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ICSE Class X Math 2013 Paper

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Question : 1 of 46
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SECTION - I
Given A=[2−620],B=[−3240],A=\begin{bmatrix}2 & -6 \\ 2 & 0\end{bmatrix}, B=\begin{bmatrix}-3 & 2 \\ 4 & 0\end{bmatrix}, C=[4002]C=\begin{bmatrix}4 & 0 \\ 0 & 2\end{bmatrix} .
Find the matrix XX such that A+2X=2B+A+2X=2B+ C.
Solution:  
Given A=[2−620],B=[−3240],A=\begin{bmatrix}2 & -6 \\ 2 & 0\end{bmatrix}, B=\begin{bmatrix}-3 & 2 \\ 4 & 0\end{bmatrix}, C=[4002]C=\begin{bmatrix}4 & 0 \\ 0 & 2\end{bmatrix} .
∵  A+2X=2B+C\because \; A+2X=2B+C
Putting the given values, we get
[2−620]+2X  =2[−3240]\begin{bmatrix}2 & -6 \\ 2 & 0\end{bmatrix}+2X \; =2\begin{bmatrix}-3 & 2 \\ 4 & 0\end{bmatrix} +[4002]+\begin{bmatrix}4 & 0 \\ 0 & 2\end{bmatrix}
2X  =[−6+44+08+00+2]2X \; =\begin{bmatrix}-6+4 & 4+0 \\ 8+0 & 0+2\end{bmatrix} −[2−620]-\begin{bmatrix}2 & -6 \\ 2 & 0\end{bmatrix}
X  =  12[−41062]X \; = \; \frac{1}{2}\begin{bmatrix}-4 & 10 \\ 6 & 2\end{bmatrix}
X  =[−2531]X \; =\begin{bmatrix}-2 & 5 \\ 3 & 1\end{bmatrix}
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