Consider the following in respect of the matrix \(\rm A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix}\) 1. Inverse of A does not exist 2. A 3= A 3. 3A = A 2 Which of the above are correct?
1 and 3 only
Analyzing Matrix Properties: Inverse and Powers
We are given a square matrix \( \rm A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix} \) and asked to determine the correctness of three statements regarding its properties, specifically its inverse and powers.
Evaluating Statement 1: Matrix Inverse Existence
The first statement is "Inverse of A does not exist". A square matrix has an inverse if and only if its determinant is non-zero. Let's calculate the determinant of matrix A.
The matrix A is:
| 1 | 1 | 1 |
| 1 | 1 | 1 |
| 1 | 1 | 1 |
We can calculate the determinant \( \det(\rm A) \) using the cofactor expansion along the first row:
\[ \det(\rm A) = 1 \cdot \det \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} - 1 \cdot \det \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} + 1 \cdot \det \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \] \[ \det(\rm A) = 1((1)(1) - (1)(1)) - 1((1)(1) - (1)(1)) + 1((1)(1) - (1)(1)) \] \[ \det(\rm A) = 1(1 - 1) - 1(1 - 1) + 1(1 - 1) \] \[ \det(\rm A) = 1(0) - 1(0) + 1(0) = 0 \]Since the determinant of A is \( 0 \), the matrix A is singular. A singular matrix does not have an inverse.
Alternatively, we can observe that the rows (and columns) of matrix A are identical, meaning they are linearly dependent. A matrix with linearly dependent rows or columns has a determinant of 0. Thus, the inverse does not exist.
Therefore, statement 1 is correct.
Evaluating Statement 2: Matrix Power A³ = A
The second statement is "A³ = A". To evaluate this, we first need to calculate A² and then A³.
Let's calculate A² = A \( \cdot \) A:
\[ \rm A^2 = \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix} \]Each element of A² is the dot product of a row from the first matrix and a column from the second matrix. For example, the element in the first row, first column is \( (1)(1) + (1)(1) + (1)(1) = 1 + 1 + 1 = 3 \).
Since all rows and columns of A are identical, every element in the resulting matrix A² will be the same value, which is 3.
| 3 | 3 | 3 |
| 3 | 3 | 3 |
| 3 | 3 | 3 |
So, \( \rm A^2 = \begin{bmatrix} 3 & 3 & 3\\ 3 & 3 & 3\\ 3 & 3 & 3 \end{bmatrix} \). We can notice that \( \rm A^2 = 3 \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix} = 3A \).
Now let's calculate A³ = A² \( \cdot \) A:
\[ \rm A^3 = \begin{bmatrix} 3 & 3 & 3\\ 3 & 3 & 3\\ 3 & 3 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix} \]Each element of A³ is the dot product of a row from A² and a column from A. For example, the element in the first row, first column is \( (3)(1) + (3)(1) + (3)(1) = 3 + 3 + 3 = 9 \).
Similar to A², every element in A³ will be 9.
| 9 | 9 | 9 |
| 9 | 9 | 9 |
| 9 | 9 | 9 |
So, \( \rm A^3 = \begin{bmatrix} 9 & 9 & 9\\ 9 & 9 & 9\\ 9 & 9 & 9 \end{bmatrix} \). We can notice that \( \rm A^3 = 9 \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix} = 9A \).
The statement is A³ = A. We found A³ = 9A. For A³ to equal A, we would need 9A = A, which implies 8A = 0. Since A is not the zero matrix, 8A is not the zero matrix.
Therefore, A³ is not equal to A. Statement 2 is incorrect.
Evaluating Statement 3: Matrix Equation 3A = A²
The third statement is "3A = A²". We have already calculated both 3A and A².
From evaluating statement 2, we found \( \rm A^2 = \begin{bmatrix} 3 & 3 & 3\\ 3 & 3 & 3\\ 3 & 3 & 3 \end{bmatrix} \).
Now let's calculate 3A:
\[ \rm 3A = 3 \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 \cdot 1 & 3 \cdot 1 & 3 \cdot 1\\ 3 \cdot 1 & 3 \cdot 1 & 3 \cdot 1\\ 3 \cdot 1 & 3 \cdot 1 & 3 \cdot 1 \end{bmatrix} = \begin{bmatrix} 3 & 3 & 3\\ 3 & 3 & 3\\ 3 & 3 & 3 \end{bmatrix} \]Comparing 3A and A², we see that:
\[ \rm 3A = \begin{bmatrix} 3 & 3 & 3\\ 3 & 3 & 3\\ 3 & 3 & 3 \end{bmatrix} \] \[ \rm A^2 = \begin{bmatrix} 3 & 3 & 3\\ 3 & 3 & 3\\ 3 & 3 & 3 \end{bmatrix} \]Since 3A is equal to A², statement 3 is correct.
Summary of Correct Statements
Based on our analysis:
- Statement 1: Inverse of A does not exist (Correct).
- Statement 2: A³ = A (Incorrect, A³ = 9A).
- Statement 3: 3A = A² (Correct).
Therefore, statements 1 and 3 are correct.
Revision Table: Matrix Properties Review
| Property | Condition | For Matrix A | Result |
|---|---|---|---|
| Inverse Exists | Determinant \( \ne 0 \) | \( \det(\rm A) = 0 \) | Inverse does not exist |
| Matrix Square (A²) | A \( \cdot \) A | \( \begin{bmatrix} 3 & 3 & 3\\ 3 & 3 & 3\\ 3 & 3 & 3 \end{bmatrix} \) | A² = 3A |
| Matrix Cube (A³) | A² \( \cdot \) A | \( \begin{bmatrix} 9 & 9 & 9\\ 9 & 9 & 9\\ 9 & 9 & 9 \end{bmatrix} \) | A³ = 9A |
Additional Information: Special Matrices and Properties
The matrix A in this question is an example of a matrix where all elements are the same. Such matrices have interesting properties related to powers and determinants.
- Determinant: For an n x n matrix where all elements are 'k', the determinant is \( k \cdot n^{n-1} \) if n=1, and \( 0 \) if n > 1. For our 3x3 matrix with k=1, the determinant is 0.
- Powers: For an n x n matrix A with all elements 1, \( \rm A^m = n^{m-1} A \) for any positive integer m. In this case, n=3 and the elements are 1. So, \( \rm A^m = 3^{m-1} A \).
- For m=2: \( \rm A^2 = 3^{2-1} A = 3^1 A = 3A \). This matches our calculation.
- For m=3: \( \rm A^3 = 3^{3-1} A = 3^2 A = 9A \). This also matches our calculation.
- Singular Matrix: A matrix with a determinant of zero is called a singular matrix. Singular matrices do not have an inverse. Non-singular matrices have a non-zero determinant and an inverse.
- Matrix Multiplication: Remember that matrix multiplication is not commutative (AB is generally not equal to BA). However, when multiplying a matrix by itself (powers), the order is implicit.
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Select the correct answer using the code given below:
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A = [m n], B = [-n -m] & \( C = \begin{bmatrix} m \\ -m\end{bmatrix} \)
1. CA = CB
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Important Questions from Operations on Matrices
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