Diagonal Matrix Operations
Question:
Let \[ A = \text{diag}(2,-5,9), \quad B = \text{diag}(1,1,-4), \quad C = \text{diag}(-6,3,4) \] Find:
(i) \(A – 2B\)
(ii) \(B + C – 2A\)
(iii) \(2A + 3B – 5C\)
Let \[ A = \text{diag}(2,-5,9), \quad B = \text{diag}(1,1,-4), \quad C = \text{diag}(-6,3,4) \] Find:
(i) \(A – 2B\)
(ii) \(B + C – 2A\)
(iii) \(2A + 3B – 5C\)
Solution:
For diagonal matrices, operations are performed on corresponding diagonal elements.
(i) \(A – 2B\)
\[ A – 2B = \text{diag}(2-2\cdot1,\,-5-2\cdot1,\;9-2(-4)) \] \[ = \text{diag}(0,\,-7,\;17) \] \[ = \begin{bmatrix} 0 & 0 & 0 \\ 0 & -7 & 0 \\ 0 & 0 & 17 \end{bmatrix} \](ii) \(B + C – 2A\)
\[ = \text{diag}(1-6-4,\;1+3+10,\;-4+4-18) \] \[ = \text{diag}(-9,\;14,\;-18) \] \[ = \begin{bmatrix} -9 & 0 & 0 \\ 0 & 14 & 0 \\ 0 & 0 & -18 \end{bmatrix} \](iii) \(2A + 3B – 5C\)
\[ = \text{diag}(4+3+30,\;-10+3-15,\;18-12-20) \] \[ = \text{diag}(37,\;-22,\;-14) \] \[ = \begin{bmatrix} 37 & 0 & 0 \\ 0 & -22 & 0 \\ 0 & 0 & -14 \end{bmatrix} \]Final Answers:
\[ (i)\ \begin{bmatrix}0&0&0\\0&-7&0\\0&0&17\end{bmatrix} \] \[ (ii)\ \begin{bmatrix}-9&0&0\\0&14&0\\0&0&-18\end{bmatrix} \] \[ (iii)\ \begin{bmatrix}37&0&0\\0&-22&0\\0&0&-14\end{bmatrix} \]