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The product of all eigenvalues of the matrix $\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{array}\right]$ is

  1. $-1$
  2. $0$
  3. $1$
  4. $2$ 
in Linear Algebra edited by
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7 Answers

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\[ A =
\begin{bmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 39 \\
\end{bmatrix}
\]

Since it is asking the product of eigen value. So, we know the property of eigen value that the product of any eigen value is equal to it's determinant. so we have to find the determinant of given matrix.

$\mid A \mid= 1(5*9-8*6)-2(9*4-7*6)+3(4*8-7*5)$

$\mid A \mid= 1(45-48)-2(36-42)+3(32-35)$

$\mid A \mid= 1(-3)-2(-6)+3(-3)$

$\mid A \mid=-3+12-9=0$

Because the given matrix determinant is zero which indicates that one of the eigenvalues is zero. So we can say that the product of eigenvalue is zero.

Option $(B)$ is correct.
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Convert the matrix to either Lower triangular matrix or Upper triangular matrix and you will get one of the diagonal element to be zero. Which indicate the product of all eigenvalues equal to be 0.
Answer:

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