GATE 2022 EE Q.42: solution and answer

GATE 2022 · EEQ.42 · 2 marks · MCQ

Consider a matrix $A = 100041 0 - 2 1$ .

The matrix $A$ satisfies the equation $6 A^{- 1} = A^{2} + c A + d I$ , where $c$ and $d$ are scalars and $I$ is the identity matrix.

Then $(c + d)$ is equal to

(A)
$5$
(B)
$17$
(C)
$- 6$
(D)
$11$

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Answer and solution

ANSWER

$c = - 6$ and $d = 11$ , so $c + d = 5$ . The answer is option (A).

Official answer: ASource: Official answer key, GATE 2022 EE (IIT Kharagpur)

$A$ is block diagonal. It is the entry $1$ plus the $2 \times 2$ block $B = [41 - 2 1]$ , so $det (A - λ I) = (1 - λ) det (B - λ I)$ .
For $B$ : $tr B = 4 + 1 = 5$ and $det B = (4) (1) - (- 2) (1) = 6$ . So $det (B - λ I) = λ^{2} - 5 λ + 6 = (λ - 2) (λ - 3)$ .
The characteristic equation $det (A - λ I) = 0$ is therefore $(λ - 1) (λ - 2) (λ - 3) = 0$ , which expands to $λ^{3} - 6 λ^{2} + 11 λ - 6 = 0$ . (The overall sign of $det (A - λ I)$ does not change the equation.)
Cayley-Hamilton theorem: every square matrix satisfies its own characteristic equation, so $A^{3} - 6 A^{2} + 11 A - 6 I = 0$ .
$det A = 1 \cdot 2 \cdot 3 = 6 \neq = 0$ , so $A^{- 1}$ exists. Multiply the equation by $A^{- 1}$ to get $A^{2} - 6 A + 11 I - 6 A^{- 1} = 0$ , that is $6 A^{- 1} = A^{2} - 6 A + 11 I .$
Compare with $6 A^{- 1} = A^{2} + c A + d I$ to get $c = - 6$ and $d = 11$ . These values are unique. If a second pair $(c^{'}, d^{'})$ also worked, then $(c - c^{'}) A + (d - d^{'}) I = 0$ . Since $A$ is not a scalar multiple of $I$ , this forces $c = c^{'}$ and $d = d^{'}$ .
$c + d = - 6 + 11 = 5$ , which is option (A).

KEY IDEA

By the Cayley-Hamilton theorem, $A$ satisfies its characteristic equation $λ^{3} - 6 λ^{2} + 11 λ - 6 = 0$ . Multiplying by $A^{- 1}$ gives $6 A^{- 1}$ as a quadratic in $A$ . Because $A$ is block diagonal, the characteristic polynomial is just $(1 - λ)$ times the $2 \times 2$ block's polynomial, $λ^{2} - (tr B) λ + det B$ .

WATCH OUT FOR

Getting the sign of $det B$ wrong: $4 \cdot 1 - (- 2) \cdot 1 = 6$ , not $2$ . With $det B = 2$ you get the wrong cubic and the wrong constant term, and a coefficient such as $- 6$ (option C) or $11$ (option D) looks like the answer. Those are the separate values of $c$ and $d$ , not their sum. Another slip is dropping the $- 6 I$ term when multiplying by $A^{- 1}$ , which loses the factor $6$ in front of $A^{- 1}$ .

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GATE 2022 EE Q.42: solution and answer

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