1. **Problem 10:** Determine if each matrix is Hermitian, Skew-Hermitian, or neither. 2. Recall definitions: - A matrix $M$ is **Hermitian** if $M = M^\dagger$, where $M^\dagger$ is the conjugate transpose. - A matrix $M$ is **Skew-Hermitian** if $M = -M^\dagger$. 3. For matrix (i): $$ M = \begin{bmatrix} 3 & 1 - 2i & 4 + 7i \\ 1 + 2i & -4 & -2i \\ 4 - 7i & 2i & 5 \end{bmatrix} $$ Calculate $M^\dagger$ by taking transpose and conjugate: $$ M^\dagger = \begin{bmatrix} 3 & 1 - 2i & 4 + 7i \\ 1 + 2i & -4 & -2i \\ 4 - 7i & 2i & 5 \end{bmatrix} $$ Note that conjugate of $1 + 2i$ is $1 - 2i$, and so on. Check if $M = M^\dagger$: - Entry $(1,2)$ in $M$ is $1 - 2i$, and entry $(2,1)$ in $M^\dagger$ is conjugate of $(1,2)$ in $M$, which is $1 + 2i$. - Since $(1,2) \neq (2,1)^*$, $M \neq M^\dagger$. Check if $M = -M^\dagger$: - Entry $(1,1)$ is 3, but $-M^\dagger_{1,1} = -3$, not equal. Therefore, matrix (i) is **neither**. 4. For matrix (ii): $$ N = \begin{bmatrix} -i & 3i & i \\ 3i & 7i & -5i \\ i & -5i & -i \end{bmatrix} $$ Calculate $N^\dagger$: $$ N^\dagger = \begin{bmatrix} i & -3i & -i \\ -3i & -7i & 5i \\ -i & 5i & i \end{bmatrix} $$ Check if $N = -N^\dagger$: - $-N^\dagger = \begin{bmatrix} -i & 3i & i \\ 3i & 7i & -5i \\ i & -5i & -i \end{bmatrix} = N$ Thus, $N = -N^\dagger$, so matrix (ii) is **Skew-Hermitian**. 5. For matrix (iii): $$ P = \begin{bmatrix} 1 & -i & -1 + i \\ i & 1 & 1 + i \\ 1 + i & -1 + i & 0 \end{bmatrix} $$ Calculate $P^\dagger$: $$ P^\dagger = \begin{bmatrix} 1 & -i & 1 - i \\ i & 1 & -1 - i \\ -1 - i & 1 - i & 0 \end{bmatrix} $$ Compare $P$ and $P^\dagger$: - $(1,3)$ in $P$ is $-1 + i$, but $(3,1)$ in $P^\dagger$ is $-1 - i$. - Since $P \neq P^\dagger$ and $P \neq -P^\dagger$, matrix (iii) is **neither**. --- 6. **Problem 11:** Find real numbers $x,y,z$ such that matrix $$ A = \begin{bmatrix} 3 & x + 2i & yi \\ 3 - 2i & 0 & 1 + zi \\ yi & 1 - xi & -1 \end{bmatrix} $$ is Hermitian. 7. For $A$ to be Hermitian, $A = A^\dagger$. This means: - Diagonal entries must be real. - Off-diagonal entries satisfy $A_{ij} = \overline{A_{ji}}$. 8. Check diagonal entries: - $3$, $0$, and $-1$ are real, so no condition here. 9. Check off-diagonal entries: - Entry $(1,2)$ is $x + 2i$, entry $(2,1)$ is $3 - 2i$. Condition: $x + 2i = \overline{3 - 2i} = 3 + 2i$. Equate real and imaginary parts: - Real: $x = 3$ - Imaginary: $2i = 2i$ (already equal) - Entry $(1,3)$ is $yi$, entry $(3,1)$ is $yi$. Condition: $yi = \overline{yi} = -yi$ (since conjugate of $yi$ is $-yi$ if $y$ is real). This implies: $$ yi = -yi \implies 2yi = 0 \implies y = 0 $$ - Entry $(2,3)$ is $1 + zi$, entry $(3,2)$ is $1 - xi$. Condition: $1 + zi = \overline{1 - xi} = 1 + xi$. Equate imaginary parts: $$ zi = xi \implies z = x $$ 10. From above, $x = 3$, $y = 0$, and $z = x = 3$. **Final answers:** - Problem 10: (i) Neither (ii) Skew-Hermitian (iii) Neither - Problem 11: $$ x = 3, y = 0, z = 3 $$