If A - B - - 1 0 2- 2 2 2- 1 1 1 - and A - B - - 1 4 4- 4 2 0- -1 -1 2 -then prove that A - - 1 2 3- 3 2 1- 0 0 2 - and B - - 0 -2 -1- -1 0 1- 1 1 0 -

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Matrix Definition and Representation

If A + B = [...

Question

If A + B = $⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 2 & 2 & 2 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$ and A - B = $⎡ ⎢ ⎣ \begin{matrix} 1 & 4 & 4 4 & 2 & 0 - 1 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦$
then prove that
A = $⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 3 & 2 & 1 0 & 0 & 2 \end{matrix} ⎤ ⎥ ⎦$ and B = $⎡ ⎢ ⎣ \begin{matrix} 0 & - 2 & - 1 - 1 & 0 & 1 1 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦$

Open in App

Solution

Suggest Corrections

Similar questions

If \sin^{- 1} \frac{2 a}{1 + a^{2}} - \cos^{- 1} \frac{1 - b^{2}}{1 + b^{2}} = \tan^{- 1} \frac{2 x}{1 - x^{2}}, then prove that x = \frac{a - b}{1 + a b}

\tan A = \frac{5}{6} and \tan B = \frac{1}{11}

A + B = \frac{π}{4}

\tan A = \frac{m}{m - 1} and \tan B = \frac{1}{2 m - 1}

A - B = \frac{π}{4}

If \sin^{- 1} \frac{2 a}{1 + a^{2}} + \sin^{- 1} \frac{2 b}{1 + b^{2}} = 2 \tan^{- 1} x, prove that x = \frac{a + b}{1 - a b} .

a, b

c

|\begin{matrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{matrix}| =

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} +

=

A + B = 45^{0},

(1 + tan A) (1 + tan B) = 2

(cot A - 1) (cot B - 1) = 2

A - B = \frac{3 π}{4},

(1 - tan A) (1 + tan B) = 2

Join BYJU'S Learning Program