Maths Matrices

Let $A = (\begin{matrix} 2 & - 1 \\ 0 & 2 \end{matrix}) .$ If $B = I -^{5} ? C_{1} (a d j A) +^{5} ? C_{2} {(a d j A)}^{2} - . . .^{5} ? C_{5} {(a d j A)}^{5},$ then the sum of all elements of the matrix B is

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Contributor-Level 10

Kindly consider the following figure

B = (I – adjA)⁵

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New answer posted

3 weeks ago

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Contributor-Level 10

Kindly consider the following figure

B = (I – adjA)⁵

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New answer posted

3 weeks ago

$\underset{x \to \frac{1}{\sqrt[]{2}}}{l i m} \frac{s i n (c o s^{- 1} x) - x}{1 - t a n (c o s^{- 1} x)}$ is equal to:

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Raj Pandey

Contributor-Level 9

$\underset{x \to \frac{1}{\sqrt[]{2}}}{l i m} \frac{s i n (c o s^{- 1} x) - x}{1 - t a n (c o s^{- 1} x)}$

Let $c o s^{- 1} x = \frac{π}{4} + θ$

= $\underset{θ \to \infty}{l i m} \frac{\sqrt[]{2} s i n θ}{- 2 t a n θ} (1 - t a n θ) = - \frac{1}{\sqrt[]{2}}$

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New answer posted

3 weeks ago

Let $A = (\begin{matrix} 2 & - 1 \\ 0 & 2 \end{matrix}) .$ If $B = I -^{5} C_{1} (a d j A) +^{5} C_{2} {(a d j A)}^{2} - . . .^{5} C_{5} {(a d j A)}^{5},$ then the sum of all elements of the matrix B is

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

B = (I – adjA)⁵

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New question posted

3 weeks ago

Let $A = (\begin{matrix} 2 & - 1 \\ 0 & 2 \end{matrix}) .$ If $B = I -^{5} ? C_{1} (a d j A) +^{5} ? C_{2} {(a d j A)}^{2} - . . .^{5} ? C_{5} {(a d j A)}^{5},$ then the sum of all elements of the matrix B is

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New answer posted

3 weeks ago

Let $M = [\begin{matrix} 0 & - α \\ α & 0 \end{matrix}],$ where a is a non-zero real number an $N = \sum_{k = 1}^{49} M^{2 k} .$ If (I – M²)N = -2I, then the positive integral value of a is…………….

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alok kumar singh

Contributor-Level 10

$N = M^{2} + M^{4} + . . . . . + M^{98}$

$= (- α^{2} I) + {(- α^{2} I)}^{2} + . . . . + {(- α^{2} I)}^{49}$

$= I (- α^{2} + α^{4} - α^{6} + . . . . - α^{98})$

N = $- I (α^{2} - α^{4} + α^{6} . . . . . . . + α^{98})$

$= - I \frac{α^{2} (1 - {(- α^{2})}^{49})}{1 - (- α^{2})}$

N = $\frac{- I α^{2} (1 + α^{98})}{1 + α^{2}}$

Now $(I - m^{2}) N = - 2 I$

$(I + α^{2} I) \frac{(- I α^{2} \cdot (1 + α^{98})}{1 + α^{2}} = - 2 I$

-> a¹⁰⁰ + a² = 2

->a = $\pm 1$

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3 weeks ago

If the system of equations

x + 6y + 6z = $λ x$

4x – y + 4z = $λ y$

$2 λ x + 2 λ y + λ z = 5 z$

has non-trivial solution then the value of $| λ |$ equals (where $λ \in$ R)

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Contributor-Level 10

$| \begin{matrix} 1 - α & 6 & 6 \\ 4 & - 1 - α & 4 \\ 2 α & 2 α & α - 5 \end{matrix} | = 0$

$α = - 5$

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3 weeks ago

Let S = ${(a, b) : A^{3} = A}$ where A is a matrix defined as A = $(\begin{matrix} \frac{1}{2} & \frac{1}{2} \\ a & b \end{matrix})$ then n(S) equals (where n(A) denotes number of elements in set A)

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Contributor-Level 10

$| A - x I | = | \begin{matrix} 1 / 2 - x & 1 / 2 \\ a & b - x \end{matrix} | = x^{2} - (b + \frac{1}{2}) x + \frac{b - a}{2}$

from Cayley – Hamilton Theorem

$A^{2} - (b + \frac{1}{2}) A + (\frac{b - a}{2}) I = 0$

$\Rightarrow A^{3} = ({(b + \frac{1}{2})}^{2} - (\frac{b - a}{2})) A - (b + \frac{1}{2}) (\frac{b - a}{2}) I$

$∴ {(b + \frac{1}{2})}^{2} - (\frac{b - a}{2}) = 1 & (b + \frac{1}{2}) (\frac{b - a}{2}) = 0$

$∴ (a, b) = (\frac{1}{2}, \frac{1}{2}), (- \frac{3}{2}, - \frac{3}{2}), (\frac{3}{2}, - \frac{1}{2})$

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3 weeks ago

Let A and B are non-singular matrices of order 3 such that $| A | = 5$ and A^-1B² + AB = 0, then A² $| A^{2} |$ - adj (adjB) is equal to

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alok kumar singh

Contributor-Level 10

$A^{2} {| A |}^{2} - {| B |}^{3.2} B$

$= 25 A^{2} - | B | . B$

$\begin{array}{l} = 25 A^{2} + | B | A^{2} \\ = 0 \end{array}$

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3 weeks ago

Let A be a square matrix of order 2 such that A² – 4A + 4I = 0, where I is an identity matrix of order 2. If B = A? + 4A? + 6A³ + 4A² + A – 162 I, then det(B) is equal to

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