| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Matrix inverse calculation |
| Difficulty | Standard +0.8 This is a Further Maths question requiring determinant calculation with a parameter (showing it's never zero), then finding a 3×3 matrix inverse with algebraic entries. While the techniques are standard (cofactor method), the algebraic manipulation with parameter x throughout makes it more demanding than a numerical 3×3 inverse, placing it moderately above average difficulty. |
| Spec | 4.03i Determinant: area scale factor and orientation4.03o Inverse 3x3 matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \( | \mathbf{A} | = x(-4+2x)-(-2+4x)+3(-4+16)\) |
| \(= 2x^2 - 8x + 38\) | A1 | Correct simplified determinant |
| \(2x^2-8x+38 = 2(x-2)^2+30\) or \(\frac{d}{dx}(2x^2-8x+38)=4x-8=0\Rightarrow x=2\), or \(b^2-4ac = 64-4\times2\times38 = \ldots\) | M1 | Starts process of showing det \(\mathbf{A}\neq 0\). E.g. completes the square, finds minimum point or finds discriminant of \(x^2-4x+19=\ldots\) |
| \(2x^2-8x+38 \geqslant 30 > 0\), or \(b^2-4ac < 0\), therefore det \(\mathbf{A}\neq 0\) which means \(\mathbf{A}\) is non-singular | A1cso | Appropriate reasoning and conclusion stating \(\mathbf{A}\) is non-singular. All 3 previous marks needed. No need to evaluate discriminant, so ISW slips in calculation provided \(64-4\times2\times38=\ldots\) or \(16-4\times19=\ldots\) seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Cofactor matrix: \(\begin{pmatrix} -4+2x & 2-4x & 12 \\ -5 & -x+12 & 2x-4 \\ x-12 & -x^2+6 & 4x-2 \end{pmatrix}\) | M1A1 | M1: Applies correct method to reach at least a matrix of cofactors — 2 correct rows or 2 correct columns needed. A1: Correct cofactor matrix |
| \(\mathbf{A}^{-1} = \frac{1}{2x^2-8x+38}\begin{pmatrix} -4+2x & -5 & x-12 \\ 2-4x & -x+12 & -x^2+6 \\ 12 & 2x-4 & 4x-2 \end{pmatrix}\) | dM1A1 | dM1: Transposes and divides by their determinant. If determinant has been divided by 2 (acceptable for (a)) and then used here it is not their determinant and so scores dM0. 2 correct rows or 2 correct columns needed from previous matrix. Depends on previous method mark. A1: Correct matrix |
# Question 6(a):
$$\mathbf{A} = \begin{pmatrix} x & 1 & 3 \\ 2 & 4 & x \\ -4 & -2 & -1 \end{pmatrix}$$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $|\mathbf{A}| = x(-4+2x)-(-2+4x)+3(-4+16)$ | M1 | Correct determinant attempt (expand by any row or column) or use Rule of Sarrus. Sign errors allowed **only within the brackets** |
| $= 2x^2 - 8x + 38$ | A1 | Correct simplified determinant |
| $2x^2-8x+38 = 2(x-2)^2+30$ or $\frac{d}{dx}(2x^2-8x+38)=4x-8=0\Rightarrow x=2$, or $b^2-4ac = 64-4\times2\times38 = \ldots$ | M1 | Starts process of showing det $\mathbf{A}\neq 0$. E.g. completes the square, finds minimum point or finds discriminant of $x^2-4x+19=\ldots$ |
| $2x^2-8x+38 \geqslant 30 > 0$, or $b^2-4ac < 0$, therefore det $\mathbf{A}\neq 0$ which means $\mathbf{A}$ is non-singular | A1cso | Appropriate reasoning and conclusion stating $\mathbf{A}$ is non-singular. **All 3 previous marks needed**. No need to evaluate discriminant, so ISW slips in calculation provided $64-4\times2\times38=\ldots$ or $16-4\times19=\ldots$ seen |
# Question 6(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Cofactor matrix: $\begin{pmatrix} -4+2x & 2-4x & 12 \\ -5 & -x+12 & 2x-4 \\ x-12 & -x^2+6 & 4x-2 \end{pmatrix}$ | M1A1 | M1: Applies correct method to reach at least a matrix of cofactors — 2 correct rows or 2 correct columns needed. A1: Correct cofactor matrix |
| $\mathbf{A}^{-1} = \frac{1}{2x^2-8x+38}\begin{pmatrix} -4+2x & -5 & x-12 \\ 2-4x & -x+12 & -x^2+6 \\ 12 & 2x-4 & 4x-2 \end{pmatrix}$ | dM1A1 | dM1: Transposes and divides by their determinant. If determinant has been divided by 2 (acceptable for (a)) and then used here it is **not** their determinant and so scores dM0. 2 correct rows or 2 correct columns needed from previous matrix. Depends on previous method mark. A1: Correct matrix |6.
$$\mathbf { A } = \left( \begin{array} { r r r }
x & 1 & 3 \\
2 & 4 & x \\
- 4 & - 2 & - 1
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathbf { A }$ is non-singular for all real values of $x$.
\item Determine, in terms of $x , \mathbf { A } ^ { - 1 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2022 Q6 [8]}}