Edexcel IAL Maths FP3.6 Further Matrix Algebra

[Maximum number: 4]

\mathbf{T}=\left(\begin{array}{lll} 2 & 3 & 7 \\ 3 & 2 & 6 \\ a & 4 & b \end{array}\right) \quad \mathbf{U}=\left(\begin{array}{rrr} 6 & -1 & -4 \\ 15 & c & -9 \\ -8 & a & 5 \end{array}\right)

where a, b and c are constants.
Given that T U=I

determine the value of a, the value of b and the value of c

\mathbf{TU}=\mathbf I

6a+60-8b=0,\quad -4a-36+5b=1

6a-8b=-60,\quad -4a+5b=37

a=2,\quad b=9

-2+3c+7a=0

c=-4

M1: obtains at least two equations with at least one correct.\ndM1: obtains values for two of a,b,c.\nA1 A1: a=2,b=9,c=-4.

[Maximum number: 11]

The matrix M is given by

\left(\begin{array}{rrr} 2 & 0 & -1 \\ k & 3 & 2 \\ -2 & 1 & k \end{array}\right)

(a)

Show that $\operatorname{det} \mathbf{M}=5 k-10$

Given that $k \neq 2$

[ 2 ]

$\det M= \begin{vmatrix} 2&0&-1\\ k&3&2\\ -2&1&k \end{vmatrix} =2\begin{vmatrix}3&2\\1&k\end{vmatrix} -0\begin{vmatrix}k&2\\-2&k\end{vmatrix} (-1)\begin{vmatrix}k&3\\-2&1\end{vmatrix}$ =2(3k-2)-(k+6)=6k-4-k-6=5k-10

M1 A1*

(b)

find $\mathbf{M}^{-1}$ in terms of k.

The points O(0,0,0), A(4,-8,3), B(-2,5,-4) and C(4,-6,8) are the vertices of a tetrahedron T.

The transformation represented by matrix M transforms T to a tetrahedron with volume 50

[ 4 ]

$\operatorname{adj}M= \left(\begin{smallmatrix} 3k-2&-1&3\\ -k^2-4&2k-2&-4-k\\ k+6&-2&6 \end{smallmatrix}\right)$ $M^{-1}=\frac{1}{5k-10} \left(\begin{smallmatrix} 3k-2&-1&3\\ -k^2-4&2k-2&-4-k\\ k+6&-2&6 \end{smallmatrix}\right)$

(c)

Determine the possible values of k.

[ 5 ]

$\text{Volume of }T=\frac16|\mathbf a\cdot(\mathbf b\times\mathbf c)| =\frac{20}{3}$ $\text{Volume image}=|\det M|\cdot\frac{20}{3} \Rightarrow \frac{20}{3}|5k-10|=50$ $k=\frac72\quad\text{or}\quad k=\frac12$

[Maximum number: 8]

\mathbf{A}=\left(\begin{array}{rrr} 1 & -1 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & a \end{array}\right) \quad a \neq 1

(a)

Find $\mathbf{A}^{-1}$ in terms of a.

\mathbf{B}=\left(\begin{array}{rrr} 1 & -1 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{array}\right)

The straight line $l_{1}$ is mapped onto the straight line $l_{2}$ by the transformation represented by the matrix B.

The equation of $l_{2}$ is

(\mathbf{r}-(12 \mathbf{i}+4 \mathbf{j}+6 \mathbf{k})) \times(-6 \mathbf{i}+2 \mathbf{j}+3 \mathbf{k})=\mathbf{0}

[ 4 ]

|A|=a-2+a-1+2-1=2a-2 $\operatorname{cof}(A)=((a-2,1-a,1),(a+2,a-1,-3),(-2,0,2))$ $\operatorname{adj}(A)=((a-2,a+2,-2),(1-a,a-1,0),(1,-3,2))$ $A^{-1}=\frac{1}{2a-2}((a-2,a+2,-2),(1-a,a-1,0),(1,-3,2))$

(b)

Find a vector equation for the line $l_{1}$

[ 4 ]

$a=4,\quad A^{-1}=\frac16((2,6,-2),(-3,3,0),(1,-3,2))$

B1ft

$\frac16((2,6,-2),(-3,3,0),(1,-3,2))(12-6\lambda,\ 4+2\lambda,\ 6+3\lambda)^T$ $=(6-\lambda,\ -4+4\lambda,\ 2-\lambda)^T$ $\mathbf r=(6,-4,2)^T+\lambda(-1,4,-1)^T$

[Maximum number: 4]

The matrix M is given by

\left(\begin{array}{rrr} 2 & 0 & -1 \\ k & 3 & 2 \\ -2 & 1 & k \end{array}\right)

find $\mathbf{M}^{-1}$ in terms of k.

The points O(0,0,0), A(4,-8,3), B(-2,5,-4) and C(4,-6,8) are the vertices of a tetrahedron T.

