Matrices

1. Check this before you revise anything

Exercise 3.4 has exactly one question in the current book — not the eighteen-question elementary-operations drill older material remembers. Section 3.7 (Invertible Matrices) now contains only a definition, a proof that an inverse is unique when it exists, and a proof that $(AB{)}^{-1}=B^{-1}{A}^{-1}$. It never shows how to compute an inverse.

There is no "elementary row/column operations" section anywhere in this chapter any more — confirmed by reading straight through from the definition of invertibility to the Miscellaneous Examples, with nothing about row-reducing $[A|I]$ in between.

The old stub taught that removed method as a full section, framing "finding the inverse using elementary operations" as one of the chapter's main skills. It isn't taught here at all in the current edition — computing an actual inverse is Chapter 4's job, via the adjoint of a matrix. This chapter only covers existence and uniqueness of the inverse, never its computation.

The old stub also collapsed all four real exercises into one invented 30-question group. The book actually has Exercise 3.1 (10 questions, notation and order), Exercise 3.2 (22 questions, the arithmetic — addition, scalar multiplication, multiplication), Exercise 3.3 (12 questions, transpose and symmetric/skew-symmetric matrices), Exercise 3.4 (1 question), and a Miscellaneous Exercise (11 questions) — 56 questions in total.

2. What this chapter covers

3. Matrix arithmetic

A matrix of order $m\times n$ has $m$ rows and $n$ columns. Two matrices can only be added if they share the same order — add entry by entry. Multiplication is stricter still: $A_{m\times n}{B}_{n\times p}$ is only defined when the inner dimensions match, and the result is $m\times p$.

Matrix multiplication is not commutative — $AB\ne BA$ in general, and one side may not even be defined when the other is. It's worth internalising with a concrete pair rather than just the rule: for $A=\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]$ and $B=\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]$, $AB=\left[\begin{array}{cc}7& 1\\ 33& 34\end{array}\right]$ while $BA=\left[\begin{array}{cc}16& 5\\ 39& 25\end{array}\right]$ — visibly different matrices, not just different-looking arithmetic.

Types of matrices (3.3). The named types recur constantly in later proofs, so they are worth fixing early.

Equality of matrices (3.3.1). Two matrices are equal only when they have the same order and every corresponding pair of entries agrees. This is what turns a matrix equation into a system of simultaneous scalar equations, which is how most Exercise 3.1 and 3.2 questions are solved.

How the product is actually formed (3.4.5). The entry in row $i$, column $j$ of $AB$ is the dot product of row $i$ of $A$ with column $j$ of $B$:

$(AB{)}_{ij}={\sum }_{k=1}^n{a}_{ik}{b}_{kj}$

This is why the inner dimensions must agree — the row and the column being paired must have the same length.

Properties that do hold (3.4.3, 3.4.4, 3.4.6). Addition is commutative and associative, with $O$ as identity and $-A$ as inverse. Scalar multiplication distributes both ways: $k(A+B)=kA+kB$ and $(k+l)A=kA+lA$.

Multiplication, despite failing commutativity, is associative, $(AB)C=A(BC)$, and distributive over addition, $A(B+C)=AB+AC$. The identity satisfies $AI=IA=A$.

Two failures that have no scalar analogue. First, $AB=O$ does not imply $A=O$ or $B=O$ — a product of two non-zero matrices can be the zero matrix. Second, cancellation fails: $AB=AC$ with $A\ne O$ does not give $B=C$.

Both catch students who carry scalar habits into matrix algebra, and both appear as explicit exercise questions.

4. Transpose, symmetric, and skew-symmetric matrices

The transpose $A^{\prime }$ (or $A^T$) swaps rows and columns. Two properties carry the most weight in proofs: $(A^{\prime }{)}^{\prime }=A$, and $(AB{)}^{\prime }=B^{\prime }{A}^{\prime }$ — the order reverses.

A square matrix is symmetric if $A^{\prime }=A$, and skew-symmetric if $A^{\prime }=-A$ (which forces every diagonal entry to be zero, since $a_{ii}=-a_{ii}$ only when $a_{ii}=0$).

Every square matrix splits into a symmetric part and a skew-symmetric part:

$A=\underset{\text{symmetric}}{\underbrace{\frac{1}{2}(A+A^{\prime })}}+\underset{\text{skew-symmetric}}{\underbrace{\frac{1}{2}(A-A^{\prime })}}$

Further transpose properties (3.5.1). Alongside $(A^{\prime }{)}^{\prime }=A$ and $(AB{)}^{\prime }=B^{\prime }{A}^{\prime }$, the book records $(kA{)}^{\prime }=k{A}^{\prime }$ for a scalar $k$, and $(A+B{)}^{\prime }=A^{\prime }+B^{\prime }$. Only the product rule reverses the order; addition and scalar multiplication pass through unchanged.

