Question

If one of the eigen values of a square matrix A order 3 $\times$ 3 is zero, then det A =

Answer 1

Let $A=\left[\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right]$
$|A|=a_1{b}_2{c}_3-a_1{b}_3{c}_2-b_1{a}_2{c}_3+b_1{c}_2{a}_3+a_1{a}_2{a}_3-c_1{a}_3{b}_2$
Now, $A-\lambda I=\left[\begin{array}{ccc}{a}_1-\lambda & b_1& c_1\\ a_2& b_2-\lambda & c_2\\ a_3& b_3& c_3-\lambda \end{array}\right]$
$\Rightarrow det(A-\lambda I)=(a_1-\lambda )\left[\right(b_2-\lambda \left)\right(c_3-\lambda )-b_3{c}_2]-b_1\left[a_2\right(c_3-\lambda )-c_2{a}_3]+c_1[a_2{b}_3-a_3(b_2-\lambda \left)\right]$
Now if one of the eigen values is zero, one root of $\lambda$ should be zero.
Therefore, constant term in the above polynomial is zero.
$\Rightarrow a_1{b}_2{c}_3-a_1{b}_3{c}_2-b_1{a}_2{c}_3+b_1{c}_2{a}_3+a_1{a}_2{a}_3-c_1{a}_3{b}_2=0$
$\Rightarrow detA=0$