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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES3 Marks
Question
Find X and Y, if:
  1. $\text {X + Y} = \begin{bmatrix}7&0\\2&5\end{bmatrix} \text {and}\ \text{X} - \text{Y} = \begin{bmatrix}3&0\\0&3\end{bmatrix}$
  2. $2\text {X }+ 3\text {Y} = \begin{bmatrix}2&3\\4&0\end{bmatrix} \text {and}\ 3\text{X }+\text{ 2Y} = \begin{bmatrix}-2&-2\\-1&5\end{bmatrix}$
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Answer

  1. $\text{Given: X }+\text{ Y}=\begin{bmatrix}7&0\\2&5\end{bmatrix}...\text{(i)}\ \text{and X} -\text{ Y}=\begin{bmatrix}3&0\\0&3\end{bmatrix}...\text{(ii)}$
Adding eq. (i) and (ii), we get

$2\text {X}=\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} $

$\Rightarrow \text{X}=\frac{1}{2}\cdot\begin{bmatrix}10&0\\2&8\end{bmatrix}=\begin{bmatrix}5&0\\1&4\end{bmatrix}$

Subtracting eq. (i) and (ii), we get

$\text{2Y}=\begin{bmatrix}7&0\\2&5\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}=\begin{bmatrix}7-3&0-0\\2-0&5-3\end{bmatrix}=\begin{bmatrix}4&0\\2&2\end{bmatrix}$

$\text{Y}=\frac{1}{2}\cdot\begin{bmatrix}4&0\\2&2\end{bmatrix}=\begin{bmatrix}2&0\\1&1\end{bmatrix}$
  1. $\text{Given} :2\text{ X} +3\text{Y}=\begin{bmatrix}2&3\\4&0\end{bmatrix}...\text{(iii)}$
$\text{and}\ 3\text{ X} + 2\text{Y}=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}...\text{(iv)}$

Multiplying eq. (iii) by 2, $4\text{X} +6\text{Y}=2\begin{bmatrix}2&3\\4&0\end{bmatrix}=\begin{bmatrix}4&6\\8&0\end{bmatrix}...\text{(v)}$

Multipiying eq. (iv) by 3, $\text{9X + 6Y}=3\begin{bmatrix}2&-2\\-1&5\end{bmatrix}=\begin{bmatrix}6&-6\\-3&15\end{bmatrix}...\text{(vi)}$

Subtract (vi) - Eq. (v) $=\text{5X}=\begin{bmatrix}6&-6\\-3&15\end{bmatrix}-\begin{bmatrix}4&6\\8&0\end{bmatrix}$

$=\begin{bmatrix}6-4&-6-6\\-3-8&15-0\end{bmatrix}=\begin{bmatrix}2&-12\\-11&15\end{bmatrix}$

$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}2&-12\\-11&15\end{bmatrix}=\begin{bmatrix}\frac{2}{5}&-\frac{12}{5}\\-\frac{11}{5}&{3}\end{bmatrix}$

Now, From eq. (iii), $\text{3Y}=\begin{bmatrix}2&3\\4&0\end{bmatrix}-2\text{X}=\begin{bmatrix}2&3\\4&0\end{bmatrix}-2\begin{bmatrix}\frac{2}{5}&-\frac{12}{5}\\-\frac{11}{5}&3\end{bmatrix}$

$\Rightarrow\text{3Y}=\begin{bmatrix}2&3\\4&0\end{bmatrix}-\begin{bmatrix}\frac{4}{5}&-\frac{24}{5}\\-\frac{22}{5}&6\end{bmatrix}$

$=\begin{bmatrix}2-\frac{4}{5}&3+\frac{24}{5}\\4+\frac{22}{5}&0-6\end{bmatrix}=\begin{bmatrix}\frac{6}{5}&\frac{39}{5}\\\frac{42}{5}&-6\end{bmatrix}$

$\Rightarrow\text{Y}=\frac{1}{3}\begin{bmatrix}\frac{6}{5}&\frac{39}{5}\\\frac{42}{5}&-6\end{bmatrix}=\begin{bmatrix}\frac{2}{5}&\frac{13}{5}\\\frac{14}{5}&-2\end{bmatrix}$

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