Matrices and Determinants

$A$ is a $3\times 3$ skew-symmetric matrix with arbitrary entries. What is $\det A$?

Zero. Always, for every odd order, whatever the entries are.

A determinant is unchanged by transposition, so $\det A=\det A^T$. Skew-symmetry says $A^T=-A$, and pulling a factor of $-1$ out of each of the three rows gives $\det (-A)=(-1{)}^3\det A$. Putting these together,

$\det A=\det A^T=\det (-A)=-\det A\Rightarrow 2\det A=0.$

Nothing about the individual entries entered the argument. That is the character of this chapter at Advanced level: the questions are about structural properties — symmetry, singularity, commutation, consistency — and they are answered by identities rather than by expansion. A candidate who reaches for the six-term expansion of a $3\times 3$ determinant on seeing this question has already lost the time the question was designed to cost.

1. Why row operations work

For a $2\times 2$ matrix the determinant is, up to sign, the area of the parallelogram spanned by the two rows. That single picture explains the rules that are usually memorised.

Multiplying a row by $k$ stretches one side of the parallelogram, so the area scales by $k$ — hence a common factor comes out of a row, not out of the whole determinant.

Swapping two rows reverses orientation, so the sign flips. Adding a multiple of one row to another slides the parallelogram along a fixed base without changing its height, so the area is untouched. The last of these is the workhorse: it is what lets you create zeros before expanding.

Illustration 1

Evaluate $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$.

Every column sums to $a+b+c$, so replacing $R_1$ by $R_1+R_2+R_3$ makes the first row constant and lets that factor out:

$(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ b& c& a\\ c& a& b\end{array}\right|.$

Now subtract the first column from the second and third to create zeros, and expanding gives $-\left(a^3+b^3+c^3-3abc\right)$. Spotting the constant row sum first is what turns a six-term expansion into two operations.

2. Determinants as polynomials: the factor theorem

If a determinant whose entries are polynomials in $x$ vanishes whenever $x=a$, then two of its rows coincide at that value, so $(x-a)$ divides the determinant as a polynomial. Comparing degrees then pins down the remaining factor up to a constant, which one value of $x$ fixes. This converts several standard determinants into a line of reasoning.

Illustration 2

Show that $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=(a-b)(b-c)(c-a)$.

Regard it as a polynomial in $a$. Setting $a=b$ makes the first two columns identical, so the determinant vanishes and $(a-b)$ is a factor; by symmetry so are $(b-c)$ and $(c-a)$. The determinant has total degree $3$ and so does the product, so they differ by a constant. Comparing the coefficient of $b{c}^2$ on both sides gives that constant as $1$.

This Vandermonde determinant is worth recognising on sight, because it is exactly the condition for three points to lie on no common parabola, and it appears whenever a question asks when three such points are distinct.

Illustration 3

Without expanding, show that $\left|\begin{array}{ccc}x& x^2& 1+x^3\\ y& y^2& 1+y^3\\ z& z^2& 1+z^3\end{array}\right|=0$ when $x,y,z$ are distinct and $xyz=-1$.

Split the third column into $1$ and $x^3$, giving two determinants. The second has columns $x,x^2,x^3$, so factoring $x$, $y$, $z$ out of the rows leaves a Vandermonde with an extra factor $xyz$. The first is the Vandermonde itself with its columns cyclically rearranged. The two therefore combine to $\left(1+xyz\right)$ times a Vandermonde, which vanishes precisely when $xyz=-1$.

3. Adjoint and inverse

Expanding a determinant along a row using the cofactors of a different row gives zero, because the result is the determinant of a matrix with two equal rows. Combining that with the ordinary expansion along the matching row gives the single identity that generates everything else:

$A(\mathrm{adj}A)=(\mathrm{adj}A)A=|A|I_n.$

Taking determinants of both sides gives $|A|\cdot \left|\mathrm{adj}A\right|=|A{|}^n$, so $\left|\mathrm{adj}A\right|=|A{|}^{n-1}$. Applying the adjoint twice and using the identity again gives $\mathrm{adj}\left(\mathrm{adj}A\right)=|A{|}^{n-2}A$. Each of these is derived in one line from the first, which is why the first is the one to remember.

