Problem 65
Question
Let \(A=\left(\begin{array}{ccc}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{array}\right)\). The only correct statement about the matrix \(A\) is (A) \(A\) is a zero matrix (B) \(A^{2}=I\) (C) \(A^{-1}\) does not exist (D) \(A=(-1) I\), where \(I\) is a unit matrix
Step-by-Step Solution
Verified Answer
(B) \(A^2 = I\).
1Step 1: Identify the Elements of Matrix A
Matrix \(A\) is given as:\[A = \begin{pmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{pmatrix}\]It is a 3x3 matrix with specific entries.
2Step 2: Evaluate Each Statement
We need to check each provided statement (A, B, C, D) one by one to determine which one is true.
3Step 3: Check if A is a Zero Matrix (Option A)
A zero matrix has all its elements as zero. Clearly, matrix \(A\) is not a zero matrix because it contains non-zero elements like \(-1\). Thus, Option A is false.
4Step 4: Check if A^2 = I (Option B)
Calculate \(A^2\) and check if it equals the identity matrix \(I\):\[A^2 = \begin{pmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} = I\]Since \(A^2 = I\), Option B is correct.
5Step 5: Check if A^-1 Does Not Exist (Option C)
Since \(A^2 = I\), it implies \(A\) is invertible and its inverse is \(A\) itself. Therefore, \(A^{-1}\) exists, making Option C false.
6Step 6: Check if A = (-1)I (Option D)
Matrix \((-1)I\) should have \(-1\) along the diagonal and zeros elsewhere:\[(-1)I = \begin{pmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{pmatrix}\]Since \(A\) looks different than \((-1)I\), with distinct non-diagonal elements, Option D is false.
Key Concepts
Inverse of a MatrixIdentity MatrixZero Matrix
Inverse of a Matrix
The inverse of a matrix is an important concept in linear algebra. For a square matrix to have an inverse, it must be invertible or non-singular.
In simple terms, if there exists another matrix that, when multiplied with the original matrix, results in the identity matrix, then that original matrix has an inverse.
Here's how it works:
In simple terms, if there exists another matrix that, when multiplied with the original matrix, results in the identity matrix, then that original matrix has an inverse.
Here's how it works:
- For matrix \(A\), there is an inverse \(A^{-1}\) such that \(A \cdot A^{-1} = I\) and \(A^{-1} \cdot A = I\).
- The identity matrix \(I\) is the equivalent of number 1 in matrices, meaning it does not change the matrix it's multiplied by.
- If no such inverse exists, then the matrix is singular.
Identity Matrix
An identity matrix is a special type of square matrix that acts as the multiplicative identity for matrices, meaning any matrix multiplied by it remains unchanged.
The identity matrix \(I\) for a 3x3 matrix looks like this:
The identity matrix \(I\) for a 3x3 matrix looks like this:
- It's a diagonal matrix with ones on its main diagonal and zeros elsewhere.
- Mathematically represented as \(I = \begin{pmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{pmatrix}\).
Zero Matrix
A zero matrix is a matrix where all elements are zero.
It serves as the additive identity in matrix analysis, which means when you add any matrix to a zero matrix, the original matrix is unchanged.
Key features of a zero matrix:
Thus, statement (A) is false. Understanding zero matrices helps emphasize the unique properties of different types of matrices in linear algebra.
It serves as the additive identity in matrix analysis, which means when you add any matrix to a zero matrix, the original matrix is unchanged.
Key features of a zero matrix:
- All entries are zero, and its size can vary.
- A zero matrix is often denoted by \(O\).
Thus, statement (A) is false. Understanding zero matrices helps emphasize the unique properties of different types of matrices in linear algebra.
Other exercises in this chapter
Problem 62
The rank of the matrix \(A=\left[\begin{array}{rrrr}1 & 3 & 4 & 3 \\ 3 & 9 & 12 & 9 \\ -1 & -3 & -4 & -3\end{array}\right]\) is (A) 1 (B) 2 (C) 3 (D) 0
View solution Problem 64
If \(A=\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]\) and \(A^{2}=\left[\begin{array}{ll}\alpha & \beta \\ \beta & \alpha\end{array}\right]\), then (
View solution Problem 66
Let \(A=\left(\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right)\) (10) \(B=\left(\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -
View solution Problem 67
If \(A^{2}-A+I=0\), then the inverse of \(A\) is (A) \(A+I\) (B) \(\bar{A}\) (C) \(A-I\) (D) \(I-A\)
View solution