Consider a 3 - 3 involutory matrix B0- -4 -3 -3- 1 0 - 4 - 3 - and the matrices Bn - adj Bn-1- n- Then the value of B-0 - B-1 - B-2 - - B-10 is

We have, B_0=[ 4 amp; 3 nbsp; amp; 3; nbsp;1 amp;0 amp;α; nbsp;4 amp;β amp;3 nbsp; ] Since B_0 is an nbsp;involutory matrix, therefore B_0^2=I. The ...

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Standard XII

Mathematics

Adjoint of a Matrix

Consider a 3 ...

Question

Consider a $3 \times 3$ involutory matrix $B_{0} = ⎡ ⎢ ⎣ \begin{matrix} - 4 & - 3 & - 3 1 & 0 & α 4 & β & 3 \end{matrix} ⎤ ⎥ ⎦$ and the matrices $B_{n} = a d j (B_{n - 1}),$ $n \in N .$ Then the value of $det (B_{0}^{α β} + B_{1}^{α β} + B_{2}^{α β} + \dots + B_{10}^{α β})$ is

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Solution

We have, $B_{0} = ⎡ ⎢ ⎣ \begin{matrix} - 4 & - 3 & - 3 1 & 0 & α 4 & β & 3 \end{matrix} ⎤ ⎥ ⎦$
Since $B_{0}$ is an involutory matrix, therefore $B_{0}^{2} = I .$
Then, $B_{0}^{2} = ⎡ ⎢ ⎣ \begin{matrix} - 4 & - 3 & - 3 1 & 0 & α 4 & β & 3 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} - 4 & - 3 & - 3 1 & 0 & α 4 & β & 3 \end{matrix} ⎤ ⎥ ⎦$
$= ⎡ ⎢ ⎣ \begin{matrix} 1 & 12 - 3 β & 3 - 3 α 4 α - 4 & α β - 3 & 3 α - 3 β - 4 & 3 β - 12 & α β - 3 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow α = 1, β = 4$ and $| B_{0} | = 1$

Also, $B_{n} = a d j (B_{n - 1})$
For $n = 1,$
$B_{1} = a d j (B_{0}) = | B_{0} | \cdot B_{0}^{- 1}$
$\Rightarrow B_{1} = B_{0}$
Similarly, $B_{2} = B_{0}$ and so on.

Now, $det (B_{0}^{α β} + B_{1}^{α β} + B_{2}^{α β} + \dots + B_{10}^{α β})$
$= det (B_{0}^{4} + B_{0}^{α β} + B_{0}^{4} + \dots + B_{0}^{4})$
$= det (11 I) = 11^{3} = 1331$

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Similar questions

Q. If

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

and

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

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..............matrix

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A = [\begin{matrix} 3 & 2 4 & 3 \end{matrix}]

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R

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Consider a 3×3 involutory matrix B0=⎡⎢⎣−4−3−310α4β3⎤⎥⎦ and the matrices Bn=adj(Bn−1), n∈N. Then the value of det(Bαβ0+Bαβ1+Bαβ2+⋯+Bαβ10) is

Consider a $3 \times 3$ involutory matrix $B_{0} = ⎡ ⎢ ⎣ \begin{matrix} - 4 & - 3 & - 3 1 & 0 & α 4 & β & 3 \end{matrix} ⎤ ⎥ ⎦$ and the matrices $B_{n} = a d j (B_{n - 1}),$ $n \in N .$ Then the value of $det (B_{0}^{α β} + B_{1}^{α β} + B_{2}^{α β} + \dots + B_{10}^{α β})$ is