If \(A = \begin{bmatrix} 2 \sin \theta & \cos \theta & 0 \\ -2\cos \theta & \sin \theta & 0 \\ -1 & 1 & 1 \end{bmatrix},\) then what is A(adj A) equal to? (where) I is the identity matrix.
2I
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.
| Concept | Description |
|---|---|
| Matrix Determinant | A scalar value calculated from the elements of a square matrix. Indicates properties like invertibility. |
| Adjugate Matrix (adj A) | The transpose of the cofactor matrix of A. |
| Identity Matrix (I) | A square matrix with ones on the main diagonal and zeros elsewhere. Acts like the number 1 in matrix multiplication. |
| Matrix Property: A(adj A) | The product of a matrix A and its adjugate is equal to the determinant of A multiplied by the identity matrix: \(A(\text{adj } A) = (\det A) I\). |
| Trigonometric Identity | \(\sin^2 \theta + \cos^2 \theta = 1\), used to simplify the determinant calculation. |
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.
Important Questions from Adjoint and Inverse of a Square Matrix
Let A be a matrix of order 3 × 3 and |A| = 4. If |2adj(3A)| = 2α 3β, then what is the value of (α + β)?
If A is a square matrix, then what is adj (A -1 ) – (adj A) -1 equal to?
The matrix \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&3&2\\ 1&{{\rm{x}} - 1}&1\\ 2&7&{{\rm{x}} - 3} \end{array}} \right]\)
Will have inverse for every real number x except forIf $A = \begin{bmatrix} 2 & 7 \\ 1 & 5 \end{bmatrix}$, then what is $A + 3A^{-1}$ equal to, where $A$ is a matrix of order 2?
Find the value of $|adj (2 \cdot adj A)|$ if matrix $A$ is of the order of $3$ and $|A| = 15$.