Illustration 4

For a $3\times 3$ matrix with $|A|=4$, find $\left|\mathrm{adj}\left(\mathrm{adj}A\right)\right|$.

Two routes agree. Directly, $\mathrm{adj}\left(\mathrm{adj}A\right)=|A{|}^{1}A=4A$, whose determinant is $4^3\cdot 4=256$. Using the exponent rule twice, $\left|\mathrm{adj}(\mathrm{adj}A)\right|={\left|\mathrm{adj}A\right|}^2={\left(|A{|}^2\right)}^2=|A{|}^4=256$. Whenever an adjoint question offers two routes, taking both costs seconds and catches an exponent slip.

Illustration 5

If $A$ is a $3\times 3$ matrix with $A^2=A$ and $A\ne I$, prove that $A$ is singular.

Rewrite the condition as $A(A-I)=O$. If $|A|\ne 0$ then $A^{-1}$ exists, and multiplying on the left gives $A-I=O$, contradicting $A\ne I$. Hence $|A|=0$.

The step to notice is that $A(A-I)=O$ does not by itself force $A=O$ or $A=I$. Matrices have zero divisors, and the deduction only becomes available once invertibility is assumed.

4. Multiplication is composition, which is why order matters

A matrix acting on a column vector is a transformation of the plane or of space, and the product $AB$ means "apply $B$ first, then $A$". Read that way, non-commutativity stops being a rule to remember and becomes obvious: turning a page and then flipping it does not land where flipping and then turning does.

The definition of the product follows from the same reading. The entry in row $i$, column $j$ of $AB$ is the $i$th component of what $A$ does to the $j$th column of $B$, which is exactly the row-times-column sum.

It also explains conformability: $AB$ exists only when the number of columns of $A$ matches the number of rows of $B$, because the output of $B$ has to be something $A$ can consume. For non-square matrices $AB$ and $BA$ can therefore differ in size, and one of them may not exist at all.

The same reading gives the reversal law. To undo "do $B$, then $A$" you must undo $A$ first, so $(AB{)}^{-1}=B^{-1}{A}^{-1}$, and the transpose obeys the identical reversal, $(AB{)}^T=B^T{A}^T$. Writing $(AB{)}^{-1}=A^{-1}{B}^{-1}$ is one of the most common single-line errors in the chapter, and it is caught instantly by asking which operation happened last.

Illustration 6

Let $R=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$ be a quarter turn anticlockwise and $M=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$ a reflection in the $x$-axis. Compute $RM$ and $MR$ and interpret them.

$RM=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],MR=\left[\begin{array}{cc}0& -1\\ -1& 0\end{array}\right].$

The first swaps the coordinates, so it is the reflection in the line $y=x$; the second sends $(x,y)$ to $(-y,-x)$, the reflection in $y=-x$. Both products are reflections, both are involutory, and they are reflections in perpendicular lines. The two orders do not merely differ by a sign — they give geometrically different maps, and no amount of algebraic rearrangement can reconcile them.

5. What matrix algebra does not inherit from numbers

Three habits imported from ordinary algebra cause most errors here, and each has a one-line counterexample.

For the second, $\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]=O$ with neither factor zero. For the third, expanding honestly gives $X^2+XY+YX+Y^2$, and the middle terms merge only when the two commute.

A small vocabulary of special matrices carries a large share of the questions. $A$ is idempotent if $A^2=A$, nilpotent if $A^k=O$ for some $k$, involutory if $A^2=I$, and orthogonal if $A^{T}A=I$. Each definition immediately constrains the determinant: an orthogonal matrix has $|A{|}^2=1$ so $|A|=\pm 1$, an involutory matrix likewise, and a nilpotent matrix has $|A{|}^k=0$ so $|A|=0$.

Illustration 7

Show that if $A$ is nilpotent then $I-A$ is invertible, and find its inverse.

