Prove that the determinant
\[ \left| \begin{matrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{matrix} \right| \]
is independent of \(\theta\).
Evaluate
\[ \left| \begin{matrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{matrix} \right| \]
1
If
\[ A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}, \]
find \((AB)^{-1}\).
\((AB)^{-1} = \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}\)
Let
\[ A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}. \]
Verify that
(i) \([\operatorname{adj}A]^{-1} = \operatorname{adj}(A^{-1})\)
(ii) \((A^{-1})^{-1} = A\)
Evaluate
\[ \left| \begin{matrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{matrix} \right| \]
\(-2(x^3 + y^3)\)
Evaluate
\[ \left| \begin{matrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{matrix} \right| \]
\(xy\)
Solve the system of equations:
\(\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4\)
\(\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1\)
\(\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2\)
\(x = 2,\; y = 3,\; z = 5\)
Choose the correct answer.
If \(x, y, z\) are nonzero real numbers, then the inverse of matrix
\[ A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \]
is
Let
\[ A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}, \quad 0 \le \theta \le 2\pi. \]
Then