Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 11

Collecting the coefficients gives the coefficient matrixA square homogeneous linear system has a nonzero solution precisely when its matrix is a singular matrix. Expanding the determinant down the first column,Hence the necessary and sufficient condition is $a=0$ or $b=c$ or $b=-c$. A zero determinant gives a nontrivial kernel of a linear map; otherwise multiplication by the matrix inverse would force the zero solution.

With $a=1$, $b=2$, $c=2$, substitute into the matrix equations. The equations reduce to $x+2z=0$ and $3y+z=0$, with the latter repeated. Taking $z=3t$ gives the complete null spaceThe two independent rows impose two constraints on three unknowns, leaving exactly one free parameter. This also includes the zero solution at $t=0$.

Use orthogonality to the image of a matrix. For $n=(1,2,1{)}^T$, the condition is $n^{T}A=0$, since $n^{T}Ar$ must vanish for every nonzero input. Direct multiplication givesEach coefficient must vanish; choosing the three coordinate basis vectors as inputs already proves necessity. Solve $2b+c=-3$, $b+2c=1$ to obtainThese values make every output have zero dot product with $n$, including any zero output. Here orthogonality is understood in this algebraic sense; no direction is assigned to the zero vector.

For $n=(1,-1,1{)}^T$, orthogonality to the image of a matrix would requireThe second component forces $c=b$, but then the third component is $2$. Equivalently, the last two required zero components always sum to $2$. Therefore there are no such values of $a,b,c$.