Question

Let the trace and determinant of a matrix A=[acbd] be 3 and 4 respectively. The eigenvalues of A are

Answer 1

Answer:

I think 12 is the answer for this

Answer 2

Answer: Eigen values are 3,1

Explanation:

The given matrix is A = $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$

We know trace of a matrix is the sum of the principal diagonal element

The trace of A is given as 4  i.e.  a + d =4

The determinant of A is given as 3  i.e. ad - bc = 3

The following steps are used to find the eigen value

Step -1 : Multiply a 2 X 2 unit matrix to a scalar x

               $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$  x  = $\left[\begin{array}{ccc}x&0\\0&x\end{array}\right]$

Step - 2 : Subtract the above matrix from A

                $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$  - $\left[\begin{array}{ccc}x&0\\0&x\end{array}\right]$  = $\left[\begin{array}{ccc}a-x &b\\c&d - x\end{array}\right]$

Step - 3: Find the determinant of the above matrix

              determinant of $\left[\begin{array}{ccc}a-x &b\\c&d - x\end{array}\right]$ = $(a - x)(d - x) - bc$

                                                                  = $ad - ax - dx + x^{2} - bc$

                                                                  = $(ad - bc) - x(a + d) + x^{2}$

                 By substituting the given values the above equation will become

  $3 - 4x + x^{2}$

Step - 4: solve the values for x that satisfy $3 - 4x + x^{2} = 0$

After solving the above quadratic equation we get the values of x as 3 and 1 which is nothing but eigen values. (Answer)