Rows, Columns and Order of a Matrix
A matrix is a set of numbers arranged in a rectangle of rows and columns. The rows run across the matrix, from left to right. The columns run down it, from top to bottom. The word matrix was first used by the English mathematician James Joseph Sylvester, and matrices are named with capital letters such as A, B and C.
The size of a matrix is called its order. A matrix with m rows and n columns has order \( m \times n \), read as "m by n". The number of rows is always written first.
Order of a matrix: a matrix with m rows and n columns is of order \( m \times n \). Rows are counted first and columns second.
Take this matrix:
\[ A = \begin{bmatrix} 3 & -5 & 7 \\ 2 & 4 & -6 \end{bmatrix} \]
It has 2 rows and 3 columns, so its order is \( 2 \times 3 \). Its first row is 3, -5, 7 and its second row is 2, 4, -6. Its first column is 3, 2. Its second column is -5, 4. Its third column is 7, -6.
Transpose of a Matrix
Now write the same six numbers again, but this time put the first row of A down the first column, and the second row of A down the second column. Nothing is added, nothing is removed and no sign is changed. Only the positions of the numbers move.
\[ \begin{bmatrix} 3 & 2 \\ -5 & 4 \\ 7 & -6 \end{bmatrix} \]
The new matrix has 3 rows and 2 columns, so its order is \( 3 \times 2 \). This new matrix is called the transpose of A.
Transpose of a matrix: the new matrix obtained by changing the rows of a given matrix into columns and its columns into rows is called the transpose of the given matrix.
The transpose of A is written \( A' \), \( A^t \) or \( A^T \). All three mean the same thing and are read "A transpose". In the same way, the transposes of matrices B, C, ... are written \( B', C', \ldots \). For the matrix above we write:
\[ A^T = \begin{bmatrix} 3 & 2 \\ -5 & 4 \\ 7 & -6 \end{bmatrix} \]
Transposing turns an \( m \times n \) matrix into an \( n \times m \) matrix. The number in row i, column j of A moves to row j, column i of \( A^T \). For example, the -6 in row 2, column 3 of A is found in row 3, column 2 of \( A^T \).
When a matrix is not square, its order changes. The transpose of a \( 2 \times 3 \) matrix is a \( 3 \times 2 \) matrix. If your answer still has 2 rows and 3 columns, the rows have not really been turned into columns.
How to Write the Transpose
These steps work for a matrix of any order.
Step 1: Count the rows and columns of the given matrix and write its order:
\[ m \times n \]
Step 2: Draw an empty matrix in which the numbers of rows and columns are swapped, so the new order is:
\[ n \times m \]
Step 3: Copy the first row of the given matrix, reading left to right, into the first column of the new matrix, writing top to bottom.
Step 4: Copy the second row into the second column, and carry on until every row has been used.
Step 5: Check one entry. The number in row i, column j of the given matrix must now stand in row j, column i.
A row matrix has only one row, so its transpose has only one column. A row matrix becomes a column matrix, and a column matrix becomes a row matrix:
\[ \begin{bmatrix} 4 & -1 & 6 \end{bmatrix}^T = \begin{bmatrix} 4 \\ -1 \\ 6 \end{bmatrix} \]
For a square matrix the order stays the same. The numbers on the leading diagonal, the line from the top left corner to the bottom right corner, stay where they are. The other numbers swap places across that diagonal:
\[ \begin{bmatrix} 1 & 5 \\ 0 & 9 \end{bmatrix}^T = \begin{bmatrix} 1 & 0 \\ 5 & 9 \end{bmatrix} \]
Properties of Transpose of Matrix
Let A and B be two matrices of the same order, and let k be any real number. The transpose then follows the rules below. Each rule is checked on an example, so you can see for yourself why it holds.
Transpose of a Sum
The transpose of the sum of two matrices is equal to the sum of their transposes:
\[ (A + B)^T = A^T + B^T \]
Check it with \( A = \begin{bmatrix} 3 & 1 \\ -2 & 5 \end{bmatrix} \) and \( B = \begin{bmatrix} 4 & 0 \\ 6 & -1 \end{bmatrix} \).
