Problem 18
Question
Let \(A\) be \(a 2 \times 2\) matrix with real entries. Let \(I\) be the \(2 \times 2\) identity matrix. Denote by \(\operatorname{tr}(A)\), the sum of diagonal entries of \(a\). Assume that \(A^{2}=I\). \(\quad\) [2008] Statement- 1 : If \(A \neq I\) and \(A \neq-I\), then \(\operatorname{det}(A)=-1\) Statement- \(2:\) If \(A \neq I\) and \(A \neq-I\), then \(\operatorname{tr}(A) \neq 0\) (a) Statement \(-1\) is false, Statement- 2 is true (b) Statement \(-1\) is true, Statement- 2 is true; Statement \(-2\) is a correct explanation for Statement- 1 (c) Statement \(-1\) is true, Statement- 2 is true; Statement \(-2\) is not a correct explanation for Statement- 1 (d) Statement \(-1\) is true, Statement- 2 is false
Step-by-Step Solution
Verified Answer
Statement 1 is true, Statement 2 is false. Option (d).
1Step 1: Analyze given conditions
We have a matrix \(A\) such that \(A^2 = I\) where \(I\) is the identity matrix. This implies that \(A\) is an involutory matrix, meaning that applying \(A\) twice results in the identity. This implies that the possible eigenvalues for \(A\) are \(+1\) and \(-1\) since \(A^2 - I = 0\).
2Step 2: Determine eigenvalues case
Given that \(A^2 = I\), the eigenvalues of \(A\) must be either \(+1\) or \(-1\). Since \(Aeq I\) and \(Aeq -I\), \(A\) cannot have both eigenvalues as \(+1\) or \(-1\) exclusively. Therefore, one eigenvalue is \(+1\) and the other is \(-1\).
3Step 3: Calculate the determinant
The determinant of a matrix is equal to the product of its eigenvalues. Therefore, if the eigenvalues of \(A\) are \(+1\) and \(-1\), then \(\text{det}(A) = (+1) \times (-1) = -1\). Statement 1 is true.
4Step 4: Calculate the trace
The trace of a matrix is the sum of its eigenvalues. With eigenvalues \(+1\) and \(-1\), \(\text{tr}(A) = 1 + (-1) = 0\). This contradicts Statement 2 which supposed \(\text{tr}(A) eq 0\). Therefore, Statement 2 is false.
5Step 5: Evaluate the correctness of explanations
Statement 1 is true based on the determinant evaluation in Step 3. Statement 2 is false as the trace was calculated to be zero. Thus, Statement -2 is not a correct explanation of Statement -1 since it is false.
Key Concepts
EigenvaluesDeterminant of matrixTrace of matrix
Eigenvalues
When dealing with square matrices in matrix algebra, eigenvalues are crucial concepts. They provide insights into a matrix's properties, such as stability and ease of computation. For a given matrix \( A \), an eigenvalue is a scalar \( \lambda \) that satisfies the equation \( A\mathbf{v} = \lambda\mathbf{v} \) where \( \mathbf{v} \) is an eigenvector of \( A \). These scalars are distinctive because they do not change the direction of the eigenvector, only its magnitude.
In the specific context where \( A^2 = I \), the matrix \( A \) is known as an involutory matrix. An important characteristic of such matrices is that their eigenvalues must be either \(+1\) or \(-1\). This is derived from the equation \( A^2 = I \), which translates to \( A^2 - I = 0 \). Solving this equation leads to potential eigenvalues of \( A \) being \(+1\) and \(-1\). Since \( A eq I \) and \( A eq -I \), \( A \) cannot have both eigenvalues as \( +1 \) or both as \( -1 \). Thus, one eigenvalue must be \(+1\) and the other \(-1\), ensuring their product results in a specific determinant value.
In the specific context where \( A^2 = I \), the matrix \( A \) is known as an involutory matrix. An important characteristic of such matrices is that their eigenvalues must be either \(+1\) or \(-1\). This is derived from the equation \( A^2 = I \), which translates to \( A^2 - I = 0 \). Solving this equation leads to potential eigenvalues of \( A \) being \(+1\) and \(-1\). Since \( A eq I \) and \( A eq -I \), \( A \) cannot have both eigenvalues as \( +1 \) or both as \( -1 \). Thus, one eigenvalue must be \(+1\) and the other \(-1\), ensuring their product results in a specific determinant value.
Determinant of matrix
The determinant of a matrix is a special number that can be calculated from its elements and has significant theoretical importance in linear algebra. For a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant \( \text{det}(A) \) is calculated using the formula \( ad - bc \).
In terms of eigenvalues, the determinant of a matrix is also the product of its eigenvalues. In our case, since the matrix \( A \) satisfies \( A^2 = I \) and has eigenvalues \( +1 \) and \( -1 \), their product is \((+1) \times (-1) = -1 \). Thus, the determinant of \( A \) is \(-1\). This confirms Statement 1 from the exercise which asserts that the determinant of the matrix is \(-1\). Determinants are critical as they are used to determine if a matrix is invertible (non-zero determinant) and are often employed in solving systems of linear equations.
In terms of eigenvalues, the determinant of a matrix is also the product of its eigenvalues. In our case, since the matrix \( A \) satisfies \( A^2 = I \) and has eigenvalues \( +1 \) and \( -1 \), their product is \((+1) \times (-1) = -1 \). Thus, the determinant of \( A \) is \(-1\). This confirms Statement 1 from the exercise which asserts that the determinant of the matrix is \(-1\). Determinants are critical as they are used to determine if a matrix is invertible (non-zero determinant) and are often employed in solving systems of linear equations.
Trace of matrix
The trace of a matrix is one of its fundamental properties, representing the sum of its eigenvalues or the sum of its diagonal elements. For a \( 2 \times 2 \) matrix \( A \), if it looks like \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), then the trace, denoted as \( \operatorname{tr}(A) \), is \( a + d \).
When using eigenvalues, the trace can be found as the sum of these values. For \( A^2 = I \), the eigenvalues are \( +1 \) and \( -1 \). Therefore, the trace is \((+1) + (-1) = 0\). This result is inconsistent with Statement 2, which incorrectly claims that the trace should not be zero. Understanding the trace is valuable as it not only provides the sum of eigenvalues but also connects to various properties of a matrix, like its overall energy and it frequently appears in characteristic equations used in advanced computations.
When using eigenvalues, the trace can be found as the sum of these values. For \( A^2 = I \), the eigenvalues are \( +1 \) and \( -1 \). Therefore, the trace is \((+1) + (-1) = 0\). This result is inconsistent with Statement 2, which incorrectly claims that the trace should not be zero. Understanding the trace is valuable as it not only provides the sum of eigenvalues but also connects to various properties of a matrix, like its overall energy and it frequently appears in characteristic equations used in advanced computations.
Other exercises in this chapter
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