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Home  >>  CBSE XII  >>  Math  >>  Matrices
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If X and Y are $2\times 2$matrices,then solve the following matrix equation for X and Y\[2X+3Y=\begin{bmatrix}2 & 3\\4 & 0\end{bmatrix},3X+2Y=\begin{bmatrix}-2 & 2\\1 & -5\end{bmatrix}.\]

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Toolbox:
  • The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
Step1:
Given:
$2X+3Y=\begin{bmatrix}2 & 3\\4 & 0\end{bmatrix}$-------(1)
$3X+2Y=\begin{bmatrix}-2 & 2\\1 & -5\end{bmatrix}$--------(2)
Multiply Equation (1) by 2 & Equation (2) by 3
Multiplying equation (1) by 2 on both side we get
$2(2X+3Y)=2\begin{bmatrix}2 & 3\\4 & 0\end{bmatrix}$
$4X+6Y=\begin{bmatrix}4 & 6\\8 & 0\end{bmatrix}$------(3)
Multiplying equation (2) by 3 on both side we get
$3(3X+2Y)=3\begin{bmatrix}-2 & 2\\1 & -5\end{bmatrix}$
$9X+6Y=\begin{bmatrix}-6 &6\\3 & -15\end{bmatrix}$------(4)
Subtracting (4) from (3)
$4X+6Y=\begin{bmatrix}4 & 6\\8 & 0\end{bmatrix}$
$9X+6Y=\begin{bmatrix}-6 &6\\3 & -15\end{bmatrix}$
______________________________
$-5X=-\begin{bmatrix}-6 &6\\3 & -15\end{bmatrix}+\begin{bmatrix}4 & 6\\8 & 0\end{bmatrix}$
$-5X=\begin{bmatrix}4 &6\\8 & 0\end{bmatrix}+(-1)\begin{bmatrix}-6 & 6\\3 & -15\end{bmatrix}$
$-5X=\begin{bmatrix}4 &6\\8 & 0\end{bmatrix}+\begin{bmatrix}6 &- 6\\-3 & 15\end{bmatrix}$
$-5X=\begin{bmatrix}4+6 &6-6\\8-3 & 0+15\end{bmatrix}$
$-5X=\begin{bmatrix}10 &0\\5 & 15\end{bmatrix}$
$X=-1/5\begin{bmatrix}10 &0\\5 & 15\end{bmatrix}$
$X=\begin{bmatrix}-2&0\\-1 & -3\end{bmatrix}$
Step2:
Substitute the value of x in equation(1)
$2X+3Y=\begin{bmatrix}2 & 3\\4 & 0\end{bmatrix}$
$2\begin{bmatrix}-2 & 0\\-1 & -3\end{bmatrix}+3Y=\begin{bmatrix}2 & 3\\4 & 0\end{bmatrix}$
$\begin{bmatrix}-4 & 0\\-2 & -6\end{bmatrix}+3Y=\begin{bmatrix}2 & 3\\4 & 0\end{bmatrix}$
$3Y=\begin{bmatrix}2 & 3\\4 & 0\end{bmatrix}+(-1)\begin{bmatrix}-4 & 0\\-2 & -6\end{bmatrix}$
$3Y=\begin{bmatrix}2 & 3\\4 & 0\end{bmatrix}+\begin{bmatrix}4 & 0\\2 & 6\end{bmatrix}$
$3Y=\begin{bmatrix}2+4 & 3+0\\4+2 & 0+6\end{bmatrix}$
$3Y=\begin{bmatrix}6 & 3\\6 & 6\end{bmatrix}$
$Y=1/3\begin{bmatrix}6 & 3\\6 & 6\end{bmatrix}$
$Y=\begin{bmatrix}2 & 1\\2 & 2\end{bmatrix}$
answered Mar 23, 2013 by sharmaaparna1
 

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