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All the transformations below fix the origin. A planar rotation preserves lengths and angles and turns every vector through one common oriented angle. Its rotation matrix isAn orthogonal reflection in a line fixes that line and reverses its perpendicular direction; reflection in the horizontal axis has reflection matrix $\mathrm{diag}(1,-1)$. A uniform dilation multiplies every vector by the same positive scale $s$, represented by $sI$; for example $2I$ doubles lengths. A shear mapping fixes one line pointwise and shifts vectors parallel to that line by an amount proportional to their transverse coordinate. A horizontal shear mapping has matrixwith $S_1$ a nontrivial example.

The first given matrix is the clockwise planar rotation through $\pi /4$, namely $A=R_{-\pi /4}$. Direct multiplication givesThus $C$ is a horizontal shear mapping of strength $2$. Because $A^{-1}=R_{\pi /4}$,So $B$ is the composition of that shear mapping followed by an anticlockwise planar rotation through $\pi /4$; order matters. It is not individually one of the four elementary types under these definitions. In particular, its determinant is $1$, but it is not an orthogonal matrix, excluding both a planar rotation and an orthogonal reflection. It is not a scalar matrix, excluding a uniform dilation; its eigenvalues $\sqrt{2}+1$ and $\sqrt{2}-1$ exclude a pure shear mapping, whose eigenvalues are both $1$. As a symmetric positive-definite matrix, $B$ can also be described as stretching two perpendicular principal directions by these reciprocal factors, preserving area.

Denote the four displayed matrices, from left to right, by $T_1,T_2,T_3,T_4$. An orthogonal matrix has an orthonormal set of columns, equivalently $T^{T}T=I$. Direct column dot products give $T_j^T{T}_j=I$ for $j=1,2,4$. The first column of $T_3$ has squared Euclidean norm $(1+6+1)/6=4/3$, so the third matrix is the unique nonorthogonal matrix.

For $T_1$, the determinant is $1$. A nonidentity orthogonal transformation of ${\mathbb{R}}^3$ with determinant $1$ is a rotation: it has a fixed axis and restricts to a plane rotation on its orthogonal complement. Consequently $T_1$ is the rotation.

For $T_2$, calculation givesA plane reflection fixes its entire plane, and therefore has eigenvalue $1$. Since $T_2-I$ is invertible, $T_2$ is not a plane reflection. Thus $T_2$ is the combination of a rotation and a reflection. More explicitly, $(1,-2,0{)}^T$ is an eigenvector with eigenvalue $-1$. On the perpendicular plane, $T_2$ is a rotation with $\cos \varphi =2/3$, because $\mathrm{tr}{T}_2=-1+2\cos \varphi =1/3$. This gives a three-dimensional improper orthogonal transformation.

For $T_4$, set $n=(1,2,2{)}^T/3$. ThenThis reflection matrix reverses the normal $n$ and fixes every vector perpendicular to $n$. Therefore $T_4$ is reflection in the plane $x+2y+2z=0$. In particular, $\det T_4=-1$ and $\det (T_4-I)=0$. Finally $T_3$ represents none of the three listed orthogonal transformations, since each preserves the Euclidean norm.

Choose the $z$-axis along the line of intersection of the two planes. Their unit normals lie in the $xy$-plane; after choosing the $x$-axis, takewhere $\theta$ is the oriented angle between the planes. The two reflection matrices restrict toon the $xy$-plane and both fix the $z$-axis. Their product isThus the composition of two plane reflections is a rotation about the intersection line through angle $2\theta$.

Writing $n$ as a column vector, the reflection isso its reflection matrix is