If A0 - - 2 -2 -4- -1 3 4- 1 -2 -3 - and B0 - -4 -3 -3- 1 0 1- 4 4 3 -Bn - adj Bn-1- n - N and I is an identity matrix order 3- then correct satement is- are

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 A0 = [ 2 -...

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 option is 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

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} ⎤ ⎥ ⎦

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

and

I

is an identity matrix order 3, then correct satement is/ are

Q. 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})

Q. 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

Q. Let

n \geq 2

be an integer,

A = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} cos (\frac{2 π}{n}) & sin (\frac{2 π}{n}) & 0 - sin (\frac{2 π}{n}) & cos (\frac{2 π}{n}) & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

and

I

is the identity matrix of order

3

., then following of which is correct

Q. If

I = p A^{3} + q A

, where

I

is an identity matrix of order

3

and

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

, then find the value of

p

and

q

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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