If 2A+3B=(2 -1 4)3 2 5and A+2B=(5 0 3) 1 6 2find A and B

1.	If 2A+3B=(2 -1 4)3 2 5and A+2B=(5 0 3) 1 6 2find A and B
Answer» Answer:- GIVEN: $\sf 2A + 3B = \begin{bmatrix} \sf <klux>2</klux> & - \sf 1 & \sf 4 \\\\ \sf 3&\sf 2& \sf 5 \end{bmatrix} \:-\: <klux>EQUATION</klux>\: (1)$ And, $\sf A + 2B = \begin{bmatrix} \sf 5& \sf 0& \sf 3 \\\\\sf 1&\sf 6&\sf2 \end{bmatrix} \:-\: equation\: (2)$ MULTIPLY equation (2) by 2 and subtract equation (1) from (2). $\implies \sf \: 2(A + 2B) - (2A + 3B)=2 \begin{bmatrix} \sf 5& \sf \: 0& \sf \: 3\ \\\\ \sf \: 1& \sf \: 6& \sf \: 2 \end{bmatrix} - \begin{bmatrix} \sf 2 & \sf - 1 & \sf \: 4 \\\\ \sf \: 3& \sf \: 2& \sf \: 5 \end{bmatrix} \\ \\ \\ \implies \sf \not{2A}+ 4B \: - \not{2A }- 3B = \begin{bmatrix} \sf 10& \sf \: 0& \sf \: 6\\\\ \sf \: 2& \sf \: 12& \sf \: 4 \end{bmatrix} - \begin{bmatrix} \sf 2 & \sf - 1 & \sf \: 4 \\\\ \sf \: 3& \sf \: 2& \sf \: 5 \end{bmatrix} \\\\ \\ \implies \sf \: B = \begin{bmatrix} \sf 10 - 2 & \sf \: 0 - ( - 1) & \sf \: 6 - 4 \\\\ \sf \: 2 - 3& \sf \: 12 - 2& \sf \:4 - 5 \end{bmatrix} \\ \\ \\ \implies \boxed{\sf \: B = \begin{bmatrix} \sf 8 & \sf \: 1& \sf \: 2 \\\\ \sf \: - 1& \sf \: 10 & \sf \:- 1\end{bmatrix}}$ Substitute the value of B in equation (2). $\: \implies \sf \: A + 2 \begin{bmatrix} \sf 8 & \sf \: 1& \sf \: 2 \\\\ \sf \: - 1& \sf \: 10& \sf \: - 1\end{bmatrix} = \begin{bmatrix} \sf 5& \sf \: 0& \sf \: 3 \\\\ \sf \: 1& \sf \: 6& \sf \: 2 \end{bmatrix} \\\\ \\ \implies \sf \: A + \begin{bmatrix} \sf 16 & \sf \: 2& \sf \: 4 \\\\ \sf \: - 2& \sf \: 20& \sf \: - 2\end{bmatrix} = \begin{bmatrix} \sf 5& \sf \: 0& \sf \: 3 \\\\ \sf \: 1& \sf \: 6& \sf \: 2 \end{bmatrix} \\\\\\\implies \sf \: A = \begin{bmatrix} \sf 5& \sf \: 0& \sf \: 3 \\\\ \sf \: 1& \sf \: 6& \sf \: 2 \end{bmatrix} - \begin{bmatrix} \sf 16 & \sf \: 2& \sf \: 4 \\\\ \sf \: - 2& \sf \: 20& \sf \: - 2\end{bmatrix} \\ \\\\ \implies \sf \: A = \begin{bmatrix} \sf 5 - 16& \sf \: 0 - 2& \sf \: 3 - 4 \\\\\sf \: 1 - ( - 2)& \sf \: 6 - 20& \sf \: 2 - ( - 2)\end{bmatrix} \\\\ \\ \implies \boxed{ \sf \: A = \begin{bmatrix} \sf - 11& \sf \: - 2& \sf \: - 1\\\\ \sf \: 3& \sf \: - 14& \sf \: 4\end{bmatrix}}$

