Understanding the Matrix and Adjugate Property

The question asks us to find the product \(A(\text{adj } A)\) for a given 3x3 matrix \(A\). The matrix is:

\(A = \begin{bmatrix} 2 \sin \theta & \cos \theta & 0 \\ -2\cos \theta & \sin \theta & 0 \\ -1 & 1 & 1 \end{bmatrix}\)

Here, \(\text{adj } A\) represents the adjugate (or adjoint) of matrix \(A\), and \(I\) represents the identity matrix of the same order as \(A\) (which is 3x3 in this case).

A fundamental property of matrices relating a matrix, its adjugate, and its determinant is:

\(A(\text{adj } A) = (\text{adj } A)A = (\det A) I\)

where \(\det A\) is the determinant of matrix \(A\).

To find \(A(\text{adj } A)\), we simply need to calculate the determinant of matrix \(A\) and multiply it by the identity matrix \(I\).

Calculating the Determinant of Matrix A

We will calculate the determinant of the given matrix \(A\).

\(A = \begin{bmatrix} 2 \sin \theta & \cos \theta & 0 \\ -2\cos \theta & \sin \theta & 0 \\ -1 & 1 & 1 \end{bmatrix}\)

We can expand the determinant along the first row for simplicity, as it contains a zero element. The determinant of a 3x3 matrix \(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\) is \(a(ei - fh) - b(di - fg) + c(dh - eg)\).

Applying this formula to matrix \(A\):

• The element \(a = 2 \sin \theta\). The corresponding minor determinant is \(\det \begin{bmatrix} \sin \theta & 0 \\ 1 & 1 \end{bmatrix} = (\sin \theta)(1) - (0)(1) = \sin \theta\). The term is \((2 \sin \theta)(\sin \theta) = 2 \sin^2 \theta\).
• The element \(b = \cos \theta\). The corresponding minor determinant is \(\det \begin{bmatrix} -2\cos \theta & 0 \\ -1 & 1 \end{bmatrix} = (-2\cos \theta)(1) - (0)(-1) = -2\cos \theta\). The term is \(-(\cos \theta)(-2\cos \theta) = 2 \cos^2 \theta\).
• The element \(c = 0\). The corresponding minor determinant is \(\det \begin{bmatrix} -2\cos \theta & \sin \theta \\ -1 & 1 \end{bmatrix} = (-2\cos \theta)(1) - (\sin \theta)(-1) = -2\cos \theta + \sin \theta\). Since the element is 0, this term is \(0 \times (-2\cos \theta + \sin \theta) = 0\).

Summing these terms to find the determinant of A:

\(\det A = (2 \sin \theta)(\sin \theta) - (\cos \theta)(-2\cos \theta) + (0)(\dots)\)

\(\det A = 2 \sin^2 \theta + 2 \cos^2 \theta\)

Factor out the common term 2:

\(\det A = 2 (\sin^2 \theta + \cos^2 \theta)\)

Using the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\):

\(\det A = 2 (1)\)

\(\det A = 2\)

Finding A(adj A)

Now that we have the determinant of matrix \(A\), which is 2, we can use the property \(A(\text{adj } A) = (\det A) I\).

Since \(\det A = 2\) and \(I\) is the 3x3 identity matrix \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\), we have:

\(A(\text{adj } A) = 2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

Multiplying the scalar 2 by the identity matrix:

\(A(\text{adj } A) = \begin{bmatrix} 2 \times 1 & 2 \times 0 & 2 \times 0 \\ 2 \times 0 & 2 \times 1 & 2 \times 0 \\ 2 \times 0 & 2 \times 0 & 2 \times 1 \end{bmatrix}\)

\(A(\text{adj } A) = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}\)

This result is equal to \(2I\).

Conclusion

The value of \(A(\text{adj } A)\) for the given matrix \(A\) is \(2I\).

Revision Table: Key Concepts

This table summarizes the key concepts used in solving this matrix problem.

Additional Information: Matrix Properties and Applications

Understanding matrix properties is crucial in linear algebra. The relationship \(A(\text{adj } A) = (\det A) I\) is particularly useful. Here's why:

• Matrix Inverse: If \(\det A \ne 0\), then the matrix \(A\) is invertible, and its inverse \(A^{-1}\) can be found using the formula \(A^{-1} = \frac{1}{\det A} (\text{adj } A)\). This property shows how the determinant and adjugate are directly involved in finding the inverse.
• Solving Linear Systems: Determinants and adjugates (via Cramer's rule) can be used to solve systems of linear equations, although Gaussian elimination is often more computationally efficient for larger systems.
• Geometric Interpretation: The determinant of a matrix represents the scaling factor of the volume when the matrix is considered as a linear transformation.

This problem demonstrates a direct application of a core matrix identity, simplifying the calculation significantly by avoiding the lengthy process of calculating the adjugate matrix explicitly.