Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 6

The determinant is $k-6$, so for $k\ne 6$ the inverse matrix isThe specified equations use $k=4$, with right-hand side $(2,3,1{)}^T$. Multiplying by $A^{-1}{|}_{k=4}$ givesIn particular the parameter in the inverse formula must be set to four, not left arbitrary for this numerical system.

For $k=1$, the required right-hand matrix has twice the first row of $A$, the same second row, and twice the third row. Left multiplication scales rows, soEquivalently, right multiply the given equation $BA=C$ by $A^{-1}$ to obtain the same answer.

At $k=6$, the second row is twice the first. The independent equations are $x+3y=0$ and $4x+z=0$. Therefore the null space isIt has dimension one, consistent with matrix rank two.