Matrices • Topic 3 of 3

Transpose, Symmetric & Invertible Matrices

The transpose of a matrix $A$, written $A^{T}$ (or $A'$), is obtained by interchanging its rows and columns — the first row becomes the first column, and so on. Formally $(A^{T})_{ij}=a_{ji}$, and if $A$ is $m\times n$ then $A^{T}$ is $n\times m$.

Properties of the transpose

  • $(A^{T})^{T}=A$ — transposing twice returns the original.
  • $(A+B)^{T}=A^{T}+B^{T}$ — transpose distributes over a sum.
  • $(kA)^{T}=k\,A^{T}$ for any scalar $k$.
  • Reversal law: $(AB)^{T}=B^{T}A^{T}$ — the order flips.

Symmetric and skew-symmetric matrices

A square matrix is symmetric if $A^{T}=A$ (equivalently $a_{ij}=a_{ji}$, so the matrix is a mirror image across its main diagonal). It is skew-symmetric if $A^{T}=-A$ (so $a_{ij}=-a_{ji}$); putting $i=j$ forces $a_{ii}=-a_{ii}$, hence every diagonal entry of a skew-symmetric matrix is $0$.

A central theorem: every square matrix can be written, in exactly one way, as the sum of a symmetric and a skew-symmetric matrix:

$$A=\underbrace{\tfrac12\!\left(A+A^{T}\right)}_{\text{symmetric } P}+\underbrace{\tfrac12\!\left(A-A^{T}\right)}_{\text{skew-symmetric } Q}.$$

You can check $P^{T}=P$ and $Q^{T}=-Q$ directly from the transpose rules — that is the whole proof. A matrix that is both symmetric and skew-symmetric satisfies $A=A^{T}=-A$, forcing $A=O$ (the zero matrix is the only one).

Elementary operations (transformations)

Three elementary row operations (and their column analogues) are allowed on a matrix:

  • Interchange two rows: $R_i \leftrightarrow R_j$.
  • Scale a row by a non-zero constant: $R_i \to k\,R_i,\ k\ne 0$.
  • Add a multiple of one row to another: $R_i \to R_i + k\,R_j$.

These are the tools for the elementary-operations method of inversion.

Invertible matrices and finding the inverse

A square matrix $A$ is invertible if there exists a square matrix $B$ of the same order with $AB=BA=I$. Then $B$ is unique, is called the inverse $A^{-1}$, and the reversal law holds: $(AB)^{-1}=B^{-1}A^{-1}$. An inverse exists precisely when $\det A\ne 0$ (a fact established in the Determinants chapter).

Inverse by elementary row operations. Write $A=IA$. Apply row operations to the left-hand $A$ to turn it into $I$, applying the same operations to the $I$ on the right. When the left side becomes $I$, the right side has become $A^{-1}$:

$$A=IA\ \xrightarrow{\ \text{row ops}\ }\ I=A^{-1}A,\ \text{ so the right factor is } A^{-1}.$$

The column version uses $A=AI$ and applies column operations instead. Important: in one inversion never mix row and column operations — choose one kind and stay with it. If, while reducing, a full row (or column) of the left matrix becomes all zeros, then $A$ is singular and $A^{-1}$ does not exist.

For the special $2\times2$ case there is a shortcut worth memorising: if $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ with $ad-bc\ne0$, then $A^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$ (swap the diagonal, negate the off-diagonal, divide by the determinant).

Deeper Insight — the reversal laws are the signature of non-commutativity. Both $(AB)^{T}=B^{T}A^{T}$ and $(AB)^{-1}=B^{-1}A^{-1}$ flip the order. This is no accident: undoing "do $A$, then $B$" means "undo $B$, then undo $A$" — last in, first out, like removing shoes before socks. Once you see transpose and inverse as ways of reversing a composition, remembering which order they take becomes automatic rather than a rule to memorise.

Transpose of a 2 by 3 matrix: rows of A become columns of A transpose, shown with mapping arrows Transpose: rows become columns 123 456 A (2×3) Aᵀ 14 25 36 Aᵀ (3×2) row 1↓col 1 (Aᵀ)ₙₚ = aₚₙ  —  interchange rows and columns A symmetric matrix with its main diagonal drawn and mirror-image entries equal across the diagonal, plus the inverse relation A times A inverse equals I Symmetric matrix: aₙₚ = aₚₙ 259 33 77 11 main diagonal mirror pairs are equal Invertible matrix: A · A⁻¹ = A⁻¹ · A = I
Solved Examples
1
Worked Example
Find $A^{T}$ for $A=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}$.
Solution

Swap rows and columns: $A^{T}=\begin{bmatrix}1&4\\2&5\\3&6\end{bmatrix}$ (a $3\times2$ matrix).

Answer: $A^{T}=\begin{bmatrix}1&4\\2&5\\3&6\end{bmatrix}$

2
Worked Example
Show $A=\begin{bmatrix}0&2\\-2&0\end{bmatrix}$ is skew-symmetric.
Solution

$A^{T}=\begin{bmatrix}0&-2\\2&0\end{bmatrix}=-A$. The diagonal entries are $0$ as required, so $A$ is skew-symmetric.

