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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Find $X$ and $Y$ if $$ \begin{array}{l \qquad l} (i) \quad X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} \text{ and } X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \\ (ii) \quad 2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \text{ and } 3X + 2Y = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix} \end{array} $$

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On addition:
(i)$(x+y)+(x-y)=\begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}+\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} $
$\Rightarrow x+y+x-y=\begin{bmatrix} 7+3 & 0+0 \\ 0+2 & 5+3 \end{bmatrix}$
$2x=\begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix}$
$x=\frac{1}{2}\begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} \frac{10}{2} & \frac{0}{2} \\ \frac{2}{2} &\frac{ 8}{2} \end{bmatrix}$
$x=\begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$
On subtraction:
$(x+y)-(x-y)=\begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} $
$(x+y)+(-1)(x-y)=\begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}+(-1)\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} $
$(x+y)+(x-y)=\begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}+\begin{bmatrix} -3 & 0 \\ 0 & -3 \end{bmatrix} $
$x+y-x+y=\begin{bmatrix}7-3 & 0+0\\2+0 & 5-3\end{bmatrix}$
$2y=\begin{bmatrix}4 & 0\\2& 2\end{bmatrix}$
$y=\begin{bmatrix}\frac{4}{2} & 0\\\frac{2}{2} & \frac{2}{2}\end{bmatrix}$
$y=\begin{bmatrix}2 & 0\\1& 1\end{bmatrix}$
(ii)Let us take:
$ 2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}$
Multiply both sides by 2.
$ 2(2X + 3Y) = 2\begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}$
$ 4X + 6Y = \begin{bmatrix} 2\times 2 & 3\times 2 \\ 4\times 2 & 0\times 2 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 4 & 6 \\ 8 & 0 \end{bmatrix}$-----(1)
Consider
$3x+2y=\begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}$
Multiply both sides by 3
$3(3x+2y)=3\begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}$
$9x+6y=\begin{bmatrix} 3\times 2 & 3\times -2 \\ 3\times -1 &3\times 5 \end{bmatrix}$
$\;\;\;=\begin{bmatrix} 6 & -6 \\ -3 & 15 \end{bmatrix}$-----(2)
Subtract the eq(1) from eq(2)
$9x+6y-(4x+6y)=\begin{bmatrix} 6 & -6 \\ -3 & 15 \end{bmatrix}+(-1)\begin{bmatrix} 4 & 6 \\ -8 & 0 \end{bmatrix}$
$9x+6y-4x-6y)=\begin{bmatrix} 6 & -6 \\ -3 & 15 \end{bmatrix}+(-1)\begin{bmatrix}- 4 & -6 \\ -8 & 0 \end{bmatrix}$
$5x=\begin{bmatrix} 6-4 & -6-6 \\ -3-8 & 15+0 \end{bmatrix}$
$5x=\begin{bmatrix} 2 & -12 \\ -11 & 15 \end{bmatrix}$
$x=\frac{1}{5}\begin{bmatrix} 2 & -12 \\ -11 & 15 \end{bmatrix}$
$x=\begin{bmatrix} \frac{2}{5} & \frac{-12}{5} \\ \frac{-11}{5} & \frac{15}{5} \end{bmatrix}$
$\Rightarrow \begin{bmatrix} \frac{2}{5} & \frac{-12}{5} \\ \frac{-11}{5} & 3 \end{bmatrix}$
Consider
$2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}$
Multiply both sides by 3
$3(2X + 3Y) =3 \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}$
$6X + 9Y = \begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix}$------(3)
Consider $3X + 2Y = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}$
Multiply both sides by 2
$2(3X + 2Y) = 2\begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}$
$6X + 4Y = \begin{bmatrix} 4 & -4 \\ -2 & 10 \end{bmatrix}$-----(4)
Subtract(4)-(5)
$6X + 9Y-(6x+4y) = \begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix}+(-1)\begin{bmatrix} 4 & -4 \\ -2 & 10 \end{bmatrix}$
$6x+9y-6x-4y=\begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix}+(-1)\begin{bmatrix}- 4 & 4 \\ 2 & -10 \end{bmatrix}$
$5y=\begin{bmatrix} 6-4 & 9+4 \\ 12+2 & 0-10 \end{bmatrix}$
$5y=\begin{bmatrix} 2 & 13 \\ 14 & -10 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & \frac{-10}{5} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix}$
answered Feb 13, 2013 by sreemathi.v
 

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