matrix inverse

(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

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

(a)

$\left( \begin{matrix} 1 & 2 & 3 \\ 1 & -2 & 1 \\ 2 & 1 & -1 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} 14 \\ 0 \\ 1 \end{matrix} \right)$

(i)

In Gaussian elimination, reduce the matrix to upper triangular form, then back-substitute.

Eliminate:

$\begin{aligned} R_2 &\to R_2 - R_1 \\ R_3 &\to R_3 - 2R_1\end{aligned}$

$\left( \begin{matrix} 1 & 2 & 3 \\ 0& -4 & -2 \\ 0 & -3 & -7 \end{matrix} \right| \left. \begin{matrix} 14 \\ -14 \\ -27 \end{matrix} \right)$

$R_3 \to R_3 - (-3)/(-4)R_2$

$\left( \begin{matrix} 1 & 2 & 3 \\ 0& -4 & -2 \\ 0 & 0 & -11/2 \end{matrix} \right| \left. \begin{matrix} 14 \\ -14 \\ -33/2 \end{matrix} \right)$

Back-substitute:

$R_3 \Rightarrow z = 3$

$R_2 \Rightarrow \begin{aligned} -4y -2(3) &= -14 \\ y &= 2 \end{aligned}$

$R_1 \Rightarrow \begin{aligned} x +2(2) +3(3) &= 14 \\ x &= 1 \end{aligned}$

The solution is

$\left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right)$

NST First Year Maths

Some unofficial solutions to Cambridge NST IA maths papers

2008 Paper 2 Question 16

2013 Paper 1 Question 11