matrices – NST First Year Maths

2008 Paper 2 Question 16

14th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

(i)

As an intermediate result

$A^2 = \left( \begin{matrix}0 & 0 & 1\\ 1& 0 &0\\ 0& 1& 0\end{matrix} \right)$

(ii)

$A^{3n+2} = (A^3)^n A^2 = (I)^n A^2 = A^2$

and $A^2$ is given explicitly in (i).

(iii)

Use

$M^{-1} = \frac{1}{\text{det}M }\left[ \text{cof }M \right]^T$

Now

$\text{det}A^2 = 1$

and the cofactor matrix is

$\text{cof }A^2 = \left( \begin{matrix}0 & 0 & 1\\ 1& 0 &0\\ 0& 1& 0\end{matrix} \right)$

Thus

$(A^2)^{-1}=\left( \begin{matrix}0 & 1 & 0\\ 0& 0 &1\\ 1& 0& 0\end{matrix} \right)$

This can easily be verified by multiplying by $A^2$ .

(iv)

$\begin{aligned} A^{3n+1} &= (A^3)^n A = (I)^n A = A \\ \therefore (A^{3n+1})^{-1} &= A^{-1}\end{aligned}$

We know $A^3=I$ , so

$\begin{aligned} A^2A &= I\\ A^2 = A^{-1} \end{aligned}$

so finally

$(A^{3n+1} )^{-1} = A^2$

2016 Paper 2 Question 5

11th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

Yes. A real symmetric matrix can be (orthogonally diagonalised).

(b)

No. This is a rotation matrix. The eigenvalues are non-real except for special values of $\theta$ .

2014 Paper 1 Question 11

5th Oct 20201st Nov 2020nstmathsupervisorLeave a comment

(a)

$|A| = \left| \begin{matrix} 1 & -4 & 7 \\ -4 & 4 & -4 \\ 7 & -4 & 1 \end{matrix} \right|$

Clearly, you can just expand this as is, but it can be worth simplifying with some elementary row/column operations.

$R_2 \to R_2 + R_3$

$|A| = \left| \begin{matrix} 1 & -4 & 7 \\ 3 & 0 & -3 \\ 7 & -4 & 1 \end{matrix} \right|$

$C_3 \to C_3 + C_1$

$|A| = \left| \begin{matrix} 1 & -4 & 8 \\ 3 & 0 & 0 \\ 7 & -4 & 8 \end{matrix} \right|$

$C_3 \to C_3 + 2C_2$

$|A| = \left| \begin{matrix} 1 & -4 & 0 \\ 3 & 0 & 0 \\ 7 & -4 & 0 \end{matrix} \right|$

With a column of zeros in the determinant

$|A| = 0$ .

For the trace

$\text{Tr}(A) = 1 + 4 + 1 = 6$

The determinant is equal to the product of the eigenvalues, so (at least) one of the eigenvalues equals zero.