Question

Let
$M = (\begin{array}{cccccccccc} 0 & 0 & 0 & 0 & - 1 \\ 2 & 0 & 0 & 0 & - 4 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 3 \\ 0 & 0 & 0 & 2 & 2 \end{array}) .$
If 𝑝(𝑥) is the characteristic polynomial of $𝑀,$ then 𝑝(2) − 1 equals _____________

Answer: $\underset{_____________}{31}$

Solution :-

\begin{array}{cccl} Given that & M & = & {(\begin{array}{cccccccccc} 0 & 0 & 0 & 0 & - 1 \\ 2 & 0 & 0 & 0 & - 4 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 3 \\ 0 & 0 & 0 & 2 & 2 \end{array})}_{5 \times 5} \\ ⟹ & M & = & 2 {(\begin{array}{cccccccccc} 0 & 0 & 0 & 0 & - \frac{1}{2} \\ 1 & 0 & 0 & 0 & - 2 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & \frac{3}{2} \\ 0 & 0 & 0 & 1 & 1 \end{array})}_{5 \times 5} \\ ⟹ & M & = & 2 A, where A = {(\begin{array}{cccccccccc} 0 & 0 & 0 & 0 & - \frac{1}{2} \\ 1 & 0 & 0 & 0 & - 2 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & \frac{3}{2} \\ 0 & 0 & 0 & 1 & 1 \end{array})}_{5 \times 5} \end{array}

Now, characteristic polynomial of M is denoted by $p (x)$ and

\begin{array}{crcl} p (x) & = & det (x I_{5} - M) \\ ⟹ & p (x) & = & det (x I_{5} - 2 A) \\ ⟹ & p (x) & = & det (2 (\frac{x}{2} I_{5} - A)) \\ ⟹ & p (x) & = & 2^{5} det (\frac{x}{2} I_{5} - A) [since, det (c A_{n \times n}) = c^{n} det (A_{n \times n})] \\ ∴ & p (x) & = & 2^{5} C_{A} (\frac{x}{2}) \dots \dots (1) \end{array}

Here, $C_{A} (\frac{x}{2})$ is the characteristic polynomial of $A$ in the variable $\frac{x}{2} .$

We have, $A = {(\begin{array}{cccccccccc} 0 & 0 & 0 & 0 & - \frac{1}{2} \\ 1 & 0 & 0 & 0 & - 2 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & \frac{3}{2} \\ 0 & 0 & 0 & 1 & 1 \end{array})}_{5 \times 5}$ is a companion matrix. So, the characteristic polynomial of this matrix is

\begin{array}{rcl} C_{A} (x) & = & \frac{1}{2} + 2 x + 0 \cdot x^{2} - \frac{3}{2} \cdot x^{3} - 1 \cdot x^{4} + x^{5} \\ = & \frac{1}{2} + 2 x - \frac{3}{2} \cdot x^{3} - x^{4} + x^{5} \end{array}

Replace $x$ by $\frac{x}{2}$ , we get

C_{A} (\frac{x}{2}) = \frac{1}{2} + 2 (\frac{x}{2}) - \frac{3}{2} \cdot {(\frac{x}{2})}^{3} - {(\frac{x}{2})}^{4} + {(\frac{x}{2})}^{5} .

Now, from eq(1), we have

\begin{array}{crcl} p (x) & = & 2^{5} [\frac{1}{2} + 2 (\frac{x}{2}) - \frac{3}{2} \cdot {(\frac{x}{2})}^{3} - {(\frac{x}{2})}^{4} + {(\frac{x}{2})}^{5}] \\ ∴ & p (2) & = & 2^{5} [\frac{1}{2} + 2 (\frac{2}{2}) - \frac{3}{2} \cdot {(\frac{2}{2})}^{3} - {(\frac{2}{2})}^{4} + {(\frac{2}{2})}^{5}] \\ ⟹ & p (2) & = & 2^{5} (\frac{1}{2} + 2 - \frac{3}{2} - 1 + 1) \\ ⟹ & p (2) & = & 2^{5} (\frac{1 + 4 - 3}{2}) \\ ⟹ & p (2) & = & 2^{5} (\frac{2}{2}) = 2^{5} \cdot (1) = 32. \end{array}

∴ p (2) - 1 = 32 - 1 = 31.

So, the correct answer is $\to$ $\underset{_____________}{31}$

#Linear Algebra #IIT JAM PYQ 2024 #Mathematics #IIT JAM PYQs

Question

Let M=(0000−12000−4020000020300022). If 𝑝(𝑥) is the characteristic polynomial of 𝑀, then 𝑝(2) − 1 equals _____________

Solution :-

Let
$M = (\begin{array}{cccccccccc} 0 & 0 & 0 & 0 & - 1 \\ 2 & 0 & 0 & 0 & - 4 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 3 \\ 0 & 0 & 0 & 2 & 2 \end{array}) .$
If 𝑝(𝑥) is the characteristic polynomial of $𝑀,$ then 𝑝(2) − 1 equals _____________