GATE 2011 MA Matrices

Question

Comments

$C?$

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$D)?$

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D..

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$A=\begin{bmatrix} 3&3 &0 \\3 &3 &0 \\0 &0 &6 \end{bmatrix}$

How to write polynomial equation of $3\times3$matrix$?$

$\lambda^{3}-$[Sum of leading diagonal elements]$\lambda^{2}+\left[A_{11}+A_{22}+A_{33}\right]\lambda-det(A)=0$

Where $A_{11}=$Cofactor of $a_{11}=18$

             $A_{22}=$Cofactor of $a_{22}=18$

              $A_{33}=$Cofactor of $a_{33}=0$

Now$,\lambda^{3}-(3+3+6)\lambda^{2}+(18+18+0)\lambda-0=0$

   $\lambda^{3}-12\lambda^{2}+36\lambda=0$

 $\lambda(\lambda^{2}-12\lambda+36)=0$

$\lambda\left[(\lambda-6)(\lambda-6)\right]$

So$,\lambda=0,6,6$

Answer 1

ghGiven that  $A=\begin{bmatrix} 3&3 &0 \\ 3& 3& 0\\0 &0 &6 \end{bmatrix}$

We can write characteristic equation $|A-\lambda I|=0$

                                                        $\begin{vmatrix} 3-\lambda&3&0 \\ 3&3-\lambda &0 \\ 0& 0& 6-\lambda\end{vmatrix}=0$

$(3-\lambda)[(3-\lambda)(6-\lambda)-0]-3[3(6-\lambda)-0]+0=0$

 $(3-\lambda)(3-\lambda)(6-\lambda)-9(6-\lambda)=0$

 $(3-\lambda)^{2}(6-\lambda)-9(6-\lambda)=0$

 $(6-\lambda)[(3-\lambda)^{2}-9]=0$

 $(\lambda-6)[(\lambda-3)^{2}-9]=0$

 $(\lambda-6)[\lambda^{2}+9-6\lambda-9]=0$

 $(\lambda-6)[\lambda^{2}-6\lambda]=0$

   $\lambda(\lambda-6)(\lambda-6)=0$

  So, we got characteristic roots  $\lambda=0,6,6$

Now,$f(x)=x(x-6)^{2}$

   So$,g(x)=x(x-6)$

By Cayley Hamilton theorem:-Every square matrix satisfy its characteristic equation.

        $f(x)=0$

            $(or)$

           $f(A)=0$

Lets verify $g(A)$ is zero or not$?$

    $g(x)=x(x-6)=x^{2}-6x$

     $g(A)=A^{2}-6A$

    $g(A)=\begin{bmatrix} 18&18 &0 \\18 & 18& 0\\0 &0 &36 \end{bmatrix}-6\begin{bmatrix} 3&3 &0 \\ 3& 3& 0\\0 &0 &6 \end{bmatrix}$

    $g(A)=\begin{bmatrix} 18&18 &0 \\18 & 18& 0\\0 &0 &36 \end{bmatrix}-\begin{bmatrix} 18&18 &0 \\ 18& 18& 0\\0 &0 &36 \end{bmatrix}$

    $g(A)=0$

    $A^{2}-6A=0$

also $h(x)=x$  $;A\neq0$

        $p(x)=(x-6)$  $;A-6\neq0$

g(x) is the monic polynomial of lowest degree such that $g(x)=0$

So,$g(x)=x(x-6)$ is the minimal polynomial.

Comments

@Balaji Jegan 
please go through solution once, this is the actual procedure.