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ICSE Class X Math 2015 Paper

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Question : 46 of 52
Marks: +1, -0
If A(3724),B=(0253)A\begin{pmatrix} 3 & 7 \\ 2 & 4 \end{pmatrix}, B=\begin{pmatrix} 0 & 2 \\ 5 & 3 \end{pmatrix} and C=(1−5−46)C=\begin{pmatrix} 1 & -5 \\ -4 & 6 \end{pmatrix}
Find AB−5CAB-5C .
Solution:  
   Given :   \;\text{ Given : }\; A  =(3724),B=(0253),CA\;=\begin{pmatrix} 3 & 7 \\ 2 & 4 \end{pmatrix}, B=\begin{pmatrix} 0 & 2 \\ 5 & 3 \end{pmatrix}, C =(1−5−46)=\begin{pmatrix} 1 & -5 \\ -4 & 6 \end{pmatrix}
AB  =(3724)(0253)AB\;=\begin{pmatrix} 3 & 7 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} 0 & 2 \\ 5 & 3 \end{pmatrix}
  =(0+356+210+204+12)=(35272016)\;=\begin{pmatrix} 0+35 & 6+21 \\ 0+20 & 4+12 \end{pmatrix}=\begin{pmatrix} 35 & 27 \\ 20 & 16 \end{pmatrix}
5C  =5(1−5−46)5C\;=5\begin{pmatrix} 1 & -5 \\ -4 & 6 \end{pmatrix}
  =(5−25−2030)\;=\begin{pmatrix} 5 & -25 \\ -20 & 30 \end{pmatrix}
AB−5C  =(35272016)−(5−25−2030)AB-5C\;=\begin{pmatrix} 35 & 27 \\ 20 & 16 \end{pmatrix}-\begin{pmatrix} 5 & -25 \\ -20 & 30 \end{pmatrix}
=(305240−14)=\begin{pmatrix} 30 & 52 \\ 40 & -14 \end{pmatrix}
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