Let P - - 1 1 -1- 2 -3 4- 3 -2 3 - and Q - - 1 -2 -1- 6 12 6- 5 10 5 - be two matrices- Then the rank of P - Q is

P + Q = [ 0 amp; 1 nbsp; amp;12; 8 nbsp; nbsp; amp;9 nbsp; amp;10; 8 nbsp; nbsp; amp;8 nbsp; amp;8 nbsp; ] ∵ | P + Q | = 0 ⇒ nbsp; ...

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Nilpotent Matrix

Let P = [ 1 1...

Question

Let P = $⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & - 1 2 & - 3 & 4 3 & - 2 & 3 \end{matrix} ⎤ ⎥ ⎦$ and Q = $⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & - 1 6 & 12 & 6 5 & 10 & 5 \end{matrix} ⎤ ⎥ ⎦$ be two matrices. Then the rank of P + Q is
2

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A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & - 1 2 & - 3 & 4 3 & - 2 & 3 \end{matrix} ⎤ ⎥ ⎦,

B = ⎡ ⎢ ⎣ \begin{matrix} - 1 & - 2 & - 1 6 & 12 & 6 5 & 10 & 5 \end{matrix} ⎤ ⎥ ⎦

C = ⎡ ⎢ ⎣ \begin{matrix} - 1 & - 1 & 1 2 & 2 & - 2 - 3 & - 3 & 3 \end{matrix} ⎤ ⎥ ⎦,

P

Q

2 \times 2

(P + Q) (P - Q) = P^{2} - Q^{2}

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & - 1 2 & - 3 & 4 3 & - 2 & 3 \end{matrix} ⎤ ⎥ ⎦

B = ⎡ ⎢ ⎣ \begin{matrix} - 1 & - 2 & - 1 6 & 12 & 6 5 & 10 & 5 \end{matrix} ⎤ ⎥ ⎦

3 α 3

P \neq Q

P^{3} = Q^{3}

P^{2} Q = Q^{2} P

P^{2} + Q^{2} | =

P

Q

3 \times 3

P \neq Q

P^{3} = Q^{3}

P^{2} Q = Q^{2} P

(P^{2} + Q^{2})

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