Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 18

Expanding the determinant along the first row gives

The matrix of cofactors is $\left(\begin{array}{ccc}-1& 2& -3\\ 2& 0& -2\\ -3& -2& 3\end{array}\right)$. Transposing and dividing by the nonzero determinant gives the inverse matrixMultiplication by $S$ verifies the identity matrix.

Take the diagonal matrix $T=\mathrm{diag}(1,0,0)$. Multiplying on the left retains the first row of $S$, while multiplying on the right retains its first column. ThusTheir $(1,3)$ entries differ, so this is an explicit example of noncommuting matrices.

The Leibniz formula for determinants gives $\det (A^T)={\sum }_{\sigma \in S_3}\mathrm{sgn}(\sigma ){\prod }_i{a}_{\sigma (i),i}$. Relabel the product by $j=\sigma (i)$ and then the sum by $\tau ={\sigma }^{-1}$. The inverse permutation has the same sign, soThe proof applies to any finite square matrix.

Expanding the trace of the product and exchanging the two finite summation indices givesScalar entries commute even though the matrices need not.

Apply the preceding trace identity first to the pair $AB,C$ and then to $CA,B$:This is invariance under cyclic permutation. It does not assert invariance under arbitrary rearrangement of the matrices.

The coefficient matrix has determinantA homogeneous linear system has a nonzero solution exactly when this is a singular matrix. ThereforeAt this value all solutions are $(x,y,z)=t(1,-2,1)$, with $t\ne 0$ for a nonzero solution.