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]