Moderate -0.5 This is a straightforward application of the singular matrix condition (determinant = 0). Students need to calculate det(M) = x(4x-11) - (x-2)(3x-6), set it equal to zero, and solve the resulting quadratic. While it requires careful algebraic manipulation, it's a standard textbook exercise with no conceptual difficulty beyond knowing that singular means det = 0.
1.
$$\mathbf { M } = \left( \begin{array} { c c }
x & x - 2 \\
3 x - 6 & 4 x - 11
\end{array} \right)$$
Given that the matrix \(\mathbf { M }\) is singular, find the possible values of \(x\).
Their \(3TQ = 0\) and attempts to solve relevant quadratic using factorisation or completing the square or correct quadratic formula leading to \(x =\)
\(x = -4, \ x = 3\)
A1
Both values correct
Additional Notes:
- \(x(4x-11) = (3x-6)(x-2)\) award first M1
- \(\pm(x^2 + x - 12)\) seen award first M1A1
- Both correct with no working: 4/4; only one correct: 0/4
Total: 4 marks
## Question 1:
Matrix $\mathbf{M} = \begin{pmatrix} x & x-2 \\ 3x-6 & 4x-11 \end{pmatrix}$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\det\mathbf{M} = x(4x-11) - (3x-6)(x-2)$ | M1 | Correct attempt at determinant |
| $x^2 + x - 12 \ (=0)$ | A1 | Correct 3 term quadratic |
| $(x+4)(x-3) \ (=0) \rightarrow x = \ldots$ | M1 | Their $3TQ = 0$ and attempts to solve relevant quadratic using factorisation or completing the square or correct quadratic formula leading to $x =$ |
| $x = -4, \ x = 3$ | A1 | Both values correct |
**Additional Notes:**
- $x(4x-11) = (3x-6)(x-2)$ award first M1
- $\pm(x^2 + x - 12)$ seen award first M1A1
- Both correct with no working: 4/4; only one correct: 0/4
**Total: 4 marks**
1.
$$\mathbf { M } = \left( \begin{array} { c c }
x & x - 2 \\
3 x - 6 & 4 x - 11
\end{array} \right)$$
Given that the matrix $\mathbf { M }$ is singular, find the possible values of $x$.\\
\hfill \mbox{\textit{Edexcel FP1 2013 Q1 [4]}}