Solveeit Logo
Login

Question

Mathematics Question on Matrices

Find X and Y,if (i)X+Y=[70\25](i)X+Y=\begin{bmatrix}7&0\\\2&5\end{bmatrix} and X-Y=\begin{bmatrix}3&0\\\0&3\end{bmatrix}$$(ii)2X+3Y=\begin{bmatrix}2&3\\\4&0\end{bmatrix}and 3X+2Y=[2−2−15]3X+2Y=\begin{bmatrix}2&-2\\\\-1&5\end{bmatrix}

Answer

(i)X+Y=[70\25]−(1)(i)X+Y=\begin{bmatrix}7&0\\\2&5\end{bmatrix}-(1)
X−Y=[30\03]−(2)X-Y=\begin{bmatrix}3&0\\\0&3\end{bmatrix}-(2)
Adding equations (1) and (2), we get:
2X=\begin{bmatrix}7&0\\\2&5\end{bmatrix}+\begin{bmatrix}3&0\\\0&3\end{bmatrix}$$=\begin{bmatrix}7+3& 0+0\\\ 2+0& 5+3\end{bmatrix}$$=\begin{bmatrix}10& 0\\\ 2& 8\end{bmatrix}
X=12[100 28]=[50\14]X=\frac{1}{2}\begin{bmatrix}10& 0\\\ 2& 8\end{bmatrix}=\begin{bmatrix}5&0\\\1&4\end{bmatrix}
Now X+Y=[70\25]X+Y=\begin{bmatrix}7&0\\\2&5\end{bmatrix}
  ⟹  [50\14]+Y=[70\25]\implies\begin{bmatrix}5&0\\\1&4\end{bmatrix}+Y=\begin{bmatrix}7&0\\\2&5\end{bmatrix}
  ⟹  Y=[70\25]−[50\14]\implies Y=\begin{bmatrix}7&0\\\2&5\end{bmatrix}-\begin{bmatrix}5&0\\\1&4\end{bmatrix}
  ⟹  Y=[7−50−0 2−15−4]\implies Y=\begin{bmatrix}7-5& 0-0\\\ 2-1& 5-4\end{bmatrix}
∴Y=[20\11]\therefore Y=\begin{bmatrix}2&0\\\1&1\end{bmatrix}
(ii)2X+3Y=[23\40]−(3)(ii)2X+3Y=\begin{bmatrix}2&3\\\4&0\end{bmatrix}-(3)
3X+2Y=[2−2−15]−(4)3X+2Y=\begin{bmatrix}2&-2\\\\-1&5\end{bmatrix}-(4)
Multiplying equation (3) with (2), we get:
2(2X+3Y)=2[23\40]2(2X+3Y)=2\begin{bmatrix}2&3\\\4&0\end{bmatrix}
  ⟹  4X+6Y=[46\80]....(5)\implies 4X+6Y=\begin{bmatrix}4&6\\\8&0\end{bmatrix}....(5)
Multiplying equation (4) with (3), we get:
3(3X+2Y)=3[2−2−15]3(3X+2Y)=3\begin{bmatrix}2&-2\\\\-1&5\end{bmatrix}
  ⟹  9X+6Y=[6−6−315]....(6)\implies 9X+6Y=\begin{bmatrix}6&-6\\\\-3&15\end{bmatrix}....(6)
From (5) and (6), we have:
(4X+6Y)−(9X+6Y)(4X+6Y)-(9X+6Y)
=[46\80]−[6−6−315]=\begin{bmatrix}4&6\\\8&0\end{bmatrix}-\begin{bmatrix}6&-6\\\\-3&15\end{bmatrix}
  ⟹  −5X=[4−66−(−6) 8−(−3)0−15]=[−212 11−15]\implies-5X=\begin{bmatrix}4-6& 6-(-6)\\\ 8-(-3)& 0-15\end{bmatrix}=\begin{bmatrix}-2& 12\\\ 11& -15\end{bmatrix}
∴X=−15[−212 11−15]=[25−125 −1153]\therefore X=\frac{-1}{5}\begin{bmatrix}-2& 12\\\ 11& -15\end{bmatrix}=\begin{bmatrix}\frac{2}{5}& \frac{-12}{5}\\\ \frac{-11}{5}& 3\end{bmatrix}
Now,2X+3Y=[23\40]2X+3Y=\begin{bmatrix}2&3\\\4&0\end{bmatrix}
  ⟹  2[25−125 −1153]+3Y=[23\40]\implies 2\begin{bmatrix}\frac{2}{5}& \frac{-12}{5}\\\ \frac{-11}{5}& 3\end{bmatrix}+3Y=\begin{bmatrix}2&3\\\4&0\end{bmatrix}
  ⟹  [45−245 −2256]+3Y=[23\40]\implies \begin{bmatrix}\frac{4}{5}& \frac{-24}{5}\\\ \frac{-22}{5}& 6\end{bmatrix}+3Y=\begin{bmatrix}2&3\\\4&0\end{bmatrix}
  ⟹  3Y=[23\40]−[45−245 −2256]\implies 3Y=\begin{bmatrix}2&3\\\4&0\end{bmatrix}-\begin{bmatrix}\frac{4}{5}& \frac{-24}{5}\\\ \frac{-22}{5}& 6\end{bmatrix}

  ⟹  3Y=[2−453+245 4+2250−6]=[65395 425−6]\implies 3Y=\begin{bmatrix}2-\frac{4}{5}& 3+\frac{24}{5}\\\ 4+\frac{22}{5}& 0-6\end{bmatrix}=\begin{bmatrix}\frac{6}{5}& \frac{39}{5}\\\ \frac{42}{5}& -6\end{bmatrix}

∴Y=13[65395 425−6]==[25135 145−2]\therefore Y=\frac{1}{3}\begin{bmatrix}\frac{6}{5}&\frac{ 39}{5}\\\ \frac{42}{5}&-6\end{bmatrix}==\begin{bmatrix}\frac{2}{5}& \frac{13}{5}\\\ \frac{14}{5}& -2\end{bmatrix}