Some help with matrices please

[pHlawless]

My professor gave us this problem and said its a trick question. I think I have an answer but don't want to submit it in case I am totally off. Anyone mind double checking this for me to make sure I'm not way off base?

I think the answer is the following:
Z = 2
X = 4
y = 3Am I way off base? Seemed too easy lolThanks

 

[MarkFL]

If the problem had read:

\(\displaystyle \left[\begin{array}{c}1 & 2 \\ x & 5 \\ x+y & 8 \end{array}\right]=\left[\begin{array}{c}1 & z \\ 3 & 5 \\ 7 & 8 \end{array}\right]\)

or:

\(\displaystyle \left[\begin{array}{c}1 & 2 \\ x & 6 \\ x+y & 8 \end{array}\right]=\left[\begin{array}{c}1 & z \\ 3 & 6 \\ 7 & 8 \end{array}\right]\)

then I would say:

\(\displaystyle (x,y,z)=(3,4,2)\)

However, the circled elements changes things:

\(\displaystyle \left[\begin{array}{c}1 & 2 \\ x & \enclose{circle}[mathcolor="red"]{\color{black}{5}} \\ x+y & 8 \end{array}\right]=\left[\begin{array}{c}1 & z \\ 3 & \enclose{circle}[mathcolor="red"]{\color{black}{6}} \\ 7 & 8 \end{array}\right]\)

 

[pHlawless]

Okay, so because of that does that mean there is no possible solutions?

 

[MarkFL]

Okay, so because of that does that mean there is no possible solutions?

Yes, I would be inclined to say there is no solution, however, I am no expert in linear algebra, and you may want to hear from someone who is before drawing any conclusions yet. :D

 

[Jameson]

I agree with Mark. Two matrices are equal if and only if they have the same dimensions $(m \times n$ for example) and the same entries in all positions. My guess would be that this is a typo and that Mark's solution is the one that is intended, but it would be good to check with your professor.