Determine the matrices A and B when A+2 B= [ cc 1 & 2 6 & -3 ] and 2 A - B = [ cc 2 & -1 2 & -1 ]

& A+2 B= [ cc 1 & 2 6 & -3 ] ( (i) & 2 A-B= [ ll 2 & -1 2 & -1 ] (iii) Multiplying (i) by 1 and (ii) by 2 & A+2 B= [ cc 1 & 2 6 & -3 ] & 4 A-2 B=2 [ ll 2 & -1 2 & -1 ]= [ ll 4 & -2 4 & -2 ] Adding, we get & 5 A= [ cc 1 & 2 6 & -3 ]+ [ cc 4 & -2 4 & -2 ]= [ cc 5 & 0 10 & -5 ] & A=1 5 [ cc 5 & 0 10 & 5 ]= [ cc 1 & 0 2 & -1 ] From (i) A+2 B= [ cc 1 & 2 6 & -3 ] & = [ cc 1 & 0 2 & -1 ]+2 B = [ cc 1 & 2 6 & -3 ] & 2 B = [ cc 1 & 2 6 & -3 ]- [ cc 1 & 0 2 & -1 ]= [ cc 0 & 2 4 & -2 ] & B =1 2 [ cc 0 & 2 4 & -2 ]= [ cc 0 & 1 2 & -1 ] Hence A= [ cc 1 & 0 2 & -1 ] and B= [ cc 0 & 1 2 & -1 ].

Determine the matrices A and B when A+2 B= [ cc 1 & 2 6 & -3 ] an…

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Question

Determine the matrices $A$ and $B$ when $A+2 B=$
$
\left[\begin{array}{cc}
1 & 2 \\
6 & -3
\end{array}\right] \text { and } 2 A - B =\left[\begin{array}{cc}
2 & -1 \\
2 & -1
\end{array}\right]
$

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Answer

$
\begin{aligned}
& A+2 B=\left[\begin{array}{cc}
1 & 2 \\
6 & -3
\end{array}\right] \ldots \ldots(\text { (i) } \\
& 2 A-B=\left[\begin{array}{ll}
2 & -1 \\
2 & -1
\end{array}\right] \ldots \ldots \text { (iii) }
\end{aligned}
$
Multiplying (i) by 1 and (ii) by 2
$
\begin{aligned}
& A+2 B=\left[\begin{array}{cc}
1 & 2 \\
6 & -3
\end{array}\right] \\
& 4 A-2 B=2\left[\begin{array}{ll}
2 & -1 \\
2 & -1
\end{array}\right]=\left[\begin{array}{ll}
4 & -2 \\
4 & -2
\end{array}\right]
\end{aligned}
$
Adding, we get
$
\begin{aligned}
& 5 A=\left[\begin{array}{cc}
1 & 2 \\
6 & -3
\end{array}\right]+\left[\begin{array}{cc}
4 & -2 \\
4 & -2
\end{array}\right]=\left[\begin{array}{cc}
5 & 0 \\
10 & -5
\end{array}\right] \\
& A=\frac{1}{5}\left[\begin{array}{cc}
5 & 0 \\
10 & 5
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
2 & -1
\end{array}\right]
\end{aligned}
$
From (i) $A+2 B=\left[\begin{array}{cc}1 & 2 \\ 6 & -3\end{array}\right]$
$
\begin{aligned}
& =\left[\begin{array}{cc}
1 & 0 \\
2 & -1
\end{array}\right]+2 B =\left[\begin{array}{cc}
1 & 2 \\
6 & -3
\end{array}\right] \\
& 2 B =\left[\begin{array}{cc}
1 & 2 \\
6 & -3
\end{array}\right]-\left[\begin{array}{cc}
1 & 0 \\
2 & -1
\end{array}\right]=\left[\begin{array}{cc}
0 & 2 \\
4 & -2
\end{array}\right] \\
& \therefore B =\frac{1}{2}\left[\begin{array}{cc}
0 & 2 \\
4 & -2
\end{array}\right]=\left[\begin{array}{cc}
0 & 1 \\
2 & -1
\end{array}\right]
\end{aligned}
$
Hence $A=\left[\begin{array}{cc}1 & 0 \\ 2 & -1\end{array}\right]$ and $B=\left[\begin{array}{cc}0 & 1 \\ 2 & -1\end{array}\right]$.