# Transpose of a matrix – Matrices | Class 12 Maths

The knowledge of matrices is necessary for various branches of mathematics. Matrices are one of the most powerful tools in mathematics. Now see one of the features of the matrix in this article.

### Transpose of a matrix

This one of the main properties of the matrix. The meaning of transpose is to exchange places of two or more things. In the case of the matrix, transpose meaning changes the index of the elements. In this case, we swap the row-element with the column-element or vise versa.

Let, A is a matrix of size **m × n** and **A ^{t}**

^{ }is the transpose of matrix A,

where [a(ij)] of A = [a(ji)] of A

^{t}, here 1 ≤ i ≤ m and 1 ≤ j ≤ n

**Example:**

Let, the size of matrix** A** is **2 × 3**,

Therefore,

**Transpose of A or **

After transpose it becomes **3 × 2**

## Properties of Transpose

### Transpose of Product of Matrices

This property says that, **(AB) ^{t} = B^{t}A^{t}**

**Proof**

Here A and B are two matrices of size m × n and n × p respectively

and At and Bt are their transpose form of size n × m and p × n respectively (from the product rule of matrices).

It implies, if A = [a(ij)], and At = [c(ji)]

Then, [c(ji)] = [a(ij)]

and,

If B = [b(jk)], and Bt = [d(kj)]

Then, [d(kj)] = [b(jk)]

Now, from the product rule of matrices we can write,

AB is m × p matrix and (AB)t is p × m matrix.

Also, Bt is a p × n matrix and At is a n × m matrix.

This implies that,

(Bt)(At) is a p × m matrix.

Therefore,

(AB)t and (Bt)(At) are both p × m matrices.

Now we can write,

(k, i)th element of (AB)t = (i, k)th element of AB

(k, i)thelement of(B^{t})(A^{t})Therefore,

the elements of

(AB)and^{t}(B^{t})(A^{t}) are equal.Therefore,

(AB)=^{t }(B^{t})(A^{t})

**Example:**

**Let, **

**and **

**So prove that for these matrices, (AB) ^{t }= (B^{t})(A^{t})**

**Solution:**

Here A and B are

2 × 3and3 × 2matrices respectively. So, by product rule of a matrix, we can find their product

and the final matrices would be of2 × 2matrix.

Find L.H.S –Now,

So, Transpose of AB or

Find R.H.S –and

So,

Therefore,

(AB)^{t}= B^{t}.A^{t}

### Transpose of Addition of Matrices

This property says that **(A + B) ^{t} = A^{t} + B^{t}**

**Prove:**

Here A and B are two matrices of size

m × nLet,

A = [a(ij)]andB = [b(ij)]of sizem × n.So,

(A + B)is also anm × nmatrixAlso,

Aand^{t}Bare^{t}n × mmatrices.So that, the

Transpose of (A + B)or(A + B)is an^{t}n × mmatrix.Now we can say,

Ais also an^{t}+ B^{t}n × mmatrix.Now, from the transpose rule,

(j, i)thelement of(A + B)=^{t}(i, j)thelement of(A + B)=

(i, j)thelement ofA+(i, j)thelement ofB

=(j, i)thelement ofA+^{t}(j, i)thelement ofB^{t}

=(j, i)thelement of(A^{t}+ B^{t})Therefore,

(A + B)^{t}= A^{t}+ B^{t}

**Example:**

**Let, **

**and **

**Prove that for these matrices, (A + B) ^{t} = A^{t} + B^{t}**

**Solution:**

Find L.H.S,

So,

Find R.H.S

and,

Now,

Therefore,

(A + B)^{t}= A^{t}+ B^{t}