Answer: $A^{T}=-A$, so $A$ is skew-symmetric.

3
Worked Example
Express $A=\begin{bmatrix}2&3\\1&4\end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.
Solution

$A^{T}=\begin{bmatrix}2&1\\3&4\end{bmatrix}$. Symmetric part $\tfrac12(A+A^{T})=\begin{bmatrix}2&2\\2&4\end{bmatrix}$; skew part $\tfrac12(A-A^{T})=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$. Their sum is $A$.

Answer: $A=\begin{bmatrix}2&2\\2&4\end{bmatrix}+\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ (symmetric + skew-symmetric).

4
Worked Example
If $A=\begin{bmatrix}2&0\\0&3\end{bmatrix}$, find $A^{-1}$.
Solution

For a diagonal matrix invert each diagonal entry: $A^{-1}=\begin{bmatrix}\tfrac12&0\\0&\tfrac13\end{bmatrix}$. Check: $AA^{-1}=I$.

Answer: $A^{-1}=\begin{bmatrix}\tfrac12&0\\0&\tfrac13\end{bmatrix}$

5
Worked Example
Verify the reversal law $(AB)^{T}=B^{T}A^{T}$ for $A=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $B=\begin{bmatrix}1&0\\3&1\end{bmatrix}$.
Solution

$AB=\begin{bmatrix}1&2\\0&1\end{bmatrix}\begin{bmatrix}1&0\\3&1\end{bmatrix}=\begin{bmatrix}7&2\\3&1\end{bmatrix}$, so $(AB)^{T}=\begin{bmatrix}7&3\\2&1\end{bmatrix}$. Now $B^{T}A^{T}=\begin{bmatrix}1&3\\0&1\end{bmatrix}\begin{bmatrix}1&0\\2&1\end{bmatrix}=\begin{bmatrix}7&3\\2&1\end{bmatrix}$. Both sides agree.

Answer: Both sides equal $\begin{bmatrix}7&3\\2&1\end{bmatrix}$, so $(AB)^{T}=B^{T}A^{T}$.

6
Worked Example
Find $A^{-1}$ for $A=\begin{bmatrix}2&3\\1&4\end{bmatrix}$ using the $2\times2$ shortcut, then confirm $AA^{-1}=I$.
Solution

Here $\det A=2\cdot4-3\cdot1=5\ne0$, so $A^{-1}=\tfrac15\begin{bmatrix}4&-3\\-1&2\end{bmatrix}$. Check: $A A^{-1}=\tfrac15\begin{bmatrix}2&3\\1&4\end{bmatrix}\begin{bmatrix}4&-3\\-1&2\end{bmatrix}=\tfrac15\begin{bmatrix}5&0\\0&5\end{bmatrix}=I$.

Answer: $A^{-1}=\tfrac15\begin{bmatrix}4&-3\\-1&2\end{bmatrix}$

7
Worked Example
Find the inverse of $A=\begin{bmatrix}2&1\\7&4\end{bmatrix}$ by elementary row operations.
Solution

Start from $A=IA$: $\begin{bmatrix}2&1\\7&4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A$. Apply $R_1\to\tfrac12R_1$: $\begin{bmatrix}1&\tfrac12\\7&4\end{bmatrix}=\begin{bmatrix}\tfrac12&0\\0&1\end{bmatrix}A$. Then $R_2\to R_2-7R_1$: $\begin{bmatrix}1&\tfrac12\\0&\tfrac12\end{bmatrix}=\begin{bmatrix}\tfrac12&0\\-\tfrac72&1\end{bmatrix}A$. Now $R_2\to 2R_2$: $\begin{bmatrix}1&\tfrac12\\0&1\end{bmatrix}=\begin{bmatrix}\tfrac12&0\\-7&2\end{bmatrix}A$. Finally $R_1\to R_1-\tfrac12R_2$: $\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}4&-1\\-7&2\end{bmatrix}A$. Hence $A^{-1}=\begin{bmatrix}4&-1\\-7&2\end{bmatrix}$.

Answer: $A^{-1}=\begin{bmatrix}4&-1\\-7&2\end{bmatrix}$

8
Worked Example
Find the inverse of $A=\begin{bmatrix}1&2&3\\2&5&7\\-2&-4&-5\end{bmatrix}$ by elementary row operations.
Solution

Begin with $A=IA$ and reduce the left side to $I$, mirroring each step on the right. Using $R_2\to R_2-2R_1$ and $R_3\to R_3+2R_1$ gives left $\begin{bmatrix}1&2&3\\0&1&1\\0&0&1\end{bmatrix}$ with right $\begin{bmatrix}1&0&0\\-2&1&0\\2&0&1\end{bmatrix}$. Back-substitute with $R_2\to R_2-R_3$, $R_1\to R_1-3R_3$, then $R_1\to R_1-2R_2$. The right side becomes $A^{-1}=\begin{bmatrix}3&-2&-1\\-4&1&-1\\2&0&1\end{bmatrix}$. (Quick check: the middle row of $A\,A^{-1}$ is $\begin{bmatrix}0&1&0\end{bmatrix}$.)

