Question

Given A = [ x 3
y 3 ]
If A2 = 3I , where I is the identity matrix of order 2 , find x and y.

Answer 1

x is -3 and y is -2.

The matrix A is

$\left[\begin{array}{ccc}x&3\\y&3\\\end{array}\right]$

To find A², we multiply the matrix with itself.

A² = $\left[\begin{array}{ccc}x&3\\y&3\\\end{array}\right]$ * $\left[\begin{array}{ccc}x&3\\y&3\\\end{array}\right]$

= $\left[\begin{array}{ccc}x^{2}+3y &3x+9\\xy+3y&3y+9\\\end{array}\right]$

The unit matrix is I = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

So, 3I would be = $\left[\begin{array}{ccc}3&0\\0&3\end{array}\right]$

As, A² = 3I, so

$\left[\begin{array}{ccc}x^{2}+3y &3x+9\\xy+3y&3y+9\\\end{array}\right]$ = $\left[\begin{array}{ccc}3&0\\0&3\end{array}\right]$

3x + 9 = 0

⇒ 3x = -9

⇒ x = -3

3y + 9 = 3

⇒ 3y = -6

⇒ y = -2

Answer 2

Answer:

x is -3 and y is -2.

The matrix A is

To find A², we multiply the matrix with itself.

A² =  *  

=  

The unit matrix is I =  

So, 3I would be =  

As, A² = 3I, so

=  

3x + 9 = 0

⇒ 3x = -9

⇒ x = -3

3y + 9 = 3

⇒ 3y = -6

⇒ y = -2

Hope it helped :)