Ma · Mathematics

Examples of 3×3 Matrices With Common Properties

Published By the 9to5Science editorial team

At a glance

  1. 1. Specify the property

    The title does not name particular conditions, so first identify which matrix properties you need.

  2. 2. Choose a standard example

    Use the examples below for symmetry, invertibility, rank, orthogonality, idempotence, nilpotence, or noncommutativity.

  3. 3. Check each condition

    Verify a property by computing the relevant transpose, determinant, rank, power, or matrix products.

  4. 4. Check compatibility

    Some properties cannot hold together, so test the conditions as a group rather than separately.

The phrase “meet properties” does not identify which properties are required. The examples below cover several common conditions on real 3×33\times3 matrices, so you can select the ones that fit your problem. Each example includes a check that shows why it works.

If you have a specific list of conditions, apply them together: a matrix satisfying each condition separately does not necessarily satisfy all of them at once.

Common examples

Here are useful examples, written by rows. All entries are real.

PropertyExampleQuick check
IdentityI=(100010001)I=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}IA=AI=AIA=AI=A for every 3×33\times3 matrix AA.
SymmetricS=(12023−10−14)S=\begin{pmatrix}1&2&0\\2&3&-1\\0&-1&4\end{pmatrix}ST=SS^T=S.
Skew-symmetricK=(01−2−1032−30)K=\begin{pmatrix}0&1&-2\\-1&0&3\\2&-3&0\end{pmatrix}KT=−KK^T=-K.
InvertibleA=(12003100−2)A=\begin{pmatrix}1&2&0\\0&3&1\\0&0&-2\end{pmatrix}It is triangular and det⁡(A)=1⋅3⋅(−2)=−6≠0\det(A)=1\cdot3\cdot(-2)=-6\ne0.
Singular, rank twoB=(100010110)B=\begin{pmatrix}1&0&0\\0&1&0\\1&1&0\end{pmatrix}The third column is zero, and the first two columns are independent; hence rank⁡(B)=2\operatorname{rank}(B)=2.
IdempotentP=(100010000)P=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}P2=PP^2=P.
NilpotentN=(010001000)N=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}N3=0N^3=0, although N2≠0N^2\ne0.
OrthogonalQ=(0−10100001)Q=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix}QTQ=IQ^TQ=I.

How to verify a property

Choose the test that matches the definition. To check symmetry, transpose the matrix and compare it with the original. To check skew-symmetry, compare the transpose with the negative of the matrix. For invertibility, a nonzero determinant is a quick test; for triangular matrices, multiply the diagonal entries to obtain the determinant.

For rank, count the independent rows or columns after row reduction. For idempotence, multiply the matrix by itself. For nilpotence, keep taking powers until a power is zero. For orthogonality, verify QTQ=IQ^TQ=I; equivalently, the columns of QQ are mutually perpendicular unit vectors.

Worked check: an idempotent matrix

For P=(100010000)P=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}, direct multiplication gives

P2=(100010000)(100010000)=(100010000)=P.P^2=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix} \begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix} =\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}=P.

So PP is idempotent. It is also singular, since its determinant is zero. In fact, its rank is two. This illustrates that one matrix can have several properties at once.

A pair of matrices that do not commute

If you need matrices AA and BB for which AB≠BAAB\ne BA, use

A=(010000000),B=(000100000).A=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix},\qquad B=\begin{pmatrix}0&0&0\\1&0&0\\0&0&0\end{pmatrix}.

Multiplication gives AB=(100000000)AB=\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}, while BA=(000010000)BA=\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}. Thus AB≠BAAB\ne BA. Matrix multiplication is not commutative in general, even for square matrices of the same size.

Common mistakes and incompatible conditions

  • Checking only one condition: If a task asks for a symmetric, invertible matrix, the example must satisfy both requirements. The symmetric matrix in the table is not automatically a suitable example for every other property.
  • Confusing singularity with rank zero: A singular 3×33\times3 matrix has rank less than three, but it can still have rank one or two. The zero matrix has rank zero.
  • Assuming a nilpotent matrix is invertible: A nilpotent matrix has determinant zero, so it cannot be invertible. For example, NN above is nonzero but N3=0N^3=0.
  • Overlooking special cases: A matrix that is both symmetric and skew-symmetric over the real numbers must be zero. Indeed, AT=AA^T=A and AT=−AA^T=-A imply A=−AA=-A, hence A=0A=0.
  • Treating all idempotent matrices as invertible: An invertible idempotent matrix must be the identity. If P2=PP^2=P and PP is invertible, multiply by P−1P^{-1} to get P=IP=I.

These checks help distinguish properties that can be combined from conditions that rule one another out.

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