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Question 43 of 52

Q.If 2A^T + B = (2 5; 10 2) and 2B^T + A = (1 8; 4 1), find the value of matrix A.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2024Subjective· 2mImportance★★★★★
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Transpose the first equation and combine it with the second to eliminate BTB^T, then solve for AA.

Let M1=(25102)M_1 = \begin{pmatrix}2&5\\10&2\end{pmatrix} and M2=(1841)M_2=\begin{pmatrix}1&8\\4&1\end{pmatrix}.

Given: 2AT+B=M12A^T+B = M_1 ... (i), and 2BT+A=M22B^T+A=M_2 ... (ii).

Transpose (i) (using (AT)T=A(A^T)^T=A): 2A+BT=M1T2A+B^T = M_1^T ... (iii), where M1T=(21052)M_1^T = \begin{pmatrix}2&10\\5&2\end{pmatrix}.

From (ii): BT=M2−A2B^T = \dfrac{M_2-A}{2}. Substitute into (iii):

2A+M2−A2=M1T2A + \frac{M_2-A}{2} = M_1^T

Multiply through by 22: 4A+M2−A=2M1T⇒3A=2M1T−M24A + M_2 - A = 2M_1^T \Rightarrow 3A = 2M_1^T - M_2.

Now 2M1T=(420104)2M_1^T = \begin{pmatrix}4&20\\10&4\end{pmatrix}, so 2M1T−M2=(4−120−810−44−1)=(31263)2M_1^T - M_2 = \begin{pmatrix}4-1&20-8\\10-4&4-1\end{pmatrix} = \begin{pmatrix}3&12\\6&3\end{pmatrix}.

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