Answer: $A^{-1}=\begin{bmatrix}3&-2&-1\\-4&1&-1\\2&0&1\end{bmatrix}$

9
Worked Example
Express $A=\begin{bmatrix}1&2&4\\6&8&1\\3&5&7\end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.
Solution

$A^{T}=\begin{bmatrix}1&6&3\\2&8&5\\4&1&7\end{bmatrix}$. Symmetric part $P=\tfrac12(A+A^{T})=\begin{bmatrix}1&4&\tfrac72\\4&8&3\\\tfrac72&3&7\end{bmatrix}$; skew part $Q=\tfrac12(A-A^{T})=\begin{bmatrix}0&-2&\tfrac12\\2&0&-2\\-\tfrac12&2&0\end{bmatrix}$. Then $P+Q=A$, with $P^{T}=P$ and $Q^{T}=-Q$.

Answer: $A=P+Q$ with $P=\begin{bmatrix}1&4&\tfrac72\\4&8&3\\\tfrac72&3&7\end{bmatrix}$ (symmetric) and $Q=\begin{bmatrix}0&-2&\tfrac12\\2&0&-2\\-\tfrac12&2&0\end{bmatrix}$ (skew-symmetric).

10
Worked Example
If $A$ is skew-symmetric, prove that $A^{2}$ is symmetric.
Solution

Skew-symmetry means $A^{T}=-A$. Using the reversal law, $(A^{2})^{T}=(A\,A)^{T}=A^{T}A^{T}=(-A)(-A)=A^{2}$. Since $(A^{2})^{T}=A^{2}$, the matrix $A^{2}$ is symmetric.

Answer: $(A^{2})^{T}=A^{2}$, so $A^{2}$ is symmetric.

11
Worked Example
If $A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$, prove by induction that $A^{n}=\begin{bmatrix}1&n\\0&1\end{bmatrix}$ for all $n\in\mathbb{N}$.
Solution

Base step: for $n=1$, $A^{1}=\begin{bmatrix}1&1\\0&1\end{bmatrix}$, which matches. Inductive step: assume $A^{k}=\begin{bmatrix}1&k\\0&1\end{bmatrix}$. Then $A^{k+1}=A^{k}A=\begin{bmatrix}1&k\\0&1\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}1&k+1\\0&1\end{bmatrix}$, the claim for $n=k+1$. By induction the formula holds for every natural $n$.

Answer: $A^{n}=\begin{bmatrix}1&n\\0&1\end{bmatrix}$ for all $n\in\mathbb{N}$ (proved by induction).

12
Worked Example
Solve the matrix equation $AX=B$ for $X$, where $A=\begin{bmatrix}2&5\\1&3\end{bmatrix}$ and $B=\begin{bmatrix}4\\1\end{bmatrix}$.
Solution

Since $\det A=2\cdot3-5\cdot1=1\ne0$, $A$ is invertible with $A^{-1}=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}$. Then $X=A^{-1}B=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}\begin{bmatrix}4\\1\end{bmatrix}=\begin{bmatrix}7\\-2\end{bmatrix}$. So $X=\begin{bmatrix}7\\-2\end{bmatrix}$.

Answer: $X=\begin{bmatrix}7\\-2\end{bmatrix}$

Key Points

  • $A^{T}$ swaps rows and columns: $(A^{T})_{ij}=a_{ji}$, and $(A^{T})^{T}=A$, $(A+B)^{T}=A^{T}+B^{T}$, $(kA)^{T}=kA^{T}$.
  • Reversal laws flip the order: $(AB)^{T}=B^{T}A^{T}$ and $(AB)^{-1}=B^{-1}A^{-1}$.
  • Symmetric: $A^{T}=A$. Skew-symmetric: $A^{T}=-A$, which forces all diagonal entries to be $0$.
  • Every square matrix $=\tfrac12(A+A^{T})$ (symmetric) $+\ \tfrac12(A-A^{T})$ (skew-symmetric), uniquely.
  • Three elementary operations: interchange two rows/columns, scale by a non-zero constant, add a multiple of one to another.
  • $A$ is invertible iff $\det A\ne0$; the inverse is unique and satisfies $AA^{-1}=A^{-1}A=I$.
  • Inverse by elementary operations: from $A=IA$, row-reduce the left $A$ to $I$ while mirroring on the right (never mix row and column operations).
  • $2\times2$ shortcut: $\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$ when $ad-bc\ne0$.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.$(AB)^{T}$ equals:
Explanation: Reversal law for transpose.
Q2.A skew-symmetric matrix must have diagonal entries:
Explanation: $a_{ii}=-a_{ii}\Rightarrow a_{ii}=0$.
Q3.The symmetric part of $A$ is:
Explanation: $\tfrac12(A+A^{T})$ is symmetric; $\tfrac12(A-A^{T})$ is skew.
Q4.$A$ is invertible if and only if:
Explanation: A non-zero determinant is exactly the invertibility condition.
Practice more MCQs → 📝 Assignment · 35 marks