Suppose $A^k=O$. The finite geometric sum telescopes:

$(I-A)\left(I+A+A^2+\cdots +A^{k-1}\right)=I-A^k=I.$

So the inverse is that finite sum. Nothing here needs $A$ to be small or the series to converge, because the sum stops on its own — the reason the argument works for matrices and not for numbers.

6. Powers of a matrix without induction

When a matrix differs from the identity by a nilpotent part, write $A=I+N$. Since $I$ commutes with everything, the binomial theorem applies, and because $N^k=O$ the expansion terminates.

Illustration 8

Find $A^n$ for $A=\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]$.

Here $N=A-I$ has ones on the first superdiagonal, $N^2$ has a single one in the top right corner, and $N^3=O$. So

$A^n=I+nN+\left(\genfrac{}{}{0ex}{}{n}{2}\right)N^2=\left[\begin{array}{ccc}1& n& \frac{n(n-1)}{2}\\ 0& 1& n\\ 0& 0& 1\end{array}\right].$

Checking at $n=2$ against a direct multiplication confirms the top-right entry $1$, which is the entry an induction argument most often gets wrong.

For $2\times 2$ matrices there is a second route. Expanding directly shows that every $2\times 2$ matrix satisfies

$A^2-(\mathrm{tr}A)A+|A|I=O,$

which lets any high power be reduced step by step to a combination of $A$ and $I$. The general theorem behind this sits outside the stated syllabus, but the $2\times 2$ case is a two-line verification and is worth carrying.

7. Systems of equations, and what three planes can do

For $n$ equations in $n$ unknowns with coefficient determinant $D$, Cramer's rule gives $x_i=D_i/D$ whenever $D\ne 0$: a unique solution. The interesting cases are all at $D=0$.

If $D=0$ and some $D_i\ne 0$, the system is inconsistent and has no solution. If $D=0$ and every $D_i=0$, the system is consistent but the solution is not unique — either a line or a plane of solutions, and Cramer's rule says nothing about which.

Geometrically each equation is a plane. Three planes meet in a single point when $D\ne 0$. When $D=0$ the normals are coplanar, and several pictures become possible: the planes can share a common line, they can be parallel, or — the case most often overlooked — they can form a triangular prism, meeting pairwise in three parallel lines with no point common to all three.

Illustration 9

Show that $x+y=1$, $y+z=1$ and $x-z=1$ have no solution, even though no two of the planes are parallel.

The three normals are $(1,1,0)$, $(0,1,1)$ and $(1,0,-1)$, and the third is the difference of the first two, so $D=0$. Subtracting the second equation from the first gives $x-z=0$, which contradicts the third. The three planes cut each other in three parallel lines forming a prism. No two normals are proportional, so no two planes are parallel — the usual quick test for inconsistency fails here.

Illustration 10

For which $\lambda$ is the system $x+y+z=1$, $x+2y+4z=\lambda$, $x+4y+10z={\lambda }^2$ consistent?

The coefficient determinant is $1(20-16)-1(10-4)+1(4-2)=0$, so consistency is not automatic for any $\lambda$. Subtracting the first equation from the second gives $y+3z=\lambda -1$; subtracting the second from the third gives $2y+6z={\lambda }^2-\lambda$. The second is twice the first only when

$2(\lambda -1)={\lambda }^2-\lambda \Rightarrow {\lambda }^2-3\lambda +2=0,$

so $\lambda =1$ or $\lambda =2$, and in each case there are infinitely many solutions. For every other $\lambda$ the planes form a prism.

8. Homogeneous systems

A homogeneous system $AX=O$ always has the trivial solution $X=O$, so the only question ever asked is whether it has another. It does exactly when $|A|=0$, and then the solutions form a line or a plane through the origin.

Illustration 11

Find all $\lambda$ for which $x+\lambda y+z=0$, $\lambda x+y+z=0$ and $x+y+\lambda z=0$ have a non-trivial solution.

