CUET Preparation Today
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\(B = \begin{bmatrix}-1 & 2 & 1\\1 & 2 & 3\end{bmatrix}\), \(B^T = \begin{bmatrix}-1 & 1\\2 & 2\\1 & 3\end{bmatrix}\) so, \(A^T - B^T = \begin{bmatrix}3 & 4\\-1 & 2\\0 & 1\end{bmatrix} - \begin{bmatrix}-1 & 1\\2 & 2\\1 & 3\end{bmatrix}\) \(A^T - B^T = \begin{bmatrix}4 & 3\\-3 & 0\\-1 & -2\end{bmatrix}\) |
