engineering maths

Question

Let $A=\begin{bmatrix} 2\\-4 \\7 \end{bmatrix}.\begin{bmatrix} 1 &9 &5 \end{bmatrix}$            

and $x,y$ and $z$ be the eigenvalue of $A$, then the value of $xyz$ is equal to?

Answer 1

Important properties of Eigen values:-

$(1)$Sum of all eigen values$=$Sum of leading diagonal(principle diagonal) elements=Trace of the matrix.

$(2)$ Product of all Eigen values$=Det(A)=|A|$

$(3)$ Any square diagonal(lower triangular or upper triangular) matrix eigen values are leading diagonal (principle diagonal)elements itself.

 

Example$:$$A=\begin{bmatrix} 1& 0& 0\\ 0&1 &0 \\ 0& 0& 1\end{bmatrix}$

    Diagonal matrix

  Eigenvalues are $1,1,1$

$B=\begin{bmatrix} 1& 9& 6\\ 0&1 &12 \\ 0& 0& 1\end{bmatrix}$

Upper triangular matrix

  Eigenvalues are $1,1,1$

$C=\begin{bmatrix} 1& 0& 0\\ 8&1 &0 \\ 2& 3& 1\end{bmatrix}$

Lower triangular matrix

  Eigenvalues are $1,1,1$

Now come to the question

$A=\begin{bmatrix} 2\\-4 \\7 \end{bmatrix}_{3\times1}.\begin{bmatrix} 1 &9 &5 \end{bmatrix}_{1\times3}$

$A=\begin{bmatrix} 2&18 &10 \\ -4&-36 &-20 \\7 &63 &35 \end{bmatrix}$  

Now find $Det(A)=|A|=\begin{vmatrix} 2&18 &10 \\ -4&-36 &-20 \\7 &63 &35 \end{vmatrix}$ 

             $|A|=2\times4\times7\begin{vmatrix} 1&9&5 \\ -1&-9 &-5 \\1 &9 &5 \end{vmatrix}$ 

               $|A|=0$ 

              so $xyz=0$