Question

Can anyone help me with this problem ?​

Answer 1

ANSWER:

Given:

$\:\:\:\:\bullet\:\:\:\:P=\left[\begin{array}{c c}{xy} & {y^2} \\\\ {-x^2} & {-xy}\end{array}\right]$

To Prove:

$\:\:\:\:\bullet\:\:\:\:P^2=\left[\begin{array}{c c}{0} & {0} \\\\ {0} & {0}\end{array}\right]$

Solution:

$\text{We are given that,}\\\\:\longrightarrow P=\left[\begin{array}{c c}{xy} & {y^2} \\\\ {-x^2} & {-xy}\end{array}\right]\\\\:\implies P^2=\left[\begin{array}{c c}{xy} & {y^2} \\\\ {-x^2} & {-xy}\end{array}\right]\times\left[\begin{array}{c c}{xy} & {y^2} \\\\ {-x^2} & {-xy}\end{array}\right]\\\\:\implies P^2=\left[\begin{array}{c c}{(xy\times xy)+(y^2\times-x^2)} & {(xy\times y^2)+(y^2\times-xy)} \\\\ {(-x^2\times xy)+(-xy\times-x^2)} & {(-x^2\times y^2)+(-xy\times-xy)}\end{array}\right]$

$\\:\implies P^2=\left[\begin{array}{c c}{xe^2y^2-x^2y^2} & {xy^3-xy^3} \\\\ {-x^3y+x^3y} & {-x^2y^2+x^2y^2}\end{array}\right]\\\\\bf{:\implies P^2=\left[\begin{array}{c c}\bf{{0}} & \bf{{0}} \\\\ \bf{{0}} & \bf{{0}}\end{array}\right] = RHS}\\\\\text{\bf{HENCE PROVED!!!}}$

Learn More:

There are 7 types of matrices:

1. Row Matrix : Has only 1 row (no restrictions on columns)
2. Column Matrix : Has only 1 columns (no restrictions on rows)
3. Square Matrix : Has equal rows and columns
4. Diagonal Matrix : A square matrix with all non-diagonal elements as zero.
5. Scalar Matrix : A diagonal matrix with same diagonal elements.
6. Identity Matrix : A scalar matrix with diagonal elements as 1.
7. Zero Matrix : A matrix with all its elements 0.

Answer 2

best answer ke liye refer the attachment.

Agar achha lage to like zarur krna.