Class 12Mathematics3 and 4 .Determinants and MatricesSpecial types of matrices, Transpose of matrices and Trace of matrices
EasyIf $A = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$,verify that $(A')' = A$.
(N/A) Given the matrix $A = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}$.
To find the transpose $A'$,we interchange the rows and columns of $A$:
$A' = \begin{bmatrix} 3 & 4 \\ \sqrt{3} & 2 \\ 2 & 0 \end{bmatrix}$.
Now,to find the transpose of $A'$,denoted as $(A')'$,we interchange the rows and columns of $A'$:
$(A')' = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}$.
Comparing this result with the original matrix $A$,we see that $(A')' = A$.
Hence,the property is verified.
To find the transpose $A'$,we interchange the rows and columns of $A$:
$A' = \begin{bmatrix} 3 & 4 \\ \sqrt{3} & 2 \\ 2 & 0 \end{bmatrix}$.
Now,to find the transpose of $A'$,denoted as $(A')'$,we interchange the rows and columns of $A'$:
$(A')' = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}$.
Comparing this result with the original matrix $A$,we see that $(A')' = A$.
Hence,the property is verified.
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