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Matrices — Mathematics STD 10 — Question

ICSE BoardEnglish MediumSTD 10MathematicsMatrices1 Mark
MCQ
If $A+B=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$ and $A-2 B=\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right]$ then $A$ is equal to
  • A
    $\frac{1}{3}\left[\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right]$
  • B
    $\frac{1}{3}\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$
  • ✓
    $\left[\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right]$
  • D
    $\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$
✓

Answer

Correct option: C.
$\left[\begin{array}{ll}1 & 1 \\ 2 & 1\end{array}\right]$
$
A+B=\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right]
$
and
$
A-2 B=\left[\begin{array}{cc}
-1 & 1 \\
0 & -1
\end{array}\right]
$
$\Rightarrow 2 A+2 B=\left[\begin{array}{ll}2 & 0 \\ 2 & 2\end{array}\right]$ (Multiplying by 2) $\ldots$ (i)
$A-2 B=\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right]$
Adding (i) and (ii), we get
$
\begin{aligned}
& 3 A=\left[\begin{array}{ll}
2 & 0 \\
2 & 2
\end{array}\right]+\left[\begin{array}{cc}
-1 & 1 \\
0 & -1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 1 \\
2 & 1
\end{array}\right] \\
& \therefore A=\frac{1}{3}\left[\begin{array}{ll}
1 & 1 \\
2 & 1
\end{array}\right] .
\end{aligned}
$

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