Does the same determinant make two matrices equal to each other?

Question

Does the same determinant make two matrices equal to each other?

If I have:

Find all values of $x$ that make

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

Would I calculate and equate the determinants of both matrices to solve this problem?

Edit: Below is the exact question. Do the style of brackets refer to the determinants?

Answer 1

If the problem is about an equality of the determinant, all you have to do is compute the determinants separately. The determinant of the $3\times 3$ matrix is $$ (2)(0)(6) + (-1)(5)(4) + (4)(3)(1) - (4)(0)(4) - (1)(5)(2) - (6)(3)(-1) = 0 - 20 + 12 - 0 - 10 + 18 = 0. $$ The $2\times 2$ determinant is just $x^{2} - 20$. Then, we arrive at the equation $$ 0 = x^{2} - 20 $$ which has two possible solutions: $x=\sqrt{20}$ or $x=-\sqrt{20}$. Thus, the answer is (D) if the question refers to determinants.

If not, then there is no solution.

Answer 2

That style of brackets usually refers to the matrix itself, rather than the determinant. The question either uses an unusual style, or is in error.

Answer 3

It is customary to use square brackets $[\ddots]$ to refer to the matrix as a matrix. It is also customary to use vertical bars $|\ddots|$ to refer to the determinant of the matrix.

It is not customary to use ordinary brackets $(\ddots)$ for anything.

However, mathematicians are fond of making their own notation. So, look in your text book and/or your lecture notes to see how they define this.

Matrices are equal to each other only if they are the same size and every member is equal to the member in the same place. So, in this case there are no solutions.

Determinants are just numbers and can be equal even if the matrices are not.

Since schools like to give exercises which actually have solutions, they are most likely to be talking about determinants, but it can possibly be a trick question.