Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 13

Compute the characteristic polynomial using $\det (\lambda I-M)$:Its three distinct real roots are $1$ and $3\pm \sqrt{15}$. For $\lambda =1$, the first row of $(M-I)v=0$ gives $z=3y$ and the second then gives $x=y$, so take $v_1=(1,1,3{)}^T$.

For either root of ${\lambda }^2-6\lambda -6=0$, take $z=1$ and solve the row equations to get $v=(\lambda -8,5-\lambda ,1{)}^T$. Substitution makes the first two row residuals opposite multiples of the quadratic equation and the third zero. Thus a complete choice isEvery nonzero scalar multiple is another valid eigenvector in the corresponding one-dimensional eigenspace.

For either sign, the dot product with $v_1$ isThe remaining pair also satisfiesHence the three eigenvectors are mutually orthogonal. This also follows from the spectral theorem for real symmetric matrices, since $M=M^T$ and its eigenvalues are distinct, but the displayed products verify the actual chosen vectors directly.

Write the linear system as $Ar=d$ withExpansion along the first row gives $\det A=(16-15)-2(12-5)+3(9-4)=2$. Since $\det A=2\ne 0$, $A$ is an invertible matrix, and $r=A^{-1}d$ is the unique solution for this right side.

The adjugate matrix formula givesMultiplication by the right-side column yieldsMultiplying the inverse by $A$ checks the identity matrix, and substitution of this solution into the three original rows gives $1,2,3$ respectively.