GATE 2018 EE Q.44: solution and answer

GATE 2018 · EEQ.44 · 2 marks · NAT

Let $A = 1 - 1 0 020 - 1 0 - 2$ and $B = A^{3} - A^{2} - 4 A + 5 I$ , where $I$ is the $3 \times 3$ identity matrix. The determinant of $B$ is ________ (up to 1 decimal place).

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

$det B = 1.0$ .

Official answer: 0.9 to 1.1Source: Official answer key, GATE 2018 EE (IIT Guwahati)

Find the characteristic polynomial $det (A - λ I)$ . Row 3 of $A - λ I$ is $[00 - 2 - λ]$ , so it has only one non-zero entry. Expand along row 3: $det (A - λ I) = (- 2 - λ) 1 - λ - 1 0 2 - λ = (- 2 - λ) (1 - λ) (2 - λ) = - (λ - 1) (λ - 2) (λ + 2)$ This gives the eigenvalues $λ = 1, 2, - 2$ .
Multiply out: $(λ - 1) (λ - 2) (λ + 2) = (λ^{2} - 3 λ + 2) (λ + 2) = λ^{3} - λ^{2} - 4 λ + 4$ . The characteristic equation is therefore $λ^{3} - λ^{2} - 4 λ + 4 = 0$ .
By the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation: $A^{3} - A^{2} - 4 A + 4 I = 0$ .
Split $5 I$ as $4 I + I$ : $B = A^{3} - A^{2} - 4 A + 5 I = (A^{3} - A^{2} - 4 A + 4 I) + I = 0 + I = I$ .
So $det B = det I = 1$ .
Check with the eigenvalues of a matrix polynomial. Let $p (λ) = λ^{3} - λ^{2} - 4 λ + 5$ . The eigenvalues of $B = p (A)$ are $p (1) = 1 - 1 - 4 + 5 = 1$ , $p (2) = 8 - 4 - 8 + 5 = 1$ and $p (- 2) = - 8 - 4 + 8 + 5 = 1$ . The determinant is the product of the eigenvalues, so $det B = 1 \cdot 1 \cdot 1 = 1.0$ .

KEY IDEA

The polynomial in $B$ differs from $A$ 's characteristic polynomial $λ^{3} - λ^{2} - 4 λ + 4$ only in the constant term. By Cayley-Hamilton that characteristic polynomial evaluated at $A$ is the zero matrix, so $B = 0 + I = I$ without computing any powers of $A$ . The same thing shows up in the eigenvalues: $p (λ_{i}) = 1$ for every eigenvalue $λ_{i}$ of $A$ .

WATCH OUT FOR

Students often take $det B$ as $det (A^{3}) - det (A^{2}) - 4 det A + 5$ . The determinant is not linear, so that is wrong. Other slips: getting a sign wrong when expanding $det (A - λ I)$ (for example, writing the eigenvalue $- 2$ as $+ 2$ ), or computing $A^{2}$ and $A^{3}$ by hand and making an arithmetic error. Also, $det (4 A)$ for a $3 \times 3$ matrix is $4^{3} det A$ , not $4 det A$ .

From the GATE 2018 EE paper. Every question of this paper · Download the full paper · All 46 Linear algebra questions, with notes

GATE 2018 EE Q.44: solution and answer

Answer and solution

More Linear algebra questions