If A = \(\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & \sin \theta & −\cos \theta\end{array}\right)\) , then which of the following are correct? 1. A + adj A is a null matrix 2. A −1 + adj A is a null matrix 3. A − A −1 is a null matrix Select the correct answer using the code given below :
1, 2 and 3
Analyzing Matrix Properties and Statements
We are given a matrix \( A = \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & \sin \theta & -\cos \theta\end{array}\right) \) and asked to check the correctness of three statements involving matrix A, its inverse (A-1), and its adjugate (adj A).
Calculating Determinant of Matrix A
First, let's find the determinant of matrix A. The determinant of a 3x3 matrix \(\left(\begin{array}{ccc}a & b & c \\ d & e & f \\ g & h & i\end{array}\right)\) is given by \( a(ei-fh) - b(di-fg) + c(dh-eg) \).
For matrix A:
\( \det(A) = 1 \times ((\cos \theta)(-\cos \theta) - (\sin \theta)(\sin \theta)) - 0 \times (...) + 0 \times (...) \)
\( \det(A) = 1 \times (-\cos^2 \theta - \sin^2 \theta) \)
\( \det(A) = -(\cos^2 \theta + \sin^2 \theta) \)
Using the trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( \det(A) = -1 \)
Since the determinant is non-zero, the inverse A-1 exists.
Calculating Adjugate of Matrix A
The adjugate of a matrix is the transpose of its cofactor matrix. The cofactor \( C_{ij} \) is calculated as \( (-1)^{i+j} M_{ij} \), where \( M_{ij} \) is the minor determinant.
Let's find the cofactor matrix C:
- \( C_{11} = (-1)^{1+1} \det \left(\begin{array}{cc} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array}\right) = 1 \times (-\cos^2 \theta - \sin^2 \theta) = -1 \)
- \( C_{12} = (-1)^{1+2} \det \left(\begin{array}{cc} 0 & \sin \theta \\ 0 & -\cos \theta \end{array}\right) = -1 \times (0 - 0) = 0 \)
- \( C_{13} = (-1)^{1+3} \det \left(\begin{array}{cc} 0 & \cos \theta \\ 0 & \sin \theta \end{array}\right) = 1 \times (0 - 0) = 0 \)
- \( C_{21} = (-1)^{2+1} \det \left(\begin{array}{cc} 0 & 0 \\ \sin \theta & -\cos \theta \end{array}\right) = -1 \times (0 - 0) = 0 \)
- \( C_{22} = (-1)^{2+2} \det \left(\begin{array}{cc} 1 & 0 \\ 0 & -\cos \theta \end{array}\right) = 1 \times (-\cos \theta - 0) = -\cos \theta \)
- \( C_{23} = (-1)^{2+3} \det \left(\begin{array}{cc} 1 & 0 \\ 0 & \sin \theta \end{array}\right) = -1 \times (\sin \theta - 0) = -\sin \theta \)
- \( C_{31} = (-1)^{3+1} \det \left(\begin{array}{cc} 0 & 0 \\ \cos \theta & \sin \theta \end{array}\right) = 1 \times (0 - 0) = 0 \)
- \( C_{32} = (-1)^{3+2} \det \left(\begin{array}{cc} 1 & 0 \\ 0 & \sin \theta \end{array}\right) = -1 \times (\sin \theta - 0) = -\sin \theta \)
- \( C_{33} = (-1)^{3+3} \det \left(\begin{array}{cc} 1 & 0 \\ 0 & \cos \theta \end{array}\right) = 1 \times (\cos \theta - 0) = \cos \theta \)
The cofactor matrix C is:
\( C = \left(\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos \theta & -\sin \theta \\ 0 & -\sin \theta & \cos \theta\end{array}\right) \)
The adjugate matrix adj(A) is the transpose of C:
\( \text{adj}(A) = C^T = \left(\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos \theta & -\sin \theta \\ 0 & -\sin \theta & \cos \theta\end{array}\right) \)
Calculating Inverse of Matrix A
The inverse of matrix A is given by the formula \( A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \).
\( A^{-1} = \frac{1}{-1} \left(\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos \theta & -\sin \theta \\ 0 & -\sin \theta & \cos \theta\end{array}\right) \)
\( A^{-1} = -1 \times \left(\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos \theta & -\sin \theta \\ 0 & -\sin \theta & \cos \theta\end{array}\right) \)
\( A^{-1} = \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & \sin \theta & -\cos \theta\end{array}\right) \)
Notice that \( A^{-1} = A \).
Also, from the formula \( A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \), we have \( \text{adj}(A) = \det(A) \times A^{-1} = -1 \times A^{-1} = -A^{-1} \).
Checking the Statements
Let's verify each given statement using the calculated values of A, A-1, and adj A.
