If $$ A = \begin{pmatrix} -5 & -8 & 0 \\3 & 5 & 0 \\1 & 2 & -1 \end{pmatrix} $$ then $ A^2 $ is:
To determine the property of \( A^2 \), we compute \( A^2 \) by performing matrix multiplication: \[ A^2 = \begin{pmatrix} -5 & -8 & 0 3 & 5 & 0 1& 2 & -1 \end{pmatrix} \times \begin{pmatrix} -5 & -8 & 0 \\3 & 5 & 0 \\1 & 2 & -1 \end{pmatrix} \] After matrix multiplication, we find that: \[ A^2 = I \quad (\text{the identity matrix}) \] This means the matrix \( A \) is involutory, as a matrix is involutory if \( A^2 = I \).
Thus, the correct answer is 4. Involutory.
The surface integral \( \int_S x^2 \, dS \) over the upper hemisphere
\[ z = \sqrt{1 - x^2 - y^2} \]
with radius 1 is ..........