Answer»

Answer:-

$\sf 2A + 3B = \begin{bmatrix} \sf <klux>2</klux> & - \sf 1 & \sf 4 \\\\ \sf 3&\sf 2& \sf 5 \end{bmatrix} \:-\: <klux>EQUATION</klux>\: (1)$

And,

$\sf A + 2B = \begin{bmatrix} \sf 5& \sf 0& \sf 3 \\\\\sf 1&\sf 6&\sf2 \end{bmatrix} \:-\: equation\: (2)$

MULTIPLY equation (2) by 2 and subtract equation (1) from (2).

$\implies \sf \: 2(A + 2B) - (2A + 3B)=2 \begin{bmatrix} \sf 5& \sf \: 0& \sf \: 3\ \\\\ \sf \: 1& \sf \: 6& \sf \: 2 \end{bmatrix} - \begin{bmatrix} \sf 2 & \sf - 1 & \sf \: 4 \\\\ \sf \: 3& \sf \: 2& \sf \: 5 \end{bmatrix} \\ \\ \\ \implies \sf \not{2A}+ 4B \: - \not{2A }- 3B = \begin{bmatrix} \sf 10& \sf \: 0& \sf \: 6\\\\ \sf \: 2& \sf \: 12& \sf \: 4 \end{bmatrix} - \begin{bmatrix} \sf 2 & \sf - 1 & \sf \: 4 \\\\ \sf \: 3& \sf \: 2& \sf \: 5 \end{bmatrix} \\\\ \\ \implies \sf \: B = \begin{bmatrix} \sf 10 - 2 & \sf \: 0 - ( - 1) & \sf \: 6 - 4 \\\\ \sf \: 2 - 3& \sf \: 12 - 2& \sf \:4 - 5 \end{bmatrix} \\ \\ \\ \implies \boxed{\sf \: B = \begin{bmatrix} \sf 8 & \sf \: 1& \sf \: 2 \\\\ \sf \: - 1& \sf \: 10 & \sf \:- 1\end{bmatrix}}$

Substitute the value of B in equation (2).

$\: \implies \sf \: A + 2 \begin{bmatrix} \sf 8 & \sf \: 1& \sf \: 2 \\\\ \sf \: - 1& \sf \: 10& \sf \: - 1\end{bmatrix} = \begin{bmatrix} \sf 5& \sf \: 0& \sf \: 3 \\\\ \sf \: 1& \sf \: 6& \sf \: 2 \end{bmatrix} \\\\ \\ \implies \sf \: A + \begin{bmatrix} \sf 16 & \sf \: 2& \sf \: 4 \\\\ \sf \: - 2& \sf \: 20& \sf \: - 2\end{bmatrix} = \begin{bmatrix} \sf 5& \sf \: 0& \sf \: 3 \\\\ \sf \: 1& \sf \: 6& \sf \: 2 \end{bmatrix} \\\\\\\implies \sf \: A = \begin{bmatrix} \sf 5& \sf \: 0& \sf \: 3 \\\\ \sf \: 1& \sf \: 6& \sf \: 2 \end{bmatrix} - \begin{bmatrix} \sf 16 & \sf \: 2& \sf \: 4 \\\\ \sf \: - 2& \sf \: 20& \sf \: - 2\end{bmatrix} \\ \\\\ \implies \sf \: A = \begin{bmatrix} \sf 5 - 16& \sf \: 0 - 2& \sf \: 3 - 4 \\\\\sf \: 1 - ( - 2)& \sf \: 6 - 20& \sf \: 2 - ( - 2)\end{bmatrix} \\\\ \\ \implies \boxed{ \sf \: A = \begin{bmatrix} \sf - 11& \sf \: - 2& \sf \: - 1\\\\ \sf \: 3& \sf \: - 14& \sf \: 4\end{bmatrix}}$