The reversal in $(AB{)}^{\prime }=B^{\prime }{A}^{\prime }$ extends to longer products: $(ABC{)}^{\prime }=C^{\prime }{B}^{\prime }{A}^{\prime }$.

This is not just an identity to memorise — it's the standard technique for an entire class of exercise questions ("express this matrix as the sum of a symmetric and a skew-symmetric matrix"), and it works because ${\left(\frac{1}{2}(A+A^{\prime })\right)}^{\prime }=\frac{1}{2}(A^{\prime }+A)=\frac{1}{2}(A+A^{\prime })$ (unchanged under transpose, hence symmetric), while ${\left(\frac{1}{2}(A-A^{\prime })\right)}^{\prime }=\frac{1}{2}(A^{\prime }-A)=-\frac{1}{2}(A-A^{\prime })$ (negated under transpose, hence skew-symmetric).

The decomposition is unique, which is why the question always has exactly one correct answer. Note also that for a skew-symmetric matrix the zero diagonal is forced, not optional — so a skew-symmetric matrix of odd order always has determinant zero, a fact Chapter 4 uses.

A worked check. For $A=\left[\begin{array}{cc}2& 4\\ 6& 8\end{array}\right]$, the transpose is $A^{\prime }=\left[\begin{array}{cc}2& 6\\ 4& 8\end{array}\right]$. Then $\frac{1}{2}(A+A^{\prime })=\left[\begin{array}{cc}2& 5\\ 5& 8\end{array}\right]$, which is symmetric, and $\frac{1}{2}(A-A^{\prime })=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$, which is skew-symmetric with a zero diagonal as required. Adding them returns $A$.

5. Invertible matrices

A square matrix $A$ of order $m$ is invertible if there exists a square matrix $B$ of the same order with $AB=BA=I$; $B$ is called $A^{-1}$. Two facts the book proves formally:

• The inverse, if it exists, is unique. (If $B$ and $C$ were both inverses of $A$, then $B=BI=B(AC)=(BA)C=IC=C$.)
• $(AB{)}^{-1}=B^{-1}{A}^{-1}$, for invertible $A,B$ of the same order.

Only square matrices can be invertible — for $AB$ and $BA$ to both be defined and equal, $A$ and $B$ must be square matrices of the same order.

What this chapter deliberately does not do. It proves the inverse is unique and states how inverses interact with products, but it never computes one. There is no elementary-row-operations method and no adjoint formula here. Actually finding $A^{-1}$ belongs to Chapter 4, which builds it from the determinant and the adjoint.

So a question in this chapter asking you to "show $B$ is the inverse of $A$" is answered by multiplying the two matrices and checking both products equal $I$ — not by computing an inverse from scratch.

The transpose of an inverse satisfies ${\left(A^{-1}\right)}^{\prime }={\left(A^{\prime }\right)}^{-1}$, so transposing and inverting can be done in either order.

Summary

• Addition needs matching order; multiplication needs matching inner dimensions; multiplication is not commutative.
• $(A^{\prime }{)}^{\prime }=A$; $(AB{)}^{\prime }=B^{\prime }{A}^{\prime }$ (order reverses).
• Symmetric: $A^{\prime }=A$. Skew-symmetric: $A^{\prime }=-A$, forcing a zero diagonal.
• Every square matrix decomposes as $\frac{1}{2}(A+A^{\prime })+\frac{1}{2}(A-A^{\prime })$, the symmetric part plus the skew-symmetric part.
• An inverse, when it exists, is unique; $(AB{)}^{-1}=B^{-1}{A}^{-1}$.
• Equality of matrices requires the same order and every corresponding entry equal — this is what converts a matrix equation into simultaneous scalar equations.
• $(AB{)}_{ij}$ is row $i$ of $A$ dotted with column $j$ of $B$, which is why the inner dimensions must agree.
• Matrix multiplication is associative and distributive, and $AI=IA=A$ — it fails only commutativity.
• $AB=O$ does not imply $A=O$ or $B=O$, and $AB=AC$ with $A\ne O$ does not imply $B=C$; both scalar habits break here.
• $(kA{)}^{\prime }=k{A}^{\prime }$ and $(A+B{)}^{\prime }=A^{\prime }+B^{\prime }$ pass through unchanged; only the product reverses, $(ABC{)}^{\prime }=C^{\prime }{B}^{\prime }{A}^{\prime }$.
• The symmetric/skew-symmetric decomposition is unique, and a skew-symmetric matrix always has a zero diagonal.
• ${\left(A^{-1}\right)}^{\prime }={\left(A^{\prime }\right)}^{-1}$ — transposing and inverting commute.
• This chapter proves an inverse is unique but never computes one — no elementary-operations method appears here. Computing an actual inverse (via the adjoint) is Chapter 4's content, not this chapter's.