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If $\begin{pmatrix} a+4 &3b \\ 8 & -6 \end{pmatrix} = \begin{pmatrix} 2a+2 &b+2 \\ 8 & a-8b \end{pmatrix}$, write the value of $a-2b$.

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If (a+43b8−6)=(2a+2b+28a−8b)" role="presentation" style="position: relative;">(a+483b6)=(2a+28b+2a8b)(a+43b8−6)=(2a+2b+28a−8b)\begin{pmatrix} a+4 &3b \\ 8 & -6 \end{pmatrix} = \begin{pmatrix} 2a+2 &b+2 \\ 8 & a-8b \end{pmatrix}:
⇒a+4=2a+2→a=2" role="presentation" style="position: relative;">a+4=2a+2a=2⇒a+4=2a+2→a=2\Rightarrow a + 4 = 2a + 2 \rightarrow a = 2
⇒3b=b+2→b=1" role="presentation" style="position: relative;">3b=b+2b=1⇒3b=b+2→b=1\Rightarrow 3b = b+2 \rightarrow b = 1
⇒a−2b=2−2×1=0" role="presentation" style="position: relative;">a2b=22×1=0⇒a−2b=2−2×1=0\Rightarrow a - 2b = 2 - 2 \times 1 = 0
answered Mar 21, 2014 by balaji.thirumalai
edited Dec 18, 2017 by meena.p
 

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