20N.1.AHL.TZ0.F_4

pestleMathematicsAIHLPaper 120N· ahl-1-14-introduction-to-matricessource ↗

The matrix $A$ is given by $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ .

By considering the determinant of a relevant matrix, show that the eigenvalues, $λ$ , of $A$ satisfy the equation

$λ^{2} - α λ + β = 0$ ,

where $α$ and $β$ are functions of $a, b, c, d$ to be determined.

[4]

Verify that

$A^{2} - α A + β I =$ 0.

[5]

b.i.

Assuming that $A$ is non-singular, use the result in part (b)(i) to show that

$A^{- 1} = \frac{1}{β} (α I - A)$ .

[2]

b.ii.

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$|\begin{matrix} a - λ & b \\ c & d - λ \end{matrix}| = 0$ M1

$(a - λ) (d - λ) - b c = 0$ M1A1

$λ^{2} - (a + d) λ + a d - b c = 0$ A1

$α = (a + d); β = a d - b c$

[4 marks]

$A^{2} = [\begin{matrix} a & b \\ c & d \end{matrix}] [\begin{matrix} a & b \\ c & d \end{matrix}] = [\begin{matrix} a^{2} + b c & a b + b d \\ a c + c d & b c + d^{2} \end{matrix}]$ (M1)A1

$A^{2} - (a + d) A + (a d - b c) I =$

$[\begin{matrix} a^{2} + b c & a b + b d \\ a c + c d & b c + d^{2} \end{matrix}] - (a + d) [\begin{matrix} a & b \\ c & d \end{matrix}] + (a d - b c) [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ M1

$= [\begin{matrix} a^{2} + b c - a (a + d) + a d - b c & a b + b d - b (a + d) \\ a c + c d - c (a + d) & b c + d^{2} - d (a + d) + a d - b c \end{matrix}]$ A2

$=$ 0 AG

Note: Award A1A0 for a single error.

[5 marks]

b.i.

multiply throughout by $A^{- 1}$ giving M1

$A - α I + β A^{- 1} =$ 0 A1

$A^{- 1} = \frac{1}{β} (α I - A)$ AG

[2 marks]

b.ii.

Examiners’ report

[N/A]

b.i.

[N/A]

b.ii.