If 3A - - 1 2 2- 2 1 -2- x 2 y - and A- AT -I- then find the value of x-y- -3 Marks-

Given that, 3A = [ 1 amp; 2 amp; 2; nbsp; 2 amp; 1 amp; 2; nbsp; x amp; 2 amp; y ] nbsp; ⇒ A = [ 1/3 amp; 2/3 amp; 2/3; nbsp; 2/3 amp; ...

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

Byju's Answer

Standard XII

Mathematics

Transpose of a Matrix

If 3A = [ 1 2...

Question

If $3 A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 2 2 & 1 & - 2 x & 2 & y \end{matrix} ⎤ ⎥ ⎦$ and $A \cdot A^{T} = I$ , then find the value of $x + y$ . [3 Marks]

Open in App

Solution

$G i v e n t h a t, 3 A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 2 2 & 1 & - 2 x & 2 & y \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow A = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \frac{2}{3} & \frac{1}{3} & - \frac{2}{3} \frac{x}{3} & \frac{2}{3} & \frac{y}{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ [0.5 M a r k]$
$∴ A^{T} = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{3} & \frac{2}{3} & \frac{x}{3} \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \frac{2}{3} & - \frac{2}{3} & \frac{y}{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ [0.5 M a r k]$
$A A^{T} = I \Rightarrow ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \frac{2}{3} & \frac{1}{3} & - \frac{2}{3} \frac{x}{3} & \frac{2}{3} & \frac{y}{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ \cdot ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{3} & \frac{2}{3} & \frac{x}{3} \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \frac{2}{3} & - \frac{2}{3} & \frac{y}{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{9} + \frac{4}{9} + \frac{4}{9} & \frac{2}{9} + \frac{2}{9} - \frac{4}{9} & \frac{x}{9} + \frac{4}{9} + \frac{2 y}{9} \frac{2}{9} + \frac{2}{9} - \frac{4}{9} & \frac{4}{9} + \frac{1}{9} + \frac{4}{9} & \frac{2 x}{9} + \frac{2}{9} - \frac{2 y}{9} \frac{x}{9} + \frac{4}{9} + \frac{2 y}{9} & \frac{2 x}{9} + \frac{2}{9} - \frac{2 y}{9} & \frac{x^{2}}{9} + \frac{4}{9} + \frac{y^{2}}{9} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow x^{2} + y^{2} + 4 = 9 \Rightarrow x^{2} + y^{2} = 5 . . . (1)$

$\Rightarrow 2 x + 2 - 2 y = 0 \Rightarrow y = x + 1 . . . (2)$

$\Rightarrow x + 4 + 2 y = 0 \Rightarrow x + 2 y = - 4 . . . (3)$

Solving Equations (1) and (2)

$x^{2} + (x + 1)^{2} = 5 \Rightarrow x^{2} + x^{2} + 2 x + 1 = 5 \Rightarrow x^{2} + x - 2 = 0 \Rightarrow x = - 2, 1 \Rightarrow (x, y) \equiv (1, 2), (- 2, - 1)$

Satisfying these in Equation (3)

$(- 2) + 2 (- 1) = - 4 (1) + 2 (2) \neq - 4$
Hence, $(- 2, - 1)$ satisfies.

So $x + y = - 3$ . [2 Marks]

Suggest Corrections

Similar questions

Q. If

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

, and

A A^{'} = I

, then

- x - y =

...................

Q. (a)

If $H = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 2 & 3 & 1 3 & 1 & 2 \end{matrix} ⎤ ⎥ ⎦, G = ⎡ ⎢ ⎣ \begin{matrix} 2 & 6 & - 10 0 & - 4 & 8 2 & - 8 & 4 \end{matrix} ⎤ ⎥ ⎦$

Find $3 H - \frac{1}{2} G .$
$[3]$

(b) If $3 A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 2 2 & 1 & - 2 x & 2 & y \end{matrix} ⎤ ⎥ ⎦$ and $A \cdot A^{T} = I$ , then find the value of $x + y$ .
$[3]$

(c) If $A = [\begin{matrix} 2 & 5 1 & 3 \end{matrix}], B = [\begin{matrix} 4 & - 2 - 1 & 3 \end{matrix}]$ and $I$ is Identity matrix of same order and $A^{t}$ is the transpose of matrix $A$ find $A^{t} . B + B I$ .
$[4]$