Find an expression for det(A) in terms of $\lambda$, simplifying your answer.

Hence show that A is singular when $\lambda  = 1$ and find the other value of $\lambda$ for which A is singular.

Explain how it can be seen immediately that B is singular without calculating its determinant.

Determine the null space of B.

Explain briefly how your results verify the rank-nullity theorem.

Prove, using mathematical induction, that

B$^n = {8^{n - 2}}$B$^2$ for $n \in {\mathbb{Z}^ + },{\text{ }}n \geqslant 3$.

det(A) $= \lambda (12 - 7\lambda ) + 3(3\lambda  - 6) + 2(14 - 12)$     M1A1

$= 12\lambda  - 7{\lambda ^2} + 9\lambda  - 18 + 4$

$=  - 7{\lambda ^2} + 21\lambda  - 14$     A1

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A is singular when $\lambda  = 1$ because the determinant is zero     R1

Note:     Do not award the R1 if the determinant has not been obtained.

the other value is 2     A1

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the third row is the sum of the first two rows     A1

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the null space satisfies

$\left[ {\begin{array}{*{20}{c}} 1&3&2 \\ 2&4&1 \\ 3&7&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \\ 0 \\ 0 \end{array}} \right]$     M1

$x + 3y + 2z = 0$

$2x + 4y + z = 0$     (A1)

$3x + 7y + 3z = 0$

the solution is (by GDC or otherwise)

$\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5 \\ { - 3} \\ 2 \end{array}} \right]\alpha$ where $\alpha  \in \mathbb{R}$     M1A1

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the rank-nullity theorem for square matrices states that

rank of matrix + dimension of null space = number of columns     A1

here, rank = 2, dimension of null space = 1 and number of columns = 3     A1

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first show that the result is true for $n = 3$

B$^2 = \left[ {\begin{array}{*{20}{c}} {13}&{29}&{11} \\ {13}&{29}&{11} \\ {26}&{58}&{22} \end{array}} \right]$     A1

B$^3 = \left[ {\begin{array}{*{20}{c}} {104}&{232}&{88} \\ {104}&{232}&{88} \\ {208}&{464}&{176} \end{array}} \right]$     A1

therefore B$^3 = 8$B$^2$ so true for $n = 3$     R1

assume the result is true for $n = k$, that is B$^k = {8^{k - 2}}$B$^2$     M1

consider B$^{k + 1} = {8^{k - 2}}$B$^2$     M1

$= {8^{k - 2}}8$B$^2$

$= {8^{k - 1}}$B$^2$     A1

therefore, true for $n = k \Rightarrow$ true for $n = k + 1$ and since the result is true for $n = 3$, it is true for $n \geqslant 3$     R1

[7 marks]