Question

Given a={x 3/y 3}
If 42=31
Where I is the identity matrix upto 2 Find x and y

Answer 1

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

Answer 2

$\huge{\boxed{\red{\bf{Answer}}}}$

x is -3 and y is -2.

The matrix A is

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

To find A², we multiply the matrix with itself.

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

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

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

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

As, A² = 3I, so

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

$= \begin{gathered}\left[\begin{array}{ccc}3&0\\0&3\end{array}\right]\end{gathered}$

3x + 9 = 0

⇒ 3x = -9

⇒ x = -3

3y + 9 = 3

⇒ 3y = -6

⇒ y = -2