Question

solve this fast ... it's urgent ... don't give any irrelevant
ansr !! ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\bigg[ \begin{matrix}x + 2&6 \\ 3&5z \end{matrix} \bigg] = \bigg[ \begin{matrix} - 5& {y}^{2} + y \\ 3& - 20 \end{matrix} \bigg]$

Since, matrices are equal.

It means, corresponding elements are same.

$\rm :\longmapsto\:x + 2 = - 5$

$\rm :\longmapsto\:x = - 5 - 2$

$\bf\implies \:\boxed{ \bf{ \:x = - 7}}$

Also,

$\rm :\longmapsto\:5z = - 20$

$\bf\implies \:\boxed{ \bf{ \:z = - \: 4}}$

Also,

$\rm :\longmapsto\: {y}^{2} + y = 6$

$\rm :\longmapsto\: {y}^{2} + y - 6 = 0$

$\rm :\longmapsto\: {y}^{2} + 3y - 2y - 6 = 0$

$\rm :\longmapsto\:y(y + 3) - 2(y + 3) = 0$

$\rm :\longmapsto\:(y + 3)(y - 2) = 0$

$\bf\implies \:\boxed{ \bf{ \:y = - 3 \: \: or \: \: y = 2}}$

Additional Information :-

Let us consider same type of example!!

Question :- Solve for x and y, if

$\rm :\longmapsto\:\bigg[ \begin{matrix}x + y&5 \\ 1&xy \end{matrix} \bigg] = \bigg[ \begin{matrix}6&5 \\ 1&8 \end{matrix} \bigg]$

Solution

Given that

$\rm :\longmapsto\:\bigg[ \begin{matrix}x + y&5 \\ 1&xy \end{matrix} \bigg] = \bigg[ \begin{matrix}6&5 \\ 1&8 \end{matrix} \bigg]$

Since, Matrices are equal.

So, corresponding elements are same

$\rm :\longmapsto\:x + y = 6$

$\rm :\longmapsto\: y = 6 - x - - - (1)$

Also,

$\rm :\longmapsto\:xy = 8$

On substituting the value of y, we get

$\rm :\longmapsto\:x(6 - x) = 8$

$\rm :\longmapsto\:6x - {x}^{2} = 8$

$\rm :\longmapsto\: {x}^{2} - 6x + 8 = 0$

$\rm :\longmapsto\: {x}^{2} - 4x - 2x + 8 = 0$

$\rm :\longmapsto\:x(x - 4) - 2(x - 4) = 0$

$\rm :\longmapsto\:(x - 4)(x - 2) = 0$

$\bf\implies \:x = 2 \: \: \: or \: \: \: x = 4$

So, corresponding values of y are as follow

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y = 6 - x \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 4 & \sf 2 \\ \\ \sf 2 & \sf 4 \end{array}} \\ \end{gathered}$