A = \( \begin{pmatrix} 2 & -1 \\ 1 & -3 \end{pmatrix} \) and B = \( \begin{pmatrix} 0 & -2 \\ 3 & -5 \end{pmatrix} \). Find (BA)⁻¹.

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

1. [B2] BA = \( \begin{pmatrix} -2 & 6 \\ 1 & 12 \end{pmatrix} \) (B1 for a 2 by 2 matrix with 2 or 3 correct elements)
2. [B2] (BA)⁻¹ = 1/(-30) \( \begin{pmatrix} -3 & -6 \\ -1 & -2 \end{pmatrix} \) oe isw (FT their non-singular BA, B1 FT for either 1/(-30) or \( \begin{pmatrix} -3 & -6 \\ -1 & -2 \end{pmatrix} \) ) OR Alternative method A⁻¹B⁻¹: B2 for A⁻¹ = 1/(-5) \( \begin{pmatrix} -3 & 1 \\ -1 & 2 \end{pmatrix} \) isw or B⁻¹ = 1/6 \( \begin{pmatrix} -5 & 2 \\ -3 & 0 \end{pmatrix} \) isw (B1 for a multiplier of 1/(-5) or for \( \begin{pmatrix} -3 & 1 \\ -1 & 2 \end{pmatrix} \) or for a multipler of 1/6 or for \( \begin{pmatrix} -5 & 2 \\ -3 & 0 \end{pmatrix} \) ) B2 FT for A⁻¹ B⁻¹ = their 1/(-30) × their \( \begin{pmatrix} 12 & -6 \\ -1 & -2 \end{pmatrix} \) or B1 FT for a 2 by 2 matrix with 2 or 3 correct elements
3. [B1] 2 × 3
4. [B2] X = \( \begin{pmatrix} 2 & -1 \end{pmatrix} \) oe isw (B1 for each correct element; must be in a 1 by 2 matrix or M1 for a full method as far as finding values for the two elements)