Class 12Mathematics3 and 4 .Determinants and MatricesSpecial types of matrices, Transpose of matrices and Trace of matrices
EasyIf $A=\left[\begin{array}{rrr}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{rrr}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{array}\right],$ then verify that $(A-B)^{\prime}=A^{\prime}-B^{\prime}$.
(A) First,calculate $A-B$:
$A-B = \left[\begin{array}{rrr}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{array}\right] - \left[\begin{array}{rrr}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{array}\right] = \left[\begin{array}{rrr}3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0\end{array}\right]$
Now,find the transpose $(A-B)^{\prime}$:
$(A-B)^{\prime} = \left[\begin{array}{rrr}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{array}\right]$
Next,find $A^{\prime}$ and $B^{\prime}$:
$A^{\prime} = \left[\begin{array}{rrr}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{array}\right]$,$B^{\prime} = \left[\begin{array}{rrr}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{array}\right]$
Now,calculate $A^{\prime}-B^{\prime}$:
$A^{\prime}-B^{\prime} = \left[\begin{array}{rrr}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{array}\right] - \left[\begin{array}{rrr}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{array}\right] = \left[\begin{array}{rrr}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{array}\right]$
Since $(A-B)^{\prime} = A^{\prime}-B^{\prime}$,the property is verified.
$A-B = \left[\begin{array}{rrr}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{array}\right] - \left[\begin{array}{rrr}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{array}\right] = \left[\begin{array}{rrr}3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0\end{array}\right]$
Now,find the transpose $(A-B)^{\prime}$:
$(A-B)^{\prime} = \left[\begin{array}{rrr}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{array}\right]$
Next,find $A^{\prime}$ and $B^{\prime}$:
$A^{\prime} = \left[\begin{array}{rrr}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{array}\right]$,$B^{\prime} = \left[\begin{array}{rrr}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{array}\right]$
Now,calculate $A^{\prime}-B^{\prime}$:
$A^{\prime}-B^{\prime} = \left[\begin{array}{rrr}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{array}\right] - \left[\begin{array}{rrr}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{array}\right] = \left[\begin{array}{rrr}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{array}\right]$
Since $(A-B)^{\prime} = A^{\prime}-B^{\prime}$,the property is verified.
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