Nullity of matrix

Question

Nullity of a matrix = Total  number columns – Rank of that matrix

But how to calculate value of x when nullity is already given(1 in this case)

Answer 1

Nullity of a matrix = Total  number columns – Rank of that matrix = n-r

So n-r = 1 (Given)

Hence r= 3-1 = 2

As rank of matrix is 2, so Determinant of A should be equal to 0.

Hence, |A| = 2*(x-18) - 3*(x-81) + 7*(2-9) = 0

So x = 158.

Comments

Thank you so much..

one more doubt

What if they gave n-r=2 ... i.e. r=1

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Rank of a matrix being 1 denotes that number of non-zero rows in Echleon form is 1. So in that case, after converting the given matrix to echleon form whereby you will get 2 rows or 2 columns of the transformed matrix as 0, you need to equate whatever element you get regarding x to zero so that the above mentioned condition is satisfied( i.e. getting two rows or two columns as 0 in the transformed matrix).

Answer 2

Given , 

Nullity of a matrix = Total  number columns – Rank of that matrix

Rank of that matrix =Total  number columns –  Nullity of a matrix

                                   = 3 -1 = 2

We will solve this by converting the given matrix into Echelon form using Gauss Elimination method.

 1)

 $\Rightarrow$ $\begin{bmatrix} 2 &3 &7 \\ 1 &1 &9 \\ 9 & 2 & x \end{bmatrix}$

R3 $\leftarrow$ R3 – R2

 $\Rightarrow$  $\begin{bmatrix} 2 &3 &7 \\ 1 &1 &9 \\ 8 & 1 & x-9 \end{bmatrix}$

 

2)

           R2 $\leftarrow$ 2R2 – R1

           R3 $\leftarrow$ R3 – 4R1

 $\Rightarrow$     $\begin{bmatrix} 2 &3 &7 \\ 0 &-1 &9 \\ 0 & -11 & x-37 \end{bmatrix}$

3)

             R3 $\leftarrow$ R3 – 11R2

       $\Rightarrow$       $\begin{bmatrix} 2 &3 &7 \\ 0 &-1 &9 \\ 0 & 0 & x-37-121\end{bmatrix}$     

 

        Since there should be only two pivot column ( since the rank is 2) 

         therefore,

         x-37-121 = 0

         $\therefore$ x = 158