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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?

1 Answer

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Expanding without any elementary operations will take a lot of time.The key is to use these operations.

$\begin{bmatrix} 0 & 4 &2 &1 \\ 3& -1&0 &2 \\ 5& 2& x & 4\\ 6& 1& -1 & 0 \end{bmatrix}$

R1 <-> R2  and  R2 <-> R3

$\begin{bmatrix} 3 & -1 &0 &2 \\ 5& 2&x &4\\ 0& 4& 2 & 1\\ 6& 1& -1 & 0 \end{bmatrix}$

R2 = R2 - 5/3 * R1 and R4 = R4 - 2*R1

$\begin{bmatrix} 3 & -1 &0 &2 \\ 0& 11/3&x &2/3\\ 0& 4& 2 & 1\\ 0& 3& -1 & -4 \end{bmatrix}$

Solving,we get

3[ (11/3)(-7) -(3x/3)(-19) + (2/3)(-10) ] =255

-77 + 57x -20 = 245

Therefore, x=6.

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