If the determinant of the below matrix is 245, what is the value of X?
$\begin{bmatrix} 0 & 4& 2 &1 \\ 3& -1 & 0 & 2\\ 5&2 & x& 4\\ 6& 1 & -1 & 0 \end{bmatrix}$
I do $C_2=C_2-4C_4,C_3=C_3-2C_4$
$\begin{bmatrix} 0 & 0& 0 &1 \\ 3& -9 & -4 & 2\\ 5&-14 & x-8& 4\\ 6& 1 & -1 & 0 \end{bmatrix}$
Now I again do, $C_1=C_1+6C_3, C_2=C_2+C_3$
$\begin{bmatrix} 0 & 0& 0 &1 \\ -21& -13 & -4 & 2\\ 6x-43&x-22 & x-8& 4\\ 0& 0 & -1 & 0 \end{bmatrix}$
Now I expand along A[1,4]
$-1\times det \Biggl(.\begin{bmatrix} -21 & -13&-4 \\ 6x-3& x-22 &x-8 \\ 0& 0 & -1 \end{bmatrix}=B(let) \Biggr)$
Now I expand along B[3,3]
$-1 \times -1 \times \begin{vmatrix}
-21 &-13 \\
6x-3 &x-22
\end{vmatrix}$
Now if I evaluate above and keep equal to 245, I get x in negative decimals.
But answer is given to be 6 and It is correct, because by keeping 6 in place of x I am getting determinant as 245(Result verified by computer program).
Please let me know where I am going wrong?