Q.Find the transpose of each of the following matrices:

Concept: Matrix Transpose — swapping rows and columns: if $A = [a_{ij}]$, then $A^T = [a_{ji}]$.

(i) The given matrix is $3 \times 1$ (column vector). Transpose swaps it to a $1 \times 3$ row vector:

(ii) This is a $2 \times 2$ square matrix. Interchange rows and columns:

(iii) This is a $2 \times 3$ matrix. Its transpose will be $3 \times 2$:

Why the transpose works

The transpose operation is one of the simplest yet most powerful ideas in matrix algebra. If you picture a matrix as a rectangular grid of numbers, taking the transpose is like rotating that grid around its main diagonal — the top-left to bottom-right line. Every row becomes a column, and every column becomes a row.

Formally, if $A$ is an $m \times n$ matrix (m rows, n columns), its transpose $A^T$ is an $n \times m$ matrix where $(A^T)_{ij} = A_{ji}$. That subscript swap is the entire rule.

Now let's apply this to each of the three matrices given.

(i) $\begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix}$

1. This is a $3 \times 1$ matrix — three rows, one column. It's a column vector.

2. Transposing it means the single column becomes a single row. So the result will be a $1 \times 3$ row vector.

3. The first (and only) column has entries $5$, $\frac12$, and $-1$, in that order from top to bottom. After transposing, these become the entries of the first (and only) row, in the same order left to right.

4. Therefore:

(ii) $\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$

1. This is a $2 \times 2$ square matrix. Its transpose will also be $2 \times 2$.

2. The element at position $(1,1)$ is $1$ — it stays at $(1,1)$ because the diagonal doesn't move.

3. The element at $(1,2)$ is $-1$. After transposing, it moves to $(2,1)$.

4. The element at $(2,1)$ is $2$. It moves to $(1,2)$.

5. The element at $(2,2)$ is $3$ — it stays at $(2,2)$.

6. Putting it together:

Notice that the off-diagonal entries simply swapped places.

(iii) $\begin{bmatrix} \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix}$

1. This is a $2 \times 3$ matrix (2 rows, 3 columns). Its transpose will be $3 \times 2$.

2. Row 1 of the original is $\begin{bmatrix} \sqrt3 & 5 & 6 \end{bmatrix}$. After transposing, this becomes column 1 of the result.

3. Row 2 of the original is $\begin{bmatrix} 2 & 3 & -1 \end{bmatrix}$. This becomes column 2 of the result.

4. So the first column of the transpose is $\begin{bmatrix} \sqrt3 \\ 5 \\ 6 \end{bmatrix}$ and the second column is $\begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}$.

5. Writing it as a matrix:

Method: Writing the transpose of a matrix

The transpose $A'$ is formed by turning each row of $A$ into a column (the $(i,j)$ entry of $A$ moves to position $(j,i)$); the order flips from $m\times n$ to $n\times m$.

Steps

Step 1: Note the new order

An $m\times n$ matrix becomes $n\times m$ — a $2\times3$ transposes to $3\times2$, and a column ($3\times1$) becomes a row ($1\times3$).

Step 2: Turn rows into columns

Write the first row of $A$ as the first column of $A'$, the second row as the second column, and so on.

Step 3: Check the diagonal is fixed

For a square matrix the diagonal entries stay put; only the off-diagonal entries swap across the diagonal.

Common Mistakes

Mistake 1: Keeping the shape the same instead of flipping it

Why it's wrong: the transpose of a $3\times1$ column vector is a $1\times3$ row vector, not another column; the order must flip. Correct approach: swap the row and column counts when transposing.

Mistake 2: Swapping only some off-diagonal entries

Why it's wrong: every entry moves from $(i,j)$ to $(j,i)$; leaving a pair unswapped gives a matrix that is neither $A$ nor $A'$. Correct approach: rewrite each full row of $A$ as the corresponding column of $A'$.