3 and 4 .Determinants and Matrices
easy

If $A^{\prime}=\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right],$ then verify that $(A-B)^{\prime}=A^{\prime}-B^{\prime}$

Option A
Option B
Option C
Option D

Solution

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

$\therefore(A-B)^{\prime}=\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]$

$A^{\prime}-B^{\prime}=\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]=\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]$

Thus, we have verified that $(A-B)^{\prime}=A^{\prime}-B^{\prime}$

Standard 12
Mathematics

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