- Home
- Standard 12
- Mathematics
જો $\mathrm{X}+\mathrm{Y}=\left[\begin{array}{ll}5 & 2 \\ 0 & 9\end{array}\right]$ અને $\mathrm{X}-\mathrm{Y}=\left[\begin{array}{cc}3 & 6 \\ 0 & -1\end{array}\right]$ હોય, તો $X$ અને $Y$ શોધો.
$X=$ $\left[\begin{array}{ll}4 & 4 \\ 0 & 4\end{array}\right]$, $Y=$ $\left[\begin{array}{rr}1 & -2 \\ 0 & 5\end{array}\right]$
$X=$ $\left[\begin{array}{ll}4 & 4 \\ 0 & 4\end{array}\right]$, $Y=$ $\left[\begin{array}{rr}1 & -2 \\ 0 & 5\end{array}\right]$
$X=$ $\left[\begin{array}{ll}4 & 4 \\ 0 & 4\end{array}\right]$, $Y=$ $\left[\begin{array}{rr}1 & -2 \\ 0 & 5\end{array}\right]$
$X=$ $\left[\begin{array}{ll}4 & 4 \\ 0 & 4\end{array}\right]$, $Y=$ $\left[\begin{array}{rr}1 & -2 \\ 0 & 5\end{array}\right]$
Solution
We have $({\text{X}} + {\text{Y}}) + ({\text{X}} – {\text{Y}}) = $ $\left[ {\begin{array}{*{20}{l}}
5&2 \\
0&9
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
3&6 \\
0&{ – 1}
\end{array}} \right]$
or $(X + X) + (Y – Y) = $ $\left[ {\begin{array}{*{20}{l}}
8&8 \\
0&8
\end{array}} \right] \Rightarrow 2X = $ $\left[ {\begin{array}{*{20}{l}}
8&8 \\
0&8
\end{array}} \right]$
or $X = \frac{1}{2}\left[ {\begin{array}{*{20}{l}}
8&8 \\
0&8
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
4&4 \\
0&4
\end{array}} \right]$
Also $(X + Y) – (X – Y) = $ $\left[ {\begin{array}{*{20}{c}}
5&2 \\
0&9
\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}
3&6 \\
0&{ – 1}
\end{array}} \right]$
or $(X – X) + (Y + Y) = $ $\left[ {\begin{array}{*{20}{c}}
{5 – 3}&{2 – 6} \\
0&{9 + 1}
\end{array}} \right] \Rightarrow $ $2{\text{Y}} = \left[ {\begin{array}{*{20}{c}}
2&{ – 4} \\
0&{10}
\end{array}} \right]$
or $Y = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
2&{ – 4} \\
0&{10}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{ – 2} \\
0&5
\end{array}} \right]$