The transformation represented by matrix M transforms T to a tetrahedron with volume 50

$\operatorname{adj}M= \left(\begin{smallmatrix} 3k-2&-1&3\\ -k^2-4&2k-2&-4-k\\ k+6&-2&6 \end{smallmatrix}\right)$ $M^{-1}=\frac{1}{5k-10} \left(\begin{smallmatrix} 3k-2&-1&3\\ -k^2-4&2k-2&-4-k\\ k+6&-2&6 \end{smallmatrix}\right)$

[Maximum number: 4]

\mathbf{A}=\left(\begin{array}{rrr} 1 & -1 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & a \end{array}\right) \quad a \neq 1

Find a vector equation for the line $l_{1}$

$a=4,\quad A^{-1}=\frac16((2,6,-2),(-3,3,0),(1,-3,2))$

B1ft

$\frac16((2,6,-2),(-3,3,0),(1,-3,2))(12-6\lambda,\ 4+2\lambda,\ 6+3\lambda)^T$ $=(6-\lambda,\ -4+4\lambda,\ 2-\lambda)^T$ $\mathbf r=(6,-4,2)^T+\lambda(-1,4,-1)^T$

[Maximum number: 4]

\mathbf{M}=\left(\begin{array}{rrr} 0 & -1 & 3 \\ -1 & 4 & -1 \\ 3 & -1 & 0 \end{array}\right)

Given that $\left(\begin{array}{r}1 \\ -2 \\ 1\end{array}\right)$ is an eigenvector of M

(a)

determine its corresponding eigenvalue.

[ 2 ]

\mathbf M\begin{smallmatrix}1\\-2\\1\end{smallmatrix}=\begin{smallmatrix}5\\-10\\5\end{smallmatrix}=5\begin{smallmatrix}1\\-2\\1\end{smallmatrix}

\lambda=5

M1: correct method leading to a value for $\lambda$ .\nA1: correct value.

(b)

Given that -3 is an eigenvalue of $\mathbf{M}$

determine a corresponding eigenvector.

[ 2 ]

(\mathbf M+3\mathbf I)\begin{smallmatrix}x\\y\\z\end{smallmatrix}=\begin{smallmatrix}0\\0\\0\end{smallmatrix}

\begin{smallmatrix}3&-1&3\\-1&7&-1\\3&-1&3\end{smallmatrix}\begin{smallmatrix}x\\y\\z\end{smallmatrix}=\begin{smallmatrix}0\\0\\0\end{smallmatrix}

A corresponding eigenvector is

k\begin{smallmatrix}1\\0\\-1\end{smallmatrix}

where $k\ne0$ .\nM1: uses $\mathbf M\mathbf x=-3\mathbf x$ or $(\mathbf M+3\mathbf I)\mathbf x=0$ to obtain a non-zero vector.\nA1: $k(1,0,-1)^T$ .

[Maximum number: 4]

\mathbf{M}=\left(\begin{array}{rrr} 0 & -1 & 3 \\ -1 & 4 & -1 \\ 3 & -1 & 0 \end{array}\right)

Given that $\left(\begin{array}{r}1 \\ -2 \\ 1\end{array}\right)$ is an eigenvector of M

Hence, given that $\begin{pmatrix}1\\1\\1\end{pmatrix}$ is also an eigenvector of $\mathbf{M}$

determine a diagonal matrix D and an orthogonal matrix P such that $\mathbf{D}=\mathbf{P}^{\mathrm{T}} \mathbf{M P}$

\lambda=2

\mathbf D=\begin{smallmatrix}-3&0&0\\0&2&0\\0&0&5\end{smallmatrix}

\begin{smallmatrix}1\\-2\\1\end{smallmatrix}\to\begin{smallmatrix}\frac{\sqrt6}{6}\\-\frac{\sqrt6}{3}\\\frac{\sqrt6}{6}\end{smallmatrix},\quad \begin{smallmatrix}1\\0\\-1\end{smallmatrix}\to\begin{smallmatrix}\frac{\sqrt2}{2}\\0\\-\frac{\sqrt2}{2}\end{smallmatrix},\quad \begin{smallmatrix}1\\1\\1\end{smallmatrix}\to\begin{smallmatrix}\frac{\sqrt3}{3}\\\frac{\sqrt3}{3}\\\frac{\sqrt3}{3}\end{smallmatrix}

\mathbf P=\begin{smallmatrix}\frac{\sqrt2}{2}&\frac{\sqrt3}{3}&\frac{\sqrt6}{6}\\0&\frac{\sqrt3}{3}&-\frac{\sqrt6}{3}\\-\frac{\sqrt2}{2}&\frac{\sqrt3}{3}&\frac{\sqrt6}{6}\end{smallmatrix}

B1: correct remaining eigenvalue 2.\nB1ft: diagonal matrix consistent with eigenvalues.\nM1: normalises at least one eigenvector.\nA1: fully correct consistent $\mathbf D$ and $\mathbf P$ .

Edexcel A-Level Mathematics A2 Fp3 6 Further Matrix Algebra Questions

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