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Question 18 of 37

Q.(f) If A=(2152)A = \begin{pmatrix} 2 & 1 \\ 5 & 2 \end{pmatrix}, B=(1−324)B = \begin{pmatrix} 1 & -3 \\ 2 & 4 \end{pmatrix}, find 3A+2B3A + 2B.

ChseodishaCHSE Odisha Plus Two (Class 12) Commerce Board 2019Subjective· 3mImportance★★★★★est
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3A+2B=(8−31914)3A + 2B = \begin{pmatrix} 8 & -3 \\ 19 & 14 \end{pmatrix}.

First multiply each matrix by its scalar:

3A=3(2152)=(63156),2B=2(1−324)=(2−648).3A = 3\begin{pmatrix} 2 & 1 \\ 5 & 2 \end{pmatrix} = \begin{pmatrix} 6 & 3 \\ 15 & 6 \end{pmatrix}, \qquad 2B = 2\begin{pmatrix} 1 & -3 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 2 & -6 \\ 4 & 8 \end{pmatrix}.

Now add the corresponding elements: …

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