Step 1: Adding the corresponding entries of A and B:
\[ A + B = \begin{bmatrix} 3+4 & 1+0 \\ -2+6 & 5-1 \end{bmatrix} = \begin{bmatrix} 7 & 1 \\ 4 & 4 \end{bmatrix} \]
Step 2: Turning the rows of the sum into columns:
\[ (A + B)^T = \begin{bmatrix} 7 & 4 \\ 1 & 4 \end{bmatrix} \]
Step 3: Transposing A and B separately:
\[ A^T = \begin{bmatrix} 3 & -2 \\ 1 & 5 \end{bmatrix} \]
\[ B^T = \begin{bmatrix} 4 & 6 \\ 0 & -1 \end{bmatrix} \]
Step 4: Adding the two transposes:
\[ A^T + B^T = \begin{bmatrix} 3+4 & -2+6 \\ 1+0 & 5-1 \end{bmatrix} = \begin{bmatrix} 7 & 4 \\ 1 & 4 \end{bmatrix} \]
Step 2 and Step 4 give the same matrix, so \( (A + B)^T = A^T + B^T \).
Transpose of a Difference
The transpose of the difference of two matrices is equal to the difference of their transposes:
\[ (A - B)^T = A^T - B^T \]
Check it with \( A = \begin{bmatrix} 8 & 2 \\ 5 & -3 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 6 \\ -4 & 2 \end{bmatrix} \).
Step 1: Subtracting each entry of B from the matching entry of A:
\[ A - B = \begin{bmatrix} 8-1 & 2-6 \\ 5+4 & -3-2 \end{bmatrix} = \begin{bmatrix} 7 & -4 \\ 9 & -5 \end{bmatrix} \]
Step 2: Transposing the difference:
\[ (A - B)^T = \begin{bmatrix} 7 & 9 \\ -4 & -5 \end{bmatrix} \]
Step 3: Transposing A and B separately:
\[ A^T = \begin{bmatrix} 8 & 5 \\ 2 & -3 \end{bmatrix} \]
\[ B^T = \begin{bmatrix} 1 & -4 \\ 6 & 2 \end{bmatrix} \]
Step 4: Subtracting the transposes in the same order, A first:
\[ A^T - B^T = \begin{bmatrix} 8-1 & 5+4 \\ 2-6 & -3-2 \end{bmatrix} = \begin{bmatrix} 7 & 9 \\ -4 & -5 \end{bmatrix} \]
Both matrices are the same, so \( (A - B)^T = A^T - B^T \).
Subtracting a negative number adds it. In Step 1, \( 5 - (-4) = 5 + 4 = 9 \). Most wrong answers in this kind of question come from this one sign.
Transpose of a Scalar Multiple
A real number k that multiplies every entry of a matrix is called a scalar. Multiplying by k and then transposing gives the same result as transposing and then multiplying by k:
\[ (kA)^T = kA^T \]
Check it with \( A = \begin{bmatrix} 1 & 4 \\ -2 & 0 \\ 3 & 5 \end{bmatrix} \) and the scalar \( k = 3 \).
Step 1: Multiplying every entry of A by 3:
\[ 3A = \begin{bmatrix} 3 & 12 \\ -6 & 0 \\ 9 & 15 \end{bmatrix} \]
Step 2: Transposing 3A:
\[ (3A)^T = \begin{bmatrix} 3 & -6 & 9 \\ 12 & 0 & 15 \end{bmatrix} \]
Step 3: Transposing A first:
\[ A^T = \begin{bmatrix} 1 & -2 & 3 \\ 4 & 0 & 5 \end{bmatrix} \]
Step 4: Multiplying every entry of the transpose by 3:
\[ 3A^T = \begin{bmatrix} 3 & -6 & 9 \\ 12 & 0 & 15 \end{bmatrix} \]
The two results match, so \( (kA)^T = kA^T \). The scalar does not move. Only the rows and columns do.
