Class 12Mathematics3 and 4 .Determinants and MatricesSpecial types of matrices, Transpose of matrices and Trace of matrices
EasyIf $A=\begin{bmatrix}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{bmatrix}$ and $B=\begin{bmatrix}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{bmatrix}$,then verify that $(A+B)^{\prime}=A^{\prime}+B^{\prime}$.
(A) We have:
$A^{\prime}=\begin{bmatrix}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{bmatrix}, B^{\prime}=\begin{bmatrix}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{bmatrix}$
$A+B=\begin{bmatrix}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{bmatrix}+\begin{bmatrix}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{bmatrix}=\begin{bmatrix}-5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2\end{bmatrix}$
$\therefore (A+B)^{\prime}=\begin{bmatrix}-5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2\end{bmatrix}$
$A^{\prime}+B^{\prime}=\begin{bmatrix}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{bmatrix}+\begin{bmatrix}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{bmatrix}=\begin{bmatrix}-5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2\end{bmatrix}$
Since $(A+B)^{\prime} = A^{\prime}+B^{\prime}$,the property is verified.
$A^{\prime}=\begin{bmatrix}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{bmatrix}, B^{\prime}=\begin{bmatrix}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{bmatrix}$
$A+B=\begin{bmatrix}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{bmatrix}+\begin{bmatrix}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{bmatrix}=\begin{bmatrix}-5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2\end{bmatrix}$
$\therefore (A+B)^{\prime}=\begin{bmatrix}-5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2\end{bmatrix}$
$A^{\prime}+B^{\prime}=\begin{bmatrix}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{bmatrix}+\begin{bmatrix}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{bmatrix}=\begin{bmatrix}-5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2\end{bmatrix}$
Since $(A+B)^{\prime} = A^{\prime}+B^{\prime}$,the property is verified.
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