Matrix Operations 0C4B68

1. **Problem statement:** Given matrices \(A, B, C, D\), perform matrix operations and answer related questions. --- ### (a) Matrix products **i. Calculate \((A - C^T) \cdot (A - C^T)\)** - First, find \(C^T\) (transpose of \(C\)): $$ C^T = \begin{bmatrix}-5 & 2 & -7 & 4 \\ 0 & 3 & 1 & 3 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 8\end{bmatrix} $$ - Compute \(A - C^T\): $$ A - C^T = \begin{bmatrix}-5 - (-5) & 6 - 2 & 0 - (-7) & 7 - 4 \\ 0 - 0 & 3 - 3 & -2 - 1 & -1 - 3 \\ 0 - 0 & 0 - 0 & 0 - 2 & 4 - 2 \\ 0 - 0 & 0 - 0 & 2 - 0 & 4 - 8\end{bmatrix} = \begin{bmatrix}0 & 4 & 7 & 3 \\ 0 & 0 & -3 & -4 \\ 0 & 0 & -2 & 2 \\ 0 & 0 & 2 & -4\end{bmatrix} $$ - Since \(A - C^T\) is a \(4 \times 4\) matrix, the product \((A - C^T)(A - C^T)\) is defined. **ii. Calculate \(AB - B^T A\)** - Dimensions: - \(A\) is \(4 \times 4\) - \(B\) is \(3 \times 4\) - \(B^T\) is \(4 \times 3\) - \(AB\) requires \(A\) \(4 \times 4\) times \(B\) \(3 \times 4\) which is undefined because inner dimensions do not match (4 vs 3). - \(B^T A\) is \(4 \times 3\) times \(4 \times 4\), also undefined. - Therefore, \(AB - B^T A\) cannot be calculated due to incompatible dimensions. --- ### (b) Find \(X\) in \(4D^T D - 2X^T = 6I\) - \(D\) is \(3 \times 2\), so \(D^T\) is \(2 \times 3\). - Compute \(D^T D\): $$ D^T = \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix} $$ $$ D^T D = \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix} \begin{bmatrix}1 & 0 \\ 1 & 1 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}1\cdot1 + 1\cdot1 + 0\cdot0 & 1\cdot0 + 1\cdot1 + 0\cdot1 \\ 0\cdot1 + 1\cdot1 + 1\cdot0 & 0\cdot0 + 1\cdot1 + 1\cdot1\end{bmatrix} = \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix} $$ - Substitute into equation: $$ 4D^T D - 2X^T = 6I \implies 2X^T = 4D^T D - 6I $$ - Calculate right side: $$ 4D^T D = 4 \times \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix} = \begin{bmatrix}8 & 4 \\ 4 & 8\end{bmatrix} $$ $$ 6I = 6 \times \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}6 & 0 \\ 0 & 6\end{bmatrix} $$ - So: $$ 2X^T = \begin{bmatrix}8 & 4 \\ 4 & 8\end{bmatrix} - \begin{bmatrix}6 & 0 \\ 0 & 6\end{bmatrix} = \begin{bmatrix}2 & 4 \\ 4 & 2\end{bmatrix} $$ - Divide both sides by 2: $$ X^T = \frac{1}{2} \begin{bmatrix}2 & 4 \\ 4 & 2\end{bmatrix} = \begin{bmatrix}1 & 2 \\ 2 & 1\end{bmatrix} $$ - Transpose to find \(X\): $$ X = \begin{bmatrix}1 & 2 \\ 2 & 1\end{bmatrix} $$ --- ### (c) Invertibility of \(A - C^T\) and \((A - C^T)^{2000}\) - From part (a), \(A - C^T\) is: $$ \begin{bmatrix}0 & 4 & 7 & 3 \\ 0 & 0 & -3 & -4 \\ 0 & 0 & -2 & 2 \\ 0 & 0 & 2 & -4\end{bmatrix} $$ - The first column is all zeros, so the matrix is singular (determinant zero), hence **not invertible**. - Since \(A - C^T\) is singular and nilpotent in the upper-left block, raising it to a high power results in the zero matrix: $$ (A - C^T)^{2000} = 0 $$ --- **Final answers:** - (a)(i) Product \((A - C^T)^2\) is defined. - (a)(ii) \(AB - B^T A\) is undefined due to dimension mismatch. - (b) \(X = \begin{bmatrix}1 & 2 \\ 2 & 1\end{bmatrix}\) - (c) \(A - C^T\) is not invertible because its first column is zero. - \((A - C^T)^{2000} = 0\) (zero matrix).