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

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?

• A $3$ by $3$ matrix can never equal a $2$ by $2$ matrix. Are you sure the problem isn't asking about the determinants being equal? – Carl Schildkraut Sep 19 '16 at 4:00
• Hmm...I'm not sure if the style of brackets mean "determinant" because in my experience the brackets should be replaced by straight lines to mean determinant. On the other hand, it doesn't make any sense to say a $3\times 3$ matrix equals a $2 \times 2$ matrix. It's like saying a shoe is the the same thing as a balloon. – layman Sep 19 '16 at 4:09
• No, many different matrices have equal determinants. And a 3x3 matrix can never equal a 2x2 matrix. I suspect the question is asking which value of x make the two determinates equal. – fleablood Sep 19 '16 at 4:09
• @StopReadingThisUsername I can only assume that this means that the determinants must be equal - the answer for that is one of the possibilities and I can't think of any other way it can be interpreted. There are no values of $x$ that could possibly make a $3\times 3$ matrix equal to a $2\times 2$ matrix. – Carl Schildkraut Sep 19 '16 at 4:09
• Whoever wrote this question was incompetent. (Not the OP; I mean the instructor or textbook author or whatever.) – R.. Sep 19 '16 at 17:12

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.

• $D$ is obviously wrong, because $\sqrt5$ is not rational. (This is not meant to criticise your answer.) – Carsten S Sep 19 '16 at 11:24
• @CarstenS: really? :D – image Sep 19 '16 at 14:00
• @Marcel, I am just annoyed by the sloppy formulation of the exercise. – Carsten S Sep 19 '16 at 14:52
• @CarstenS It is obviously an approximation. Don't be obnoxious. – jpmc26 Sep 19 '16 at 18:14

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.

• On the other hand vertical lines (as delimiters) can be used to convey the determinant rather than the matrix itself, as we might write $$\left| \begin{array}{ccc} 2 & -1 &4\\3 & 0 & 5\\4 & 1 & 6\end{array} \right|=\left| \begin{array}{cc} x & 4\\5 & x\end{array} \right|$$ – hardmath Sep 19 '16 at 7:05

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.