Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 7

The two-by-two determinant is the difference of the diagonal products:

Expand the determinant along the first row:The minus sign on the middle cofactor is essential.

The third row is twice the first, so the rows are linearly dependent. Hence

The homogeneous linear system has coefficient matrixNonzero solutions exist exactly when its determinant vanishes. Expanding along the first row givesThus $\lambda =1$ or $-6$. The first equation always gives $x=3z$. For $\lambda =1$, the second then gives $y=-2z$; the third imposes the same relation. For $\lambda =-6$, the second gives $y=3z/2$, again consistent with the third. Therefore the nonzero solutions areFor every other $\lambda$, the matrix is invertible and only the zero solution exists.

Here the vertical bars denote determinants. Since $B=B^{-1}$, taking determinants gives $(\det B{)}^2=1$. The condition $\det B>0$ selects $\det B=1$. By the multiplicative property of the determinant,The hypotheses determine this determinant, while allowing many possible matrices $A$ and $B$.