If B0- - -4 -3 -3- 1 0 1- 4 4 3 -Bn-adjBn-1- - n - and I is identity matrix of order 3- Then B1-B3-B5-B7-B9 is equal to

Given : nbsp;B_0= [ 4 amp; 3 amp; 3; 1 amp;0 amp;1; 4 amp;4 amp;3 ] Finding the adjoint of the matrix, we get ⇒adj B_0 = [ 4 amp;1 amp; ...

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Determinant

If B0= [ -4 -...

Question

If $B_{0} = ⎡ ⎢ ⎣ \begin{matrix} - 4 & - 3 & - 3 1 & 0 & 1 4 & 4 & 3 \end{matrix} ⎤ ⎥ ⎦, B_{n} = adj (B_{n - 1}), \forall n \in N$ and $I$ is identity matrix of order $3.$ Then $B_{1} + B_{3} + B_{5} + B_{7} + B_{9}$ is equal to

B_{0}

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5 B_{0}

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25 B_{0}

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B_{0} + 5 I

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Solution

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Similar questions

A_{0} = ⎡ ⎢ ⎣ \begin{matrix} 2 & - 2 & - 4 - 1 & 3 & 4 1 & - 2 & - 3 \end{matrix} ⎤ ⎥ ⎦

B_{0} ⎡ ⎢ ⎣ \begin{matrix} - 4 & - 3 & - 3 1 & 0 & 1 4 & 4 & 3 \end{matrix} ⎤ ⎥ ⎦

B_{n} = a d j (B_{n - 1}), n \in N

I

b_{n + 1} = \frac{1}{1 - b_{n}}

n \geq 1

b_{1} = b_{3}

2001 \sum r = 1 b_{r}

P = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 4 & 1 & 0 16 & 4 & 1 \end{matrix} ⎤ ⎥ ⎦

I

3

Q = [q_{i j}]

p^{50} - Q = I

\frac{q_{31} + q_{32}}{q_{21}}

P = ⎡ ⎢ ⎣ \begin{matrix} 1416 \end{matrix} \begin{matrix} 014 \end{matrix} \begin{matrix} 001 \end{matrix} ⎤ ⎥ ⎦

I

3

Q = [q_{i j}]

P^{50} - Q = I

\frac{q_{31} + q_{32}}{q_{21}}

b_{1}, b_{2}, b_{3}, . . . b_{n}

b_{1} b_{2} + b_{2} b_{3} + b_{3} b_{4} + . . . + b_{n - 1} b_{n} =

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