A matrix, which obtained from a matrix by interchanging the rows as corresponding columns and vice-versa is called transpose of a matrix.

According to meaning of Transpose in English Language

- Transfer to different place.
- Changing place of two or more things.

Transpose of matrix can be done by interchanging either rows as columns or columns as rows. The process of transpose of matrix is same to both square and rectangular matrices. However, order of a matrix is considerable factor to perform the task in less steps.

$T$ or $’$ symbols are displayed superscript to the matrix name to represent transpose of a matrix in mathematics.

1.

$A = \begin{bmatrix}

6 & 2 & -9 \\

-3 & 7 & 1 \\

4 & 5 & 0 \\

\end{bmatrix}$

$A$ is a square matrix of order $3\times3$. It contains $9$ elements. Transpose of matrix $A$ is written as $A^T$ or $A’$. It contains three rows. So, the example matrix can be transposed in three steps.

**Step: 1**

Consider first row of matrix $A$. Express first row of matrix $A$ as first column of matrix $A^T$. $6, 2$ and $-9$ are elements of first row of matrix $A$ and then they become elements of first column of matrix $A^T$.

$A = \begin{bmatrix}

6 & 2 & -9 \\ \hline

-3 & 7 & 1 \\

4 & 5 & 0 \\

\end{bmatrix} ⇒ A^T = \left[\begin{array}{r|rr}

6 & & \\

2 & & \\

-9 & &

\end{array}\right]$

**Step: 2**

Consider second row of matrix $A$. Express second row of matrix $A$ as second column of matrix $A^T$. $-3, 7$ and $1$ are elements of second row of matrix $A$ and then they become elements of second column of matrix $A^T$.

$A = \begin{bmatrix}

6 & 2 & -9 \\

-3 & 7 & 1 \\ \hline

4 & 5 & 0 \\

\end{bmatrix} ⇒ A^T = \left[\begin{array}{rr|r}

6 & -3 & \\

2 & 7 & \\

-9 & 1 &

\end{array}\right]$

**Step: 3**

Consider third row of matrix $A$. Express third row of matrix $A$ as third column of matrix $A^T$. $4, 5$ and $0$ are elements of third row of matrix $A$ and then they become elements of third column of matrix $A^T$.

$A = \begin{bmatrix}

6 & 2 & -9 \\

-3 & 7 & 1 \\

4 & 5 & 0 \\ \hline

\end{bmatrix} ⇒ A^T = \left[\begin{array}{rrr|}

6 & -3 & 4 \\

2 & 7 & 5 \\

-9 & 1 & 0

\end{array}\right]$

$A^T$ is the transpose of matrix $A$ and it is obtained by transforming rows into columns. Transpose of matrix $A$ can also obtain by interchanging columns as rows.

In the case of square matrix, transpose of matrix can be performed by interchanging either rows as columns or columns are rows because the number of rows is equal to number of columns in square matrix.

2.

$B = \begin{bmatrix}

4 & -5 & 7 & -2 \\

2 & 0 & 3 & 9 \\

1 & -3 & 5 & 6 \\

\end{bmatrix}$

$B$ is a rectangular matrix of order $3\times4$. It contains $12$ elements. Transpose of matrix $B$ is written as $B^T$ or $B’$.

In previous case, the rows are transformed into columns to obtain the transpose of a matrix. Now, transform the columns into rows to obtain the transpose of matrix $B$.

Matrix $B$ contains four columns. So, transpose of matrix $B$ can be obtained in four steps.

**Step: 1**

Consider first column of matrix $B$ and write it as first row of matrix $B^T$. $4, 2$ and $1$ are three elements in first column of matrix $B$, then they become elements of first row of matrix $B^T$.

$B = \left[\begin{array}{r|rrr}

4 & -5 & 7 & -2 \\

2 & 0 & 3 & 9 \\

1 & -3 & 5 & 6 \\

\end{array}\right] ⇒ B’ = \begin{bmatrix}

4 & 2 & 1 \\ \hline

& & \\

& & \\

& & \\

\end{bmatrix}$

**Step: 2**

Consider second column of matrix $B$ and write it as second row of matrix $B^T$. $-5, 0$ and $-3$ are three elements in second column of matrix $B$, then they become elements of second row of matrix $B^T$.

$B = \left[\begin{array}{rr|rr}

4 & -5 & 7 & -2 \\

2 & 0 & 3 & 9 \\

1 & -3 & 5 & 6 \\

\end{array}\right] ⇒ B’ = \begin{bmatrix}

4 & 2 & 1 \\

-5 & 0 & 3 \\ \hline

& & \\

& & \\

\end{bmatrix}$

**Step: 3**

Consider third column of matrix $B$ and write it as third row of matrix $B^T$. $7, 3$ and $5$ are three elements in third column of matrix $B$, then they become elements of third row of matrix $B^T$.

$B = \left[\begin{array}{rrr|r}

4 & -5 & 7 & -2 \\

2 & 0 & 3 & 9 \\

1 & -3 & 5 & 6 \\

\end{array}\right] ⇒ B’ = \begin{bmatrix}

4 & 2 & 1 \\

-5 & 0 & 3 \\

7 & 3 & 5 \\ \hline

& & \\

\end{bmatrix}$

**Step: 4**

Consider fourth column of matrix $B$ and write it as fourth row of matrix $B^T$. $-2, 9$ and $6$ are three elements in fourth column of matrix $B$, then they become elements of fourth row of matrix $B^T$.

$B = \left[\begin{array}{rrrr|}

4 & -5 & 7 & -2 \\

2 & 0 & 3 & 9 \\

1 & -3 & 5 & 6 \\

\end{array}\right] ⇒ B’ = \begin{bmatrix}

4 & 2 & 1 \\

-5 & 0 & 3 \\

7 & 3 & 5 \\

-2 & 9 & 6 \\ \hline

\end{bmatrix}$

In this example, columns of matrix $B$ are transformed as rows of transpose matrix. Actually, it took four steps to obtain the transpose of matrix $B$ but it takes only $3$ steps if rows are transformed as columns.

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