Question

Consider the following matrix A. If the
eigenvalues of A are 4 and 8, then
2 3
X Y
Select one:
a. X=-4, Y=10
b. X=-3,Y=9
Y9
C. X=4,Y=10
O 5
d. X=9,Y=5​

Answer 1

Answer:

Option (a) x = -4 and y = 10 is the correct answer.

Step-by-step explanation:

We know that $|A-\lambda I| = 0$

where A is the matrix and

$\lambda$ is the eigen value and

I is the identity matrix

Therefore, on substituting the data from the question, we get,

$\left|\left[\begin{array}{cc}2&3\\x&y\end{array}\right] - \lambda\left[\begin{array}{cc}1&0\\0&1\end{array}\right]\right| = 0$

If  $\lambda$ = 4

=> $\left|\left[\begin{array}{cc}2&3\\x&y\end{array}\right] - 4\left[\begin{array}{cc}1&0\\0&1\end{array}\right] \right|= 0$

=> $\left|\left[\begin{array}{cc}2&3\\x&y\end{array}\right] - \left[\begin{array}{cc}4&0\\0&4\end{array}\right] \right|= 0$

=>  $\left|\left[\begin{array}{cc}2-4&3-0\\x-0&y-4\end{array}\right]\right| = 0$

=>  $\left|\begin{array}{cc}-2&3\\x&y-4\end{array}\right| = 0$

=> -2(y - 4) - 3x = 0

=> -2y + 8 - 3x = 0

=> - 2y - 3x = -8

=> 2y + 3x = 8                        --(i)

If  $\lambda$ = 8

=> $\left|\left[\begin{array}{cc}2&3\\x&y\end{array}\right] - 8\left[\begin{array}{cc}1&0\\0&1\end{array}\right] \right|= 0$

=> $\left|\left[\begin{array}{cc}2&3\\x&y\end{array}\right] - \left[\begin{array}{cc}8&0\\0&8\end{array}\right] \right|= 0$

=>  $\left|\left[\begin{array}{cc}2-8&3-0\\x-0&y-8\end{array}\right]\right| = 0$

=>  $\left|\begin{array}{cc}-6&3\\x&y-8\end{array}\right| = 0$

=> -6(y - 8) - 3x = 0

=> -6y + 48 - 3x = 0

=> - 6y - 3x = -48

=> 6y + 3x = 48                       --(ii)

Solving equations (i) and (ii), we get

-4y = -40

=> y = 10

Substituting y = 10 in equation (ii),

6*10 + 3x = 48

=> 60 + 3x = 48

=> 3x = 48 - 60

=> 3x = -12

=> x = -4

Therefore, Option (a) x = -4 and y = 10 is the correct answer.