Question

If $A= \left[\begin{array}{ccc}1&3\\3&4\end{array}\right]$ , find the values of x and y, if A² – xA + yI = 0.

Answer 1

Answer:

$x = 5 \\ y = - 5$
Solution:

$A= \left[\begin{array}{ccc}1&3\\3&4\end{array}\right]$

$A^{2}= \left[\begin{array}{ccc}1&3\\3&4\end{array}\right] \times \left[\begin{array}{ccc}1&3\\3&4\end{array}\right]$

$A^{2}= \left[\begin{array}{ccc}1+9&3+12\\3+12&9+16\end{array}\right]$

$A^{2}= \left[\begin{array}{ccc}10&15\\15&25\end{array}\right]$...eq1

$xA= \left[\begin{array}{ccc}x&3x\\3x&4x\end{array}\right]$....eq2

$yI= \left[\begin{array}{ccc}y&0\\0&y\end{array}\right]$...eq3

Place all values in the equation

${A}^{2} - xA + yI = 0$

$\left[\begin{array}{ccc}10&15\\15&25\end{array}\right]-\left[\begin{array}{ccc}x&3x\\3x&4x\end{array}\right]+\left[\begin{array}{ccc}y&0\\0&y\end{array}\right] =0$

$\left[\begin{array}{ccc}10-x+y&15-3x\\15-3x&25-4x+y\end{array}\right]=\left[\begin{array}{ccc}0&0\\0&0\end{array}\right]$

So equate

$15 - 3x = 0 \\ \\ - 3x = -15 \\ \\ x = 5$

$10 - x + y = 0 \\ \\ - x + y = - 10 \\ \\ y = - 10 + 5 \\ \\ y = - 5$
Hope it helps you

Answer 2

Answer:

x=5 and y= -5

Step-by-step explanation:

Concept:

Cayley Hamilton theorem:

Every square matrix satisfies its characteristic equation.

$A=\left[\begin{array}{ccc}1&3\\3&4\end{array}\right]$

characteristic equation is |A-xI|=0

$\left|\begin{array}{ccc}1-x&3\\3&4-x\end{array}\right|=0$

expanding we get

(1-x)(4-x) - 9 = 0

(x-1)(x-4) - 9 =0

x² - 5x +4 -9=0

x² - 5x -5=0

By caley hamilton theorem,

matrix A will satisfy x² - 5x -5=0

so we have,

A² - 5A -5I=0

comparing this with A² - xA +yI=0 we get

x=5 and y=-5