Statement 1: A + adj A is a null matrix
A + adj A \( = \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & \sin \theta & -\cos \theta\end{array}\right) + \left(\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos \theta & -\sin \theta \\ 0 & -\sin \theta & \cos \theta\end{array}\right) \)
\( = \left(\begin{array}{ccc}1 + (-1) & 0 + 0 & 0 + 0 \\ 0 + 0 & \cos \theta + (-\cos \theta) & \sin \theta + (-\sin \theta) \\ 0 + 0 & \sin \theta + (-\sin \theta) & -\cos \theta + \cos \theta\end{array}\right) \)
\( = \left(\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right) \)
The result is the null matrix. So, Statement 1 is correct.
Statement 2: A-1 + adj A is a null matrix
We know that adj(A) = -A-1.
A-1 + adj A \( = A^{-1} + (-A^{-1}) \)
\( = A^{-1} - A^{-1} \)
\( = \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & \sin \theta & -\cos \theta\end{array}\right) - \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & \sin \theta & -\cos \theta\end{array}\right) \)
\( = \left(\begin{array}{ccc}1 - 1 & 0 - 0 & 0 - 0 \\ 0 - 0 & \cos \theta - \cos \theta & \sin \theta - \sin \theta \\ 0 - 0 & \sin \theta - \sin \theta & -\cos \theta - (-\cos \theta)\end{array}\right) \)
\( = \left(\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right) \)
The result is the null matrix. So, Statement 2 is correct.
Statement 3: A - A-1 is a null matrix
We found that A-1 = A.
A - A-1 \( = A - A \)
\( = \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & \sin \theta & -\cos \theta\end{array}\right) - \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & \sin \theta & -\cos \theta\end{array}\right) \)
\( = \left(\begin{array}{ccc}1 - 1 & 0 - 0 & 0 - 0 \\ 0 - 0 & \cos \theta - \cos \theta & \sin \theta - \sin \theta \\ 0 - 0 & \sin \theta - \sin \theta & -\cos \theta - (-\cos \theta)\end{array}\right) \)
\( = \left(\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right) \)
The result is the null matrix. So, Statement 3 is correct.
All three statements are correct.
Conclusion
Based on our calculations, all three statements are correct:
- A + adj A is a null matrix
- A-1 + adj A is a null matrix
- A - A-1 is a null matrix
Revision Table: Key Matrix Concepts
| Concept | Definition/Formula | Notes |
|---|---|---|
| Determinant (\(\det(A)\)) | A scalar value computed from the elements of a square matrix. | A matrix is invertible if and only if its determinant is non-zero. |
| Minor (\(M_{ij}\)) | The determinant of the submatrix obtained by deleting row i and column j. | Used in calculating cofactors. |
| Cofactor (\(C_{ij}\)) | \(C_{ij} = (-1)^{i+j} M_{ij}\) | Signed minor. Used to find the adjugate and inverse. |
| Adjugate (\(\text{adj}(A)\)) | The transpose of the cofactor matrix. | Related to the inverse by \(\text{adj}(A) = \det(A) \times A^{-1}\). |
| Inverse (\(A^{-1}\)) | A matrix such that \(A A^{-1} = A^{-1} A = I\) (Identity matrix). | Exists only for non-singular (invertible) matrices. \( A^{-1} = \frac{1}{\det(A)} \text{adj}(A) \). |
| Null Matrix | A matrix where all elements are zero. | Represented by O. A + (-A) = O. |
Additional Information: Special Types of Matrices
The given matrix A has some interesting properties. Let's explore some related matrix types:
- Orthogonal Matrix: A square matrix Q is orthogonal if \( Q^T Q = Q Q^T = I \). This implies \( Q^{-1} = Q^T \). The determinant of an orthogonal matrix is either 1 or -1.
- Involutory Matrix: A square matrix A is involutory if \( A^2 = I \). This implies \( A^{-1} = A \). Our matrix A satisfies \( A^2 = I \) because \( A^{-1} = A \). Let's verify this:
\( A^2 = A \times A = \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & \sin \theta & -\cos \theta\end{array}\right) \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \theta & \sin \theta \\ 0 & \sin \theta & -\cos \theta\end{array}\right) \)
\( = \left(\begin{array}{ccc}1 \times 1 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 0 \\ 0 + \cos^2 \theta + \sin^2 \theta & 0 + \cos \theta \sin \theta - \sin \theta \cos \theta & 0 + \cos \theta \sin \theta - \sin \theta (-\cos \theta) \\ 0 + \sin \theta \cos \theta - \cos \theta \sin \theta & 0 + \sin^2 \theta + (-\cos \theta)(-\cos \theta) & 0 + \sin \theta \sin \theta + (-\cos \theta)(-\cos \theta)\end{array}\right) \)
\( = \left(\begin{array}{ccc}1 & 0 & 0 \\ 1 & 0 & 2 \sin \theta \cos \theta \\ 0 & \sin^2 \theta + \cos^2 \theta & \sin^2 \theta + \cos^2 \theta\end{array}\right) \) - Correction in multiplication:
\( (A^2)_{22} = 0 \times 0 + \cos\theta \times \cos\theta + \sin\theta \times \sin\theta = \cos^2\theta + \sin^2\theta = 1 \)
\( (A^2)_{23} = 0 \times 0 + \cos\theta \times \sin\theta + \sin\theta \times (-\cos\theta) = \cos\theta \sin\theta - \sin\theta \cos\theta = 0 \)
\( (A^2)_{32} = 0 \times 0 + \sin\theta \times \cos\theta + (-\cos\theta) \times \sin\theta = \sin\theta \cos\theta - \cos\theta \sin\theta = 0 \)
\( (A^2)_{33} = 0 \times 0 + \sin\theta \times \sin\theta + (-\cos\theta) \times (-\cos\theta) = \sin^2\theta + \cos^2\theta = 1 \)
\( A^2 = \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right) \)
So, \( A^2 = I \), which confirms A is an involutory matrix, hence \( A^{-1} = A \).
