Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 17

Reading off coefficients gives the matrix equation

Multiplication of a matrix by a column vector forms the linear combination of its columns with coefficients from that vector. Hereso $A\mathbf{r}=x{\mathbf{c}}_1+y{\mathbf{c}}_2+z{\mathbf{c}}_3$. Equating this expression with $\mathbf{b}$ reproduces the original three linear equations.

For $\mathbf{b}=0$, a nonzero coefficient vector $(x,y,z)$ with $x{\mathbf{c}}_1+y{\mathbf{c}}_2+z{\mathbf{c}}_3=0$ is precisely a linear dependence of the three columns. Three vectors in three dimensions are linearly dependent exactly when the determinant of their column matrix, or their scalar triple product, is zero. ThusFor this square homogeneous linear system, the condition is also sufficient.

Subtract twice the second row from the third row when computing the determinant. The last row becomes $(0,0,\mu -6)$, soIts zeros give exactly the parameter values with a nontrivial null space:

At $\mu =-1$, the first equation gives $x=y$, and the second then gives $z=0$. All nonzero solutions areAt $\mu =6$, the first equation gives $x=-6y$. The second then gives $3z=7y$, and the third is twice the second. Choosing an integer direction vector, all nonzero solutions areEach full null space, including the zero solution, is a line through the origin. Geometrically, the three homogeneous constraint planes have that common intersection; in each case only two of their equations are independent.

At $\mu =-1$, ${\mathbf{c}}_2=-{\mathbf{c}}_1$, while ${\mathbf{c}}_1$ and ${\mathbf{c}}_3$ are independent. Their span is a plane. A solution requires $\mathbf{b}$ to lie in that plane, so every scalar triple product formed from two columns and $\mathbf{b}$ must vanish. The product with ${\mathbf{c}}_1,{\mathbf{c}}_2$ vanishes identically; the other two give the plane condition.

Explicitly, ${\mathbf{c}}_1\times {\mathbf{c}}_3=(-7,1,3)$, soThus $\lambda =7$ is necessary and, since the two columns span the plane, sufficient. The equations then give $x-y=1$ and $z=2$, soThe solution is an affine line parallel to the homogeneous null space from part (c), obtained by intersecting the compatible constraint planes. Every point on it is nonzero because $z=2$.