Question

[tex] Solve the Matrix equation \left[\begin{array}{ccc}x²\\y²\end{array}\right] -4 \left[\begin{array}{ccc}2x\\y\end{array}\right] =\left[\begin{array}{ccc}-7\\12\end{array}\right] [\tex]

Answer 1

Answer:

x=1 or x=7

y=-2 or y=6

Step-by-step explanation:

\left[\begin{array}{ccc}x²\\y²\end{array}\right] -4 \left[\begin{array}{ccc}2x\\y\end{array}\right] =\left[\begin{array}{ccc}-7\\12\end{array}\right] [\tex]

$\left[\begin{array}{ccc}x²\\y²\end{array}\right] -4 \left[\begin{array}{ccc}2x\\y\end{array}\right] =\left[\begin{array}{ccc}-7\\12\end{array}\right] [\tex]\\hence, the equations are'\\x^{2} -8\times x+7=0\\x^{2}-7x-x+7 =0\\(x-7)(x-1)=0\\x=1 or x=7\\and second eqation is\\y^{2} -4y-12=0\\y^{2}-4y-12=0\\y^{2}-6y+2y-12=0 \\ (y+2)(y-6)=0\\y=-2 or y=6$

Answer 2

Answer:

$x=7$  or $x=1$

$y=6$  or $y=-2$

Step-by-step explanation:

$\left[\begin{array}{ccc}x^{2}\\y^{2}\end{array}\right] -4 \left[\begin{array}{ccc}2x\\y\end{array}\right] =\left[\begin{array}{ccc}-7\\12\end{array}\right]$

⇒$\left[\begin{array}{ccc}x^{2}\\y^{2}\end{array}\right]- \left[\begin{array}{ccc}8x\\4y\end{array}\right] =\left[\begin{array}{ccc}-7\\12\end{array}\right]$

⇒$\left[\begin{array}{ccc}x^{2}-8x\\y^{2}-4y\end{array}\right]=\left[\begin{array}{ccc}-7\\12\end{array}\right]$

By comparing the matrices on both side we get:

   $x^{2}-8x =-7$        and        $y^{2}-4y = 12$

now sole the above quadratic equation

$x^{2}-8x =-7$

⇒$x^{2}-8x+7=0$                  

⇒$\left ( x-7 \right )\left ( x-1 \right )=0$    

⇒$\left ( x-7 \right )=0$  or   $\left ( x-1 \right )=0$  

⇒$x=7$    or    $x=1$

                       

$y^{2}-4y = 12$

⇒$y^{2}-4y-12=0$                  

⇒$\left ( y-6 \right )\left ( y+2 \right )=0$    

⇒$\left ( y-6 \right )=0$  or   $\left ( y+2 \right )=0$  

⇒$y=6$    or    $y=-2$