How to solve simultaneously for matrices X and Y the equations?

Sep 2011
Mumbai (Bombay),Maharashtra State,India
How to solve simultaneously for Matrices X and Y the equations
$\begin{array}{rcl} 3(X+2Y)+2(2X+3Y)+\begin{pmatrix}-4&2\\5&\frac12\\0&-1\end{pmatrix}=0\end{array}$ where 0 denotes $3\times2$ zero matrix.

Solution:- I tried to solve this question but I could not get correct answer.


MHF Helper
Apr 2005
Are you asking "how can I look at this problem and immediately know the solution without doing any calculation at all?". I can't answer that- I would have to work it out!

The first thing I would do is simplify the left side of each equation. Calling the two matrices on the right side "A" and "B" (just to simplify the writing) we have \(\displaystyle 2X+ \frac{3}{2}X- 2Y+ Y= \frac{7}{2}X- Y= A\) and \(\displaystyle 3X+ 4X+ 6Y+ 6Y= 7X+ 12Y= B\).

Now solve for X and Y just as you would such equation with number. Seeing "-Y" in one equation and "12Y" in the other, I would be inclined to add 12 times the first equation to the second: \(\displaystyle 12(\frac{7}{2}X- Y)+ (7X+ 12Y)= 42X- 12Y+ 7X+ 12Y= 49X= 12A+ B\) so that \(\displaystyle X= \frac{1}{49}(12A+ B)\).

Now \(\displaystyle A= \begin{pmatrix}-2 & 5 \\-3 & 6 \\ 0 & 2 \end{pmatrix}\) and \(\displaystyle B= \begin{pmatrix}4 & -2 \\ -5 & -\frac{1}{2} \\ 0 & 1\end{pmatrix}\) so \(\displaystyle 12A+ B= \begin{pmatrix}-20 & 58 \\ -41 & \frac{143}{2} \\ 0 & 25\end{pmatrix}\). \(\displaystyle X= \begin{pmatrix}-\frac{20}{49} & \frac{58}{49}\\ -\frac{41}{49} & \frac{143}{98} \\ 0 & \frac{25}{49}\end{pmatrix}\)

Of course, once you know X, \(\displaystyle Y= \frac{7}{2}X- A\).
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