Setting the determinant to zero and using the constant row sum $\lambda +2$,

$\left|\begin{array}{ccc}1& \lambda & 1\\ \lambda & 1& 1\\ 1& 1& \lambda \end{array}\right|=(\lambda +2)\left|\begin{array}{ccc}1& 1& 1\\ \lambda & 1& 1\\ 1& 1& \lambda \end{array}\right|=(\lambda +2)(1-\lambda )(\lambda -1),$

which vanishes at $\lambda =-2$ and $\lambda =1$. At $\lambda =1$ all three equations coincide and the solution set is a plane; at $\lambda =-2$ they are distinct and meet in a line. The two singular values give geometrically different answers, and questions frequently ask which is which.

Illustration 12

If $A$ is a $3\times 3$ matrix with $|A|=0$, must every $2\times 2$ minor of $A$ vanish?

No. The matrix $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$ has determinant zero but a top-left minor equal to $1$. Singularity says the three rows are dependent; it does not say any two of them are. This distinction is what separates a system with a line of solutions from one with a plane of them.

9. Symmetric and skew-symmetric parts

Any square matrix splits uniquely as a symmetric part plus a skew-symmetric part:

$A=\underset{\text{symmetric}}{\underbrace{\frac{1}{2}\left(A+A^T\right)}}+\underset{\text{skew}}{\underbrace{\frac{1}{2}\left(A-A^T\right)}}.$

Skew-symmetry forces every diagonal entry to satisfy $a_{ii}=-a_{ii}$, hence to vanish, which is why a $3\times 3$ skew-symmetric matrix has only three free entries and why the determinant argument in the opening had so little to work with.

Illustration 13

If $A$ is skew-symmetric of odd order, show that $\mathrm{adj}A$ is symmetric or skew-symmetric according to the order.

Since $|A|=0$ we already know $A$ is singular, so the adjoint identity gives $A\left(\mathrm{adj}A\right)=O$. For the symmetry itself, transposing a cofactor matrix transposes each minor, and each minor of $A$ picks up $(-1)$ from every row it uses. For a $3\times 3$ matrix the minors are $2\times 2$, so each is unchanged, and $\mathrm{adj}A$ comes out symmetric. The general rule is that $\mathrm{adj}A$ is symmetric when $n$ is odd and skew-symmetric when $n$ is even.

Illustration 14

Express $\left[\begin{array}{cc}2& 5\\ 1& 3\end{array}\right]$ as a symmetric plus a skew-symmetric matrix.

The symmetric part is $\frac{1}{2}\left(A+A^T\right)=\left[\begin{array}{cc}2& 3\\ 3& 3\end{array}\right]$ and the skew part is $\frac{1}{2}\left(A-A^T\right)=\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]$. Adding them returns $A$, and the skew part has the zero diagonal it must have.

Summary

Advanced treats a determinant as a structural quantity, not a number to be expanded. The area picture explains why a common factor comes out of a single row, why a swap flips the sign, and why adding a multiple of one row to another changes nothing. A determinant whose entries are polynomials can be factorised by the factor theorem: it vanishes when two rows coincide, so the corresponding difference divides it, and a degree count finishes the job.

Everything about adjoints follows from $A\left(\mathrm{adj}A\right)=|A|I$. Taking determinants gives $\left|\mathrm{adj}A\right|=|A{|}^{n-1}$, and applying the identity twice gives $\mathrm{adj}\left(\mathrm{adj}A\right)=|A{|}^{n-2}A$. Matrix algebra loses commutativity and gains zero divisors, so $XY=O$ implies nothing about $X$ or $Y$ until one of them is known to be invertible, and the square of a sum keeps its cross terms separate.

Powers are best handled by splitting off a nilpotent part and letting the binomial expansion terminate, and for $2\times 2$ matrices by the relation $A^2-(\mathrm{tr}A)A+|A|I=O$.

For systems, $D\ne 0$ gives a unique solution; $D=0$ with some $D_i\ne 0$ gives none; $D=0$ with all $D_i=0$ gives infinitely many. Geometrically the last two are the difference between a triangular prism and a common line, and no two planes need be parallel for a system to be inconsistent. A homogeneous system has a non-trivial solution exactly when the determinant vanishes, and whether the solution set is a line or a plane depends on the minors, not on the determinant alone.