Linear Algebra-METest

Question

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?

Comments

Yes ma'am, but see the reduction after expanding along [1,4], it's 6x-43 not 6x-3.

one can solve only expanding and one can solve using reduction.

I think if some rows or column make $0$ so, finding the determinant is easy.

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@srestha ma'am see this 

Given that $A=\begin{bmatrix} 0 &4 &2 &1 \\ 3 &-1 &0 &2 \\ 5 &2 &x &4 \\ 6 &1 &-1 &0 \end{bmatrix}$ and $|A|=245$

find $x=?$

Now $|A|=\begin{vmatrix} 0 &4 &2 &1 \\ 3 &-1 &0 &2 \\ 5 &2 &x &4 \\ 6 &1 &-1 &0 \end{vmatrix}$

expand the row wise$,$

$|A|=0.\begin{vmatrix} -1 &0 &2 \\ 2 &x &4 \\ 1 &-1 &0 \end{vmatrix}$$-4.\begin{vmatrix} 3&0 &2 \\ 5 &x &4 \\ 6 &-1 &0 \end{vmatrix}$$+2.\begin{vmatrix} 3&-1 &2 \\ 5 &2 &4 \\ 6 &1 &0 \end{vmatrix}$$-1.\begin{vmatrix} 3&-1 &0 \\ 5 &2 &x \\ 6 &1 &-1 \end{vmatrix}$

$|A|=0-4.\begin{vmatrix} 3&0 &2 \\ 5 &x &4 \\ 6 &-1 &0 \end{vmatrix}$$+2.\begin{vmatrix} 3&-1 &2 \\ 5 &2 &4 \\ 6 &1 &0 \end{vmatrix}$$-1.\begin{vmatrix} 3&-1 &0 \\ 5 &2 &x \\ 6 &1 &-1 \end{vmatrix}$

$|A|=-4.\begin{vmatrix} 3&0 &2 \\ 5 &x &4 \\ 6 &-1 &0 \end{vmatrix}$$+2.\begin{vmatrix} 3&-1 &2 \\ 5 &2 &4 \\ 6 &1 &0 \end{vmatrix}$$-1.\begin{vmatrix} 3&-1 &0 \\ 5 &2 &x \\ 6 &1 &-1 \end{vmatrix}$

Now,$|A|=-4|A_{1}|+2|A_{2}|-1|A_{3}|$

        $|A|=-4|A_{1}|+2|A_{2}|-|A_{3}|$--------->$(1)$

Where $|A_{1}|=\begin{vmatrix} 3&0 &2 \\ 5 &x &4 \\ 6 &-1 &0 \end{vmatrix}=3(0+4)-0(0-24)+2(-5-6x)=2-12x$

            $|A_{2}|=\begin{vmatrix} 3&-1 &2 \\ 5 &2 &4 \\ 6 &1 &0 \end{vmatrix}=3(0-4)+1(0-24)+2(5-12)=-12-24-14=-50$

               $|A_{3}|=\begin{vmatrix} 3&-1 &0 \\ 5 &2 &x \\ 6 &1 &-1 \end{vmatrix}=3(-2-x)+1(-5-6x)=-6-3x-5-6x=-9x-11$

Put these values in equation $(1),$

Now$,|A|=-4(2-12x)+2(-50)-(-9x-11)$

Given that $|A|=245$

$\Rightarrow -8+48x-100+9x+11=245$

$\Rightarrow 57x-100+3=245$

$\Rightarrow 57x-97=245$

$\Rightarrow 57x=245+97$

$\Rightarrow 57x=342$

$\Rightarrow x=\frac{342}{57}$

$\Rightarrow x=6$

So$,x=6$ is the correct answer.

But this is a time-consuming method. Try, to do some operation in rows or columns and make some rows and columns $0$, then it will be easy.   

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yes, I was wrong

thanks :)

Answer 1

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.