Solve for $x$ where $$ \det\left(\begin{bmatrix} x & -1 \\ 3 & 1-x \\ \end{bmatrix}\right) =\det\left(\begin{bmatrix} 1&0&-3 \\ 2&x&-6 \\ 1&3&x-5 \\ \end{bmatrix}\right) $$

My book doesn't explain this type of problem. Please be descriptive.


closed as unclear what you're asking by Jean Marie, user223391, C. Falcon, JonMark Perry, Juniven Mar 30 '17 at 4:18

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  • $\begingroup$ I think there is a typo, e.g. the left matrix sits in a different ring to the right matrix, they literally can't be equal $\endgroup$ – mdave16 Mar 29 '17 at 22:29
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    $\begingroup$ Two matricies with different dimensions cannot be equal. Your question is wrong $\endgroup$ – K Split X Mar 29 '17 at 22:30
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    $\begingroup$ If it's meant to be that the determinants are equal, then it could be done. $\endgroup$ – ConMan Mar 29 '17 at 22:33
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    $\begingroup$ Even without the skills to type well and use nothing but the best choices for notation, OP should be aware of the differences in the meaning of the notation and understand why people were so confused and avoid using misleading notation. That is as bad as trying to use the symbols $|x|$ to denote the floor of $x$ just because you can't find how to type $\lfloor x\rfloor$... everyone will assume $|x|$ is denoting the absolute value instead of the floor unless they are told otherwise. Without knowing how to type the symbols, using words would be preferred. $\endgroup$ – JMoravitz Mar 29 '17 at 22:51
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    $\begingroup$ I wrote the problem the way the book wrote it so I'm sorry if there is an issue. Upon further inspection my print copy of the book is different from the online version so I was not following the online version either. $\endgroup$ – Sarah S Mar 29 '17 at 23:42

So to solve this question, we need to calculate the determinant of both sides, so we have

$$ \det\left(\begin{bmatrix} x & -1 \\ 3 & 1-x \\ \end{bmatrix}\right) = \det\left(\begin{bmatrix} 1 & 0 & -3 \\ 2 & x & -6 \\ 1 & 3 & x-5 \\ \end{bmatrix}\right) $$

For those you wanting to know how to do this in $\LaTeX$, it's

\det for functions like \tan, \sin, \det ...
\left(, \right) to have it match the size
entries separated by &'s (to indicate end of entry)
and \\'s to indicate a new row
and then an \end{bmatrix}

Now, the left hand side is $(x)(1-x) + 3 = -x^2 + x + 3$.

Recall that we can compute determinants by expanding across rows and columns, with $-1$'s where appropriate and the determinant of the minors.

The right hand side is (expanding from top row)

$$ 1\times \det\left( \begin{bmatrix}x & -6 \\3 & x-5\end{bmatrix}\right) - 0 \times \det\left( \begin{bmatrix}2 & 1 \\3 & x-5\end{bmatrix}\right) + (-3) \times \det\left( \begin{bmatrix}2 & 1 \\x & 3\end{bmatrix}\right)$$

From here we can do as above, \begin{align} &= 1 \times ((x)(x-5) +18) + (-3) \times (6 - x)\\ &= x^2 - 5x + 18 - 18 + 3x\\ &= x^2 -2x \end{align} So looks to me as if you have a quadratic to solve, I'll let you take the rest.

  • $\begingroup$ to the person who downvoted me, i'd love to know why, just comment with what i did wrong $\endgroup$ – mdave16 Mar 29 '17 at 23:59
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    $\begingroup$ SWelcome to the club of victims of mysterious downvotes ! $\to +1$ for your good answer. $\endgroup$ – Claude Leibovici Mar 30 '17 at 4:02

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