2014 Paper 1 Question 11

(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.

NST First Year Maths