Question

please... please give answer​

Answer 1

Step-by-step explanation:

your ans is in this picture

Answer 2

Question :

If $\bf{x - y} = \left[\begin{array}{c c c}1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$ and $\bf{x + y} = \left[\begin{array}{c c c}3 & 5 & 1 \\ -1 & 1 & 4 \\ 11 & 8 & 0\end{array}\right]$ , find x and y.

Given :

• $\bf{x - y} = \left[\begin{array}{c c c}1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$ (Equation.i)

• $\bf{x + y} = \left[\begin{array}{c c c}3 & 5 & 1 \\ -1 & 1 & 4 \\ 11 & 8 & 0\end{array}\right]$ (Equation.ii)

To find :

The value of x and y.

Solution :

To find the the value of x and y , we need to solve it linearly.

By using elimination method , we can obtain the value of x and y .

$\bf{x - y} = \left[\begin{array}{c c c}1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$

$\bf{x + y} = \left[\begin{array}{c c c}3 & 5 & 1 \\ -1 & 1 & 4 \\ 11 & 8 & 0\end{array}\right]$

________________________[By adding]

$\bf{2x} = \left[\begin{array}{c c c}1 + 3 & 1 + 5 & 1 + 1 \\ 1 + (-1) & 1 + 1 & 0 + 4 \\ 1 + 11 & 0 + 8 & 0 + 0\end{array}\right]$

By solving it , we get :-

$:\implies \bf{2x} = \left[\begin{array}{c c c}1 + 3 & 1 + 5 & 1 + 1 \\ 1 + (-1) & 1 + 1 & 0 + 4 \\ 1 + 11 & 0 + 8 & 0 + 0\end{array}\right]$

$:\implies \bf{2x} = \left[\begin{array}{c c c}4 & 6 & 2 \\ 1 - 1 & 2 & 4 \\ 12 & 8 & 0\end{array}\right]$

$:\implies \bf{2x} = \left[\begin{array}{c c c}4 & 6 & 2 \\ 0 & 2 & 4 \\ 12 & 8 & 0\end{array}\right]$

$:\implies \bf{x} = \dfrac{1}{2} \times \left[\begin{array}{c c c}4 & 6 & 2 \\ 0 & 2 & 4 \\ 12 & 8 & 0\end{array}\right]$

$:\implies \bf{x} = \left[\begin{array}{c c c}4 \times \dfrac{1}{2} & 6 \times \dfrac{1}{2} & 2 \times \dfrac{1}{2} \\ \\ 0 \times \dfrac{1}{2} & 2 \times \dfrac{1}{2} & 4 \times \dfrac{1}{2} \\ \\ 12 \times \dfrac{1}{2} & 8 \times \dfrac{1}{2} & 0 \times \dfrac{1}{2} \end{array}\right]$

$:\implies \bf{x} = \left[\begin{array}{c c c}\not{4} \times \dfrac{1}{\not{2}} & \not{6} \times \dfrac{1}{\not{2}} & \not{2} \times \dfrac{1}{\not{2}} \\ 0 & \not{2} \times \dfrac{1}{\not{2}} & \not{4} \times \dfrac{1}{\not{2}} \\ \not{12} \times \dfrac{1}{\not{2}} & \not{8} \times \dfrac{1}{\not{2}} & 0 \end{array}\right]$

$:\implies \bf{x} = \left[\begin{array}{c c c}2 & 3 & 1 \\ 0 & 1 & 2 \\ 6 & 4 & 0 \end{array}\right]$

Hence, the value of x is $\bf{x} = \left[\begin{array}{c c c}2 & 3 & 1 \\ 0 & 1 & 2 \\ 6 & 4 & 0\end{array}\right]$

Now , putting the value of x in the equation (i) , we get :

$:\implies \bf{x - y} = \left[\begin{array}{c c c}1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$

$:\implies \left[\begin{array}{c c c}2 & 3 & 1 \\ 0 & 1 & 2 \\ 6 & 4 & 0\end{array}\right] - y = \left[\begin{array}{c c c}1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$

$:\implies \left[\begin{array}{c c c}2 & 3 & 1 \\ 0 & 1 & 2 \\ 6 & 4 & 0\end{array}\right] - \left[\begin{array}{c c c}1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0\end{array}\right] = y$

$:\implies \left[\begin{array}{c c c}2 - 1 & 3 - 1 & 1 - 1 \\ 0 - 1 & 1 - 1 & 2 - 0 \\ 6 - 1 & 4 - 0 & 0 - 0\end{array}\right] = y$

$:\implies \left[\begin{array}{c c c}1 & 2 & 0 \\ - 1 & 0 & 2 \\ 5 & 4 & 0\end{array}\right] = y$

$\therefore y = \left[\begin{array}{c c c}1 & 2 & 0 \\ - 1 & 0 & 2 \\ 5 & 4 & 0\end{array}\right]$

Hence, the value of $y = \left[\begin{array}{c c c}1 & 2 & 0 \\ - 1 & 0 & 2 \\ 6 & 4 & 0\end{array}\right]$