Transpose of a Transpose
Transposing turns rows into columns. Transposing again turns those columns back into rows, so you arrive back at the matrix you started with:
\[ (A^T)^T = A \]
Take \( A = \begin{bmatrix} 2 & -5 \\ 9 & 1 \end{bmatrix} \).
Step 1: Transposing A once:
\[ A^T = \begin{bmatrix} 2 & 9 \\ -5 & 1 \end{bmatrix} \]
Step 2: Transposing the result again:
\[ (A^T)^T = \begin{bmatrix} 2 & -5 \\ 9 & 1 \end{bmatrix} = A \]
Transpose of a Zero Matrix
Zero matrix: a matrix in which every entry is 0. It is written O.
Take the zero matrix of order \( 2 \times 2 \), \( O = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \). Turning its rows into columns gives exactly the same matrix, so:
\[ O^T = O \]
Zero matrices of other orders behave in the same way. The zero matrix \( \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \) of order \( 2 \times 3 \) transposes into \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix} \), the zero matrix of order \( 3 \times 2 \). The zero matrix of order \( 3 \times 2 \) transposes back into the one of order \( 2 \times 3 \). So the transpose of a zero matrix is always a zero matrix, and when the zero matrix is square it is exactly the same matrix.
Transpose of a Unit (Identity) Matrix
Unit or identity matrix: a square matrix with 1 in every place on the leading diagonal and 0 in every other place. It is written I.
The unit matrix of order \( 2 \times 2 \) is \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \). Its first row 1, 0 becomes the first column 1, 0, and its second row 0, 1 becomes the second column 0, 1. Nothing changes, so:
\[ I^T = I \]
The unit matrix of order \( 3 \times 3 \) behaves the same way. The 1s lie on the leading diagonal, which never moves, and every 0 swaps with another 0:
\[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}^T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Transpose of a Symmetric Matrix
Symmetric matrix: a square matrix A is called symmetric if \( A^T = A \), that is, if transposing it does not change it at all.
In a symmetric matrix the numbers are mirrored across the leading diagonal. The entry in row 1, column 2 equals the entry in row 2, column 1, and so on for every pair. For example:
\[ A = \begin{bmatrix} a & b \\ b & c \end{bmatrix}, \; A^T = \begin{bmatrix} a & b \\ b & c \end{bmatrix} \]
A symmetric matrix of order \( 3 \times 3 \) works in the same way. Notice the matching pairs 2 and 2, 3 and 3, 8 and 8 on either side of the diagonal 5, 4, 1:
\[ B = \begin{bmatrix} 5 & 2 & 3 \\ 2 & 4 & 8 \\ 3 & 8 & 1 \end{bmatrix} = B^T \]
Compare this with \( C = \begin{bmatrix} 5 & 2 \\ -2 & 4 \end{bmatrix} \). Its transpose is \( C^T = \begin{bmatrix} 5 & -2 \\ 2 & 4 \end{bmatrix} \), which is not the same as C, because 2 and -2 have swapped. So C is not symmetric.
Only a square matrix can be symmetric. A matrix that is not square changes its order when transposed, so it can never equal its own transpose.
The Properties at a Glance
| Rule | What it says |
| \( (A + B)^T = A^T + B^T \) | The transpose of a sum is the sum of the transposes. |
| \( (A - B)^T = A^T - B^T \) | The transpose of a difference is the difference of the transposes. |
| \( (kA)^T = kA^T \) | A scalar can be taken outside the transpose. |
| \( (A^T)^T = A \) | Transposing twice gives back the original matrix. |
| \( O^T = O \) | A square zero matrix is its own transpose. |
| \( I^T = I \) | A unit matrix is its own transpose. |
| \( A^T = A \) | This is true exactly when A is symmetric. |
Worked Example 1: Finding a Transpose
If \( M = \begin{bmatrix} -2 & 8 \\ -3 & 5 \end{bmatrix} \), find \( M^T \).
