# Question 4:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{vmatrix} 2 & 0 & -1 \\ k & 3 & 2 \\ -2 & 1 & k \end{vmatrix} = 2\begin{vmatrix} 3 & 2 \\ 1 & k \end{vmatrix} - 0\begin{vmatrix} k & 2 \\ -2 & k \end{vmatrix} + (-1)\begin{vmatrix} k & 3 \\ -2 & 1 \end{vmatrix} = 2(3k-2)-(k+6)=\ldots$ | M1 | Correct method for expanding determinant to reach linear expression in $k$. Expand along top row or any row/column. Sarrus gives $6+k-(6+4)$ |
| $= 6k-4-k-6 = 5k-10$ | A1* | Correct expression from correct work |

**(2 marks)**

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## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{M}^T = \begin{pmatrix} 2 & k & -2 \\ 0 & 3 & 1 \\ -1 & 2 & k \end{pmatrix}$ or minors $\begin{pmatrix} 3k-2 & k^2+4 & k+6 \\ 1 & 2k-2 & 2 \\ 3 & 4+k & 6 \end{pmatrix}$ or cofactors $\begin{pmatrix} 3k-2 & -k^2-4 & k+6 \\ -1 & 2k-2 & -2 \\ 3 & -4-k & 6 \end{pmatrix}$ | M1 | Begins finding inverse by attempting transpose, matrix of minors or cofactors. Look for at least 6 correct entries |
| Adjugate matrix is $\begin{pmatrix} 3k-2 & -1 & 3 \\ -k^2-4 & 2k-2 & -4-k \\ k+6 & -2 & 6 \end{pmatrix}$ ($\geq 6$ entries correct) | M1 | Proceeds to find adjugate (may include reciprocal determinant). Look for 6 correct entries |
| Hence $\mathbf{M}^{-1} = \dfrac{1}{5k-10}\begin{pmatrix} 3k-2 & -1 & 3 \\ -k^2-4 & 2k-2 & -4-k \\ k+6 & -2 & 6 \end{pmatrix}$ | dM1 | Full method: divides adjugate by determinant. **Depends on both previous marks** |
| Fully correct inverse | A1 | Fully correct inverse |

**(4 marks)**

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## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Images of $A$, $B$ and $C$ are $(5, 4k-18, 3k-16)$, $(0, 7-2k, 9-4k)$ and $(0, 4k-2, 8k-14)$ | M1 | Attempts to find image vectors of $A$, $B$ and $C$. May be implied by at least two correct entries in one vector |
| | A1 | Correct image vectors. Allow unsimplified |
| $(\pm)50 = \frac{1}{6}\begin{vmatrix} 5 & 4k-18 & 3k-16 \\ 0 & 7-2k & 9-4k \\ 0 & 4k-2 & 8k-14 \end{vmatrix} \Rightarrow (\pm)300 = 5(\ldots)(=200k-400) \Rightarrow k=\ldots$ | M1 | Use image vectors in scalar triple product, set volume equal to 50, attempt to solve for $k$. Must include $\frac{1}{6}$ |
| $(300=200k-400 \Rightarrow) k=\dfrac{7}{2}$ or $(-300=200k-400 \Rightarrow) k=\dfrac{1}{2}$ | A1 | One correct value of $k$ |
| $k=\dfrac{1}{2}$ and $k=\dfrac{7}{2}$ | A1 | Both values correctly found |

**(5 marks)**

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### Alt Method (Volume Scale Factor)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{a}.(\mathbf{b}\times\mathbf{c}) = \begin{vmatrix} 4 & -8 & 3 \\ -2 & 5 & -4 \\ 4 & -6 & 8 \end{vmatrix} = 4(40-24)+8(-16+16)+3(12-20)=\ldots$ | M1 | Attempts appropriate scalar triple product. Rows may be in different order |
| Volume of $T$ is $\frac{1}{6}|\mathbf{a}.(\mathbf{b}\times\mathbf{c})| = \frac{1}{6}\begin{vmatrix} 4 & -8 & 3 \\ -2 & 5 & -3 \\ 4 & 6 & -8 \end{vmatrix} = \ldots\dfrac{20}{3}$ | A1 | Correct volume for tetrahedron $T$. Need not be simplified, $\frac{40}{6}$ is fine |
| Volume image of $T = |\det\mathbf{M}|\times\dfrac{20}{3} \Rightarrow \dfrac{20}{3}|5k-10|=50 \Rightarrow k=\ldots$ | M1 | Uses determinant as volume scale factor to set up equation in $k$. $\frac{1}{6}$ may be missing |
| $\left(\dfrac{20}{3}(5k-10)=50\Rightarrow\right) k=\dfrac{7}{2}$ or $\left(\dfrac{20}{3}(10-5k)=50\Rightarrow\right) k=\dfrac{1}{2}$ | A1 | One correct value |
| $k=\dfrac{1}{2}$ and $k=\dfrac{7}{2}$ | A1 | Both values correctly found |

**(5 marks) — Total: 11 marks**

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