Last Updated on March 24, 2025 by Maged kamel
Transpose of a Matrix (definitions-solved problems).
Introduction to matrix transpose.
The next item contains the properties of a matrix’s transpose. 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 a dimension of (nm).
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 addition of (A+B), then making a transpose, it will be the 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 the B matrix is (0 2 4 & 1 5 6), after transposing it, it becomes (0 1 & 2 5 & 4 6), with a dimension of (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), the second is (3 6 9), and the matrix is 3×2. Then we estimate (A)T or the transpose of matrix A and write it here. We have estimated the transpose of matrix B and 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): = 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, we multiply the two matrices A and B. Then, we estimate the transpose of the product of A by B. We multiply the transpose of B by the transpose of A. The result matches the previously estimated transpose of the product of A*B.
The next slide shows the value of BT by AT and that it matches the transpose of the product of A by B.
Solved problems for the transposing of a matrix.
First solved problem 1.29- Transpose of a matrix.
This is the first problem 1.29 from Prof. Kuldeep Singh’s linear algebra step-by-step.
Four matrices are given: 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). The transpose must be found for each given matrix.
In part I, the transpose of matrix A is required. 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, the transpose of matrix B is required. 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, the transpose of matrix C is required. 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-transpose of a matrix.
This is the second problem 1.30 from Prof. Kuldeep Singh’s linear algebra step-by-step.
Two matrices are given: 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), we need 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, 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), finding the transpose of a matrix (A+B)T is required. 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).
As the slide image shows, the estimated matrix will be equal to (A+B)T. 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 links to the following 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.