Step 1: The given matrix is:
\[ M = \begin{bmatrix} -2 & 8 \\ -3 & 5 \end{bmatrix} \]
Step 2: Writing the first row -2, 8 as the first column and the second row -3, 5 as the second column:
\[ M^T = \begin{bmatrix} -2 & -3 \\ 8 & 5 \end{bmatrix} \]
Worked Example 2: Transposing Twice
If \( B = \begin{bmatrix} 7 & 6 \\ 5 & 4 \end{bmatrix} \), prove that \( (B^T)^T = B \).
Step 1: Transposing B, so the row 7, 6 becomes a column and the row 5, 4 becomes a column:
\[ B^T = \begin{bmatrix} 7 & 5 \\ 6 & 4 \end{bmatrix} \]
Step 2: Transposing \( B^T \) in the same way:
\[ (B^T)^T = \begin{bmatrix} 7 & 6 \\ 5 & 4 \end{bmatrix} = B \]
Hence \( (B^T)^T = B \) is proved.
Worked Example 3: Sum and Symmetry
If \( A = \begin{bmatrix} 4 & 6 \\ 2 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} -2 & -1 \\ 3 & 7 \end{bmatrix} \), answer both parts:
- (a) Prove that \( (A + B)^T = A^T + B^T \).
- (b) Can \( A + B \) be called a symmetric matrix? Give a reason.
Part (a), Step 1: Adding A and B entry by entry:
\[ A + B = \begin{bmatrix} 4-2 & 6-1 \\ 2+3 & 1+7 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix} \]
Step 2: Transposing the sum. Call this result (i):
\[ (A + B)^T = \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix} \quad \text{(i)} \]
Step 3: Transposing A and B separately:
\[ A^T = \begin{bmatrix} 4 & 2 \\ 6 & 1 \end{bmatrix} \]
\[ B^T = \begin{bmatrix} -2 & 3 \\ -1 & 7 \end{bmatrix} \]
Step 4: Adding the transposes. Call this result (ii):
\[ A^T + B^T = \begin{bmatrix} 4-2 & 2+3 \\ 6-1 & 1+7 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix} \quad \text{(ii)} \]
Results (i) and (ii) are the same matrix, so \( (A + B)^T = A^T + B^T \) is proved.
Part (b): From Step 1 and Step 2, \( A + B \) and \( (A + B)^T \) are both \( \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix} \). Since \( (A + B)^T = A + B \), the matrix \( A + B \) is a symmetric matrix. You can also see the matching pair 5 and 5 on either side of the diagonal.
Worked Example 4: Finding Unknown Entries
If \( A = \begin{bmatrix} 6 & 7 \\ 8 & 2 \end{bmatrix} \), \( B = \begin{bmatrix} 8 & -2 \\ 3 & 5 \end{bmatrix} \), \( C = \begin{bmatrix} p & 11 \\ 5 & q \end{bmatrix} \) and \( (A + B)^T = C \), find the values of p and q.
Step 1: Adding A and B:
\[ A + B = \begin{bmatrix} 6+8 & 7-2 \\ 8+3 & 2+5 \end{bmatrix} = \begin{bmatrix} 14 & 5 \\ 11 & 7 \end{bmatrix} \]
Step 2: Transposing the sum:
\[ (A + B)^T = \begin{bmatrix} 14 & 11 \\ 5 & 7 \end{bmatrix} \]
Step 3: Using the condition \( (A + B)^T = C \):
\[ \begin{bmatrix} 14 & 11 \\ 5 & 7 \end{bmatrix} = \begin{bmatrix} p & 11 \\ 5 & q \end{bmatrix} \]
Step 4: Two equal matrices have equal corresponding entries. Matching the top left entries and the bottom right entries:
\[ p = 14 \]
\[ q = 7 \]
So \( p = 14 \) and \( q = 7 \). The other entries, 11 and 11, 5 and 5, also match, which shows the working is consistent.
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