Find the last row of the Matrix

Question

For the first one I have got -28 and 11 as the answer but for the matrix C I am not able to get the value of element c_32.

c₃₁ = c₃₃ = 6.

Comments

C32 can have any value because when we solve the characteristics equation by putting different eigen value then we get different value of b

Answer 1

Sum of eigen values = trace of the matrix

Product of eigen matrix = determinant of the matrix

yes, for matrix A, values will be -28, 11 and for matrix C, values will be 6, x, 6  where x can be any real number.

Comments

Yeah I think the value of x can be anything since it will not be going to affect the trace and the determinant of the matrix.

But the answer is -11 and it is coming from the characteristic equation of Matrix i.e.

$-\lambda^{3}+6\lambda^{2}-11\lambda+6$

Is there any relation between rows of the matrix and the characteristic equation of the matrix?

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@manu thakur I am getting c32 as -11.  
As you said x can take any real value, I checked by placing the values 6,-12,6 in the matrix and solved it for the 3 eigen values. |A-I|, |A-2I| and |A-3I| should have been equal to zero but it is not so.

Taking a=c31, b=c32, c=c33

a+b+c=1, a+2b+4c=8 and a+3b+9c=27 are you getting these equations?

Can you please check? Thank you

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However if we choose shortcut like

Trace = 0+0+c=1+2+3 which gives c=6

And, product of eigen values= determinant of the matrix, 1*2*3=b so 6= c ,

Then we cannot get value of b.

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Nice explanation