# 5-Transpose of a Matrix (definitions-solved problems).

## Transpose of a Matrix (definitions-solved problems).

### Introduction to matrix transpose.

The next item contains the properties of the transpose of a matrix, the transpose lets us make rows as columns, and vice versa. the transpose of a matrix with a dimension of (mxn), will give another matrix A T with the dimension of (nxm).

The letter T is a symbol of the transpose, and to be located as a power symbol, the first row will be the first column, the second row will be the second column, and so on.

For a given matrix A of dimension (2×3) after transpose will be with a dimension of (3×2), the first sequence was(1 2 3 & 2 1 3 ) after transpose will be (A)T=(1 2 & 2 1 & 3 3) with a dimension of 3×1.

Again, if we make a transpose of the matrix, it will give a matrix of dimension (2×3). So the definition is correct. The transpose will return the matrix to its original value of A. For the adding of (A+B), then make a transpose, will be= transpose of A+transpose of B.

Matrix A has the same values as given earlier, a new matrix we call B of dimension 2×3.
While B matrix  is (0 2 4 & 1  5 6), after transpose it becomes(0 1 & 2  5& 4  6), with the dimension is (3×2). After that, we add (A+B), with all elements added together, each element added to the corresponding element (1+0=1&2+2=4&3+4=7), while for the second row (2+1=3&1+5=6&3+6=9).
For the A+B, we will get the transpose matrix. The A+B are shown in red color.

The first column is (1 4 7) and the second column is (3 6 9), and the matrix is 3×2. Then we estimate (A)T or the transpose of matrix A, and we write it here. We have estimated the transpose of matrix B, and we will add the transpose of A plus the transpose of B.

The result matches the value of the transpose of (A+B). This calculation proves the validity of the transpose of (A+B) is=Transpose of A +transpose of B.

This is the third property, which states that the transpose of the product of(A),(b) equals the multiplication of the transpose of b by the transpose of A.
First, a multiplication of the two matrices A and b is done. Then we have estimated the transpose of the product of A by B. We performed a multiplication of the transpose of b by the transpose of A. The result matches the previously estimated transpose of the product of A*B.

### Solved problems for the transposing of a matrix.

#### First solved problem 1.29.

This is the first problem 1.29 from Prof. Kuldeep Singh’s linear algebra step-by-step.

Four matrices are given as follows: Matrix A is (-9 2 3& 7 -2 9& 6 -1 5) with a dimension of (3×3).
Matrix B is(1 0 0& 0 2 0& 0 0 3) with a dimension of (3×3). Matrix C is (-1 3 4 &7 9 0) with a dimension of (2×3).
Matrix D is (1 & 2& 3) with a dimension of (3×1). For each given matrix it is required to find the transpose.

In part I, it is required to find the transpose of matrix A. We will transfer columns into rows following the proper sequence. We will get the value of (A)T as (-9 7 6&12 -2 -1& 3 9 5) of dimension 3×3.

In part ii, it is required to find the transpose of matrix B. We will transfer columns into rows following the proper sequence. Secondly, the transpose of b is evaluated, as well as the transpose of A. We will get the value of (B)T as ( 1 0 0& 0 2 0& 0 0 3) of dimension (3×3).

In part IIi, it is required to find the transpose of matrix C. We will transfer columns into rows following the proper sequence. We will get the value of (C)T as (-1 7& 3 9 & 4 6) of dimension (3×2). In part iV, it is required to find the transpose of matrix DB. We will transfer columns into rows following the proper sequence.

We will get the value of (D)T as ( 1 0 3) of dimension (1×3).

#### The second solved problem 1.30.

This is the second problem 1.30 from Prof. Kuldeep Singh’s linear algebra step-by-step.

Two matrices are given as follows: Matrix A is (3 -4 1& 5 2 6) with a dimension of (2×3).
Matrix B is( -2 7 5& 1 3 -9) with a dimension of (2×3).

In part a), it is required to find the transpose of a matrix (A)T. We will transfer columns into rows following the proper sequence. We will get the value of (A)T as (3 5& -4 2 & 1 6) of dimension (3×2).

Again we get the matrix which is the transpose of (A)T. The result will produce a matrix that is the same as matrix A.

In part b), it is required to find the transpose of a matrix (2*A)T-(3B)T. We will multiply the given matrix- A by 2. The next step is to transfer columns into rows following the proper sequence. We will get the value of (2A)T as ( 6 10& -8 4 & 2 12) of dimension (3×2).

We will multiply the given matrix- B by 3. The next step is to transfer columns into rows following the proper sequence. We will get the value of (-3B)T as ( 6 -3& -21 -9&-15 27)of dimension (3×2).

We will add -(3B)T to 2*A)T. The final matrix will be a matrix of dimension(3×2) as follows:(12 7&-29 -5 &-13 39). Please refer to the next slide image for more details.

In part c), it is required to find the transpose of a matrix (A+B)T. We will add matrix A to matrix B. The next step is to estimate the transpose of both matrices A&B. We will get the value of (A)T + (B)T as ( 1 6& 3 5& 6 -3) of dimension (3×2).

The estimated matrix will be equal to (A+B)T as we can see from the slide image.

Part d is similar to part c.

In part e), it is required to find the transpose of the product of the two matrices A&b as (A*B)T.

Since matrix A is of dimension (2×3), while matrix b is of dimension (2×3), these two matrices can not be multiplied since the number of columns of matrix B exceeds the number of rows in Matrix A.

This is a link to the next post: Introduction to Types of Linear systems-Two variables.

This is a link to the matrix calculator.

For a useful external link, math is fun for the matrix part.

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