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.
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.
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
\begin{bmatrix}
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.