Standard +0.8 This is a Further Maths question requiring matrix multiplication, setting equal to a scalar multiple of the identity matrix, and solving a system of equations. While the algebraic manipulation is moderately involved, the approach is standard: multiply AB, equate off-diagonal elements to zero, then find k from diagonal elements. The 3-mark allocation suggests straightforward execution once the method is identified, placing it slightly above average difficulty.
The matrices \(\mathbf{A}\) and \(\mathbf{B}\) are defined as follows:
$$\mathbf{A} = \begin{bmatrix} x + 1 & 2 \\ x + 2 & -3 \end{bmatrix}$$
$$\mathbf{B} = \begin{bmatrix} x - 4 & x - 2 \\ 0 & -2 \end{bmatrix}$$
Show that there is a value of \(x\) for which \(\mathbf{AB} = k\mathbf{I}\), where \(\mathbf{I}\) is the \(2 \times 2\) identity matrix and \(k\) is an integer to be found.
[3 marks]
Question 4:
4 | Multiplies matrices A and
B to form the product AB
with at least one element
of the product correct.
Condone BA | 1.1a | M1 | 2
𝑥𝑥 −3𝑥𝑥−4 (𝑥𝑥−3)(𝑥𝑥+2)
𝐀𝐀𝐀𝐀 = � 2 �
(𝑥𝑥−4)(𝑥𝑥+2) 𝑥𝑥 +2
𝑥𝑥 = −2
1 0
𝐀𝐀𝐀𝐀 = 6� � ∴ 𝑘𝑘 = 6
0 1
Forms the correct
product AB (may be
unsimplified) | 1.1b | A1
Deduces from correct
product that AB ,
when | 2.2a | R1
= 6𝐈𝐈
Total | 3
Q | Marking Instructions | AO | Marks | Typical Solution
The matrices $\mathbf{A}$ and $\mathbf{B}$ are defined as follows:
$$\mathbf{A} = \begin{bmatrix} x + 1 & 2 \\ x + 2 & -3 \end{bmatrix}$$
$$\mathbf{B} = \begin{bmatrix} x - 4 & x - 2 \\ 0 & -2 \end{bmatrix}$$
Show that there is a value of $x$ for which $\mathbf{AB} = k\mathbf{I}$, where $\mathbf{I}$ is the $2 \times 2$ identity matrix and $k$ is an integer to be found.
[3 marks]
\hfill \mbox{\textit{AQA Further Paper 2 2020 Q4 [3]}}