Question

$A=\left[\begin{array}{ccc}2& 3& 1\\ 4& 1& 5\\ 3& 9& 7\end{array}\right]$. Then the additive inverse of $A$ is

• A. $\left[\begin{array}{ccc}-2& -3& 1\\ 4& -1& -5\\ -3& 9& -7\end{array}\right]$
• B. $\left[\begin{array}{ccc}-2& -3& -1\\ -4& -1& -5\\ -3& -9& -7\end{array}\right]$
• C. $\left[\begin{array}{ccc}2& -3& -1\\ -4& 1& -5\\ -3& -9& 7\end{array}\right]$
• D. $\left[\begin{array}{ccc}-2& -3& -1\\ -4& -1& -5\\ -3& 9& -7\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{ccc}-2& -3& -1\\ -4& -1& -5\\ -3& -9& -7\end{array}\right]$

Given : $A=\left[\begin{array}{ccc}2& 3& 1\\ 4& 1& 5\\ 3& 9& 7\end{array}\right]$

Additive inverse of any element $a$ is the element which when added to $a$ should give the addition as $0$.

Additive inverse of any matrix $A$ of order $n\times n$ is the matrix which when added to $A$ gives zero matrix of order $n\times n$.

Let $B$ represents additive inverse of $A$.

$⟹A+B=(0{)}_{3\times 3}$

$⟹\left[\begin{array}{ccc}2& 3& 1\\ 4& 1& 5\\ 3& 9& 7\end{array}\right]+B=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

$⟹B=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]-\left[\begin{array}{ccc}2& 3& 1\\ 4& 1& 5\\ 3& 9& 7\end{array}\right]$

$\therefore B=\left[\begin{array}{ccc}-2& -3& -1\\ -4& -1& -5\\ -3& -9& -7\end{array}\right]$.

Hence, option B is correct.