If A - 0- begin bmatrix 2 -2 -4 1-1 - 3 - 4111 -2 -3-4 -3 -3 11 - 0 - 1114 - 4 - 3lend bmatrix B n -adj B n -1- n - N and I is an identity matrix order 3 - then correct satement is-areA- Determinant of A0-A02 B02-A03-A04 B04- -10- terms is equal to zero-B- B 1- B 2- B 49 is equal to 49 B 0C- For a variable matrix X- the equation A 0 X - B 0 will have no solution-D- None of these

A_0^2 = A_0, B^2_0 = I So, Determinant of (A_0 +A^2_0 B^2_0 +A_0^3 +A_0^4 B_0^4+....10 terms) = Determinant of (A_0 + A_0 I +A_0 +A_0 I^2+....10 terms) =(A_0 ...

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Byju's Answer

Standard XII

Mathematics

Adjoint of a Matrix

If A - 0=. be...

Question

If $A_{0} = ⎡ ⎢ ⎣ \begin{matrix} 2 & - 2 & - 4 - 1 & 3 & 4 1 & - 2 & - 3 \end{matrix} ⎤ ⎥ ⎦$ and $B_{0} ⎡ ⎢ ⎣ \begin{matrix} - 4 & - 3 & - 3 1 & 0 & 1 4 & 4 & 3 \end{matrix} ⎤ ⎥ ⎦$
$B_{n} = a d j (B_{n - 1}), n \in N$ and $I$ is an identity matrix order 3, then correct satement is/ are

Determinant of

(A_{0} + A_{0}^{2} B_{0}^{2} + A_{0}^{3} + A_{0}^{4} B_{0}^{4} + . . . .10 t e r m s

) is equal to zero.

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B_{1} + B_{2} + . . . . B_{49}

is equal to 49

B_{0}

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For a variable matrix X, the equation

A_{0} X = B_{0}

will have no solution.

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None of these

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Solution

The correct options are
A Determinant of $(A_{0} + A_{0}^{2} B_{0}^{2} + A_{0}^{3} + A_{0}^{4} B_{0}^{4} + . . . .10 t e r m s$ ) is equal to zero.
B $B_{1} + B_{2} + . . . . B_{49}$ is equal to 49 $B_{0}$
C For a variable matrix X, the equation $A_{0} X = B_{0}$ will have no solution.
$A_{0}^{2} = A_{0}, B_{0}^{2} = I$
So, Determinant of $(A_{0} + A_{0}^{2} B_{0}^{2} + A_{0}^{3} + A_{0}^{4} B_{0}^{4} + . . . .10 t e r m s$ )
= Determinant of $(A_{0} + A_{0} I + A_{0} + A_{0} I^{2} + . . . .10 t e r m s)$
$= (A_{0} + A_{0} + . . . . .10 t e r m s) = | 10 A_{0} |$
$= 1000 \times ∣ ∣ ∣ ∣ \begin{matrix} 2 & - 2 & - 4 - 1 & 3 & 4 1 & - 2 & - 3 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$B_{0}^{2} = I \Rightarrow B_{0} = B_{0}^{- 1}$
$B_{1} = a d j (B_{0}), a d j (B_{0}) = | B_{0} | B_{0}^{- 1} = B_{0} [A s | B_{0} | = 1]$
$∴ B_{1} = B_{0}$
$B_{2} = a d j (B_{1}) = B_{0} . . . s o o n$
So $B_{1} + B_{2} + . . . . + B_{49} = 49 B_{0}$
$A_{0} X = B_{0} \Rightarrow | A_{0} | | X | = | B_{0} |$
$| A_{0} | = 0 a n d | B_{0} | = 1$
$0 = 1 \to$ no solution.

Suggest Corrections

Similar questions

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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

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Q. For any

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If A0=⎡⎢⎣2−2−4−1341−2−3⎤⎥⎦ and B0⎡⎢⎣−4−3−3101443⎤⎥⎦ Bn=adj(Bn−1),n∈N and I is an identity matrix order 3, then correct satement is/ are