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Find values of a and b if A = B, where

\begin{array}{l} A=\left[\begin{array}{cc} a+4 & 3 b \\ 8 & -6 \end{array}\right] \\ B=\left[\begin{array}{cc} 2 a+2 & b^{2}+2 \\ 8 & b^{2}-5 b \end{array}\right] \end{array}

 

Answers (1)

We have the matrices A and B, where

\begin{array}{l} A=\left[\begin{array}{cc} a+4 & 3 b \\ 8 & -6 \end{array}\right] \\ B=\left[\begin{array}{cc} 2 a+2 & b^{2}+2 \\ 8 & b^{2}-5 b \end{array}\right] \end{array}

We need to find the values of a and b.

We know that, if

\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]=\left[\begin{array}{ll} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array}\right]

Then, 

\\a\textsubscript{11} = b\textsubscript{11} \\a\textsubscript{12} = b\textsubscript{12} \\a\textsubscript{21} = b\textsubscript{21} \\a\textsubscript{22} = b\textsubscript{22}

Also, A = B.

\Rightarrow\left[\begin{array}{cc} \mathrm{a}+4 & 3 \mathrm{~b} \\ 8 & -6 \end{array}\right]=\left[\begin{array}{cc} 2 \mathrm{a}+2 & \mathrm{~b}^{2}+2 \\ 8 & \mathrm{~b}^{2}-5 \mathrm{~b} \end{array}\right]

This means,

 \\a + 4 = 2a + 2 \ldots (i) \\3b = b\textsuperscript{2} + 2 \ldots (ii) \\ 8 = 8 \\ -6 = b^2 -5b \ldots (iii)

From equation (i), we can find the value of a.

 \\a + 4 = 2a + 2 \\ \Rightarrow 2a - a = 4 - 2 \\$ \Rightarrow $ a = 2

From equation (ii), we can find the value of b\textsuperscript{2}.

\\3b = b\textsuperscript{2} + 2 \\$ \Rightarrow $ b\textsuperscript{2}= 3b -2

By substituting the value of b\textsuperscript{2} in equation (iii), we get

 \\-6 = b\textsuperscript{2} - 5b \\$ \Rightarrow $ -6 = (3b - 2) - 5b \\$ \Rightarrow $ -6 = 3b - 2 - 5b \\$ \Rightarrow $ -6 = 3b - 5b - 2 \\$ \Rightarrow $ -6 = -2b - 2 \\$ \Rightarrow $ 2b = 6 - 2 \\$ \Rightarrow $ 2b = 4

\begin{aligned} &\Rightarrow \mathrm{b}=\frac{4}{2}\\ &\Rightarrow b=2\\ &\text { Hence, } a=2 \text { and } b=2 \end{aligned}

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infoexpert22

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