Question

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

Answer 1

Answer:

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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