3·3=11
a14=2·1+3·4=14
a21=2·2+3·1=7
a22=2·2+3·2=10
a23=2·2+3·3=13
a24=2·2+3·4=16
a31=2·3+3·1=9
a32=2·3+3·2=12
a33=2·3+3·3=15
a34=2·3+3·4=18
![Illustration 1: Construct a 3×4 matrix A = [aij], whose elements are given by aij = 2i + 3j.](https://cn.edurev.in/ApplicationImages/Temp/4870720_f492c95c-ff6e-4bea-8b56-97e397475091_lg.png)

Sol:
The method is identical to Illustration 1: substitute respective row and column indices i and j into the given formula for aij to obtain each entry.


Hence the required matrix is:

Let A = [aij]n×n and B = [bij]n×n, and let λ be a scalar. Then

The transpose of a matrix A is obtained by turning its rows into columns (or columns into rows). If A is m × n then AT is n × m. For example, the transpose of the following matrix is shown:

Illustration 3: If

then prove that (AB)T = BTAT.
Solution:
Obtain the product AB explicitly (using the usual rule: row of A times column of B) and then form the transpose of the resulting matrix to get (AB)T.
Obtain BT and AT individually by transposing B and A respectively.
Multiply BT and AT in that order to get BTAT.
Compare the entries of (AB)T and BTAT; they are equal, which proves (AB)T=BTAT.


Illustration 4: If

Then what is (B')'A' equal to?
Sol:
Use the properties of transpose. Note that (B')' means the transpose of the transpose of B, that is (BT)T=B.
Therefore (B')'A' = BAT.

| 1. What is a matrix? | ![]() |
| 2. How do you add two matrices together? | ![]() |
| 3. What is the determinant of a matrix? | ![]() |
| 4. How do you multiply two matrices together? | ![]() |
| 5. What is the inverse of a matrix? | ![]() |