Find the additive inverse of matrix A = \(\left[\begin{array}{cc}2 & 1 \\ −3 & 0\end{array}\right]\)

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Bihar STET Paper I: Mathematics (Held In 2019 - Shift 1)
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  1. A = \(\left[\begin{array}{cc}2 & 1 \\ −3 & 0\end{array}\right]\)
  2. −A = \(\left[\begin{array}{cc}−2 & −1 \\ 3 & 0\end{array}\right]\)
  3. 2A = \(\left[\begin{array}{cc}4 & 2 \\ −6 & 0\end{array}\right]\)
  4. none of these

Answer (Detailed Solution Below)

Option 2 : −A = \(\left[\begin{array}{cc}−2 & −1 \\ 3 & 0\end{array}\right]\)
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Detailed Solution

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Concept use:

Additive inverse of matrix of 2 × 2 order \(\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]\) =\(-\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]\)

Calculation:

We have the matrix A = \(\left[\begin{array}{cc}2 & 1 \\ −3 & 0\end{array}\right]\)

Now additive inverse of matrix A  = \(-\left[\begin{array}{cc}2 & 1 \\ −3 & 0\end{array}\right]\) = \(\left[\begin{array}{cc}−2 & −1 \\ 3 & 0\end{array}\right]\)

(2) is correct

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