- Relationship between Adjugate and Inverse: The property \( \text{adj}(A) = \det(A) \times A^{-1} \) is a fundamental relationship. In our case, \( \det(A) = -1 \), so \( \text{adj}(A) = -A^{-1} \). This relationship simplifies verifying statement 2 once statement 1 or the property \( A^{-1} = A \) is established.
Understanding these matrix properties helps in quickly verifying the given statements.
Similar Questions
Consider the following in respect of matrices A, B and C of same order:
1) (A + B + C)' = A' + B’ + C’
2) (AB)’ = A’B’
3) (ABC)’ = C’B’A’
Where A’ is the transpose of the matrix A.
Which of the above are correct?If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 2}\\ 2&{ - 3}&4 \end{array}} \right]\) , then the matrix X for which 2X + 3A = 0 holds true is
Let \(A = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}\) and (mI + nA) 2= A where m, n are positive real numbers and I is the identify matrix. What is (m + n) equal to?
Let \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {\rm x + y}& \rm y\\ {\rm 2x}&{\rm x - y} \end{array}} \right],\;\rm B = \left[ {\begin{array}{*{20}{c}} 2\\ { - 1} \end{array}} \right]\) and \(\rm C = \left[ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right]\) . If AB = C, then what is the value of the determinant of the matrix A?
If \(A = \left[ {\begin{array}{c} 1&{ - 1}\\ { - 1}&1 \end{array}} \right],\) then the expression A 3- 2A 2is
If \(A = \left( {\begin{array}{*{20}{c}} 1&2\\ 2&3\\ 3&4 \end{array}} \right)\) and \(B = \left( {\begin{array}{*{20}{c}} 1&2\\ 2&1 \end{array}} \right),\) then which one of the following is correct?
A square matrix A is called orthogonal if_______ where A’ is the transpose of A.
Consider the following in respect of matrices A and B of same order:
1) A 2– B 2= (A + B) (A – B)
2) (A – I) (I + A) = O ⇔ A 2= I
Where I is the identity matrix and O is the null matrix.
Which of the above is/are correct?If \({\rm{f}}\left( {\rm{x}} \right) = \left[ {\begin{array}{c} {\cos {\rm{x}}}&{ - \sin {\rm{x}}}&0\\ {\sin {\rm{x}}}&{\cos {\rm{x}}}&0\\ 0&0&1 \end{array}} \right]\) , then which of the following are correct?
1. f(θ) × f(ϕ) = f(θ + ϕ)
2. The value of the determinant of the matrix f(θ) × f(ϕ) is 1
3. The determinant of f(x) is an even function.
Select the correct answer using the code given below:
Consider the following in respect of the matrices:
A = [m n], B = [-n -m] & \( C = \begin{bmatrix} m \\ -m\end{bmatrix} \)
1. CA = CB
2. AC = BC
3. C(A + B) = CA + CB
Which of the above statements is/are correct?
Important Questions from Operations on Matrices
The adjoint of matrix \(\left[ {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right]\)is
Consider the following in respect of matrices A, B and C of same order:
1) (A + B + C)' = A' + B’ + C’
2) (AB)’ = A’B’
3) (ABC)’ = C’B’A’
Where A’ is the transpose of the matrix A.
Which of the above are correct?If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 2}\\ 2&{ - 3}&4 \end{array}} \right]\) , then the matrix X for which 2X + 3A = 0 holds true is
Find a matrix X such that 2A + B + X = 0 , where
\(A=\begin{bmatrix} -1 & 2 \\\ 3 & 4 \end{bmatrix} \ \text{and} \;\rm B =\ \begin{bmatrix} 3 & -2 \\\ 1 & 5 \end{bmatrix} \ ?\)
The solution of the matrix equation \(\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&3\\ 1&1&1\\ 1&{ - 1}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 9\\ 6\\ 2 \end{array}} \right]\) is:
