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
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.
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.
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.
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.
Based on our analysis:
Therefore, statements 1 and 3 are correct.
| 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 |
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.
If \(A=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\), then what is the value of det(I + AA'), where I is the 3 × 3 identity matrix?
If \(A=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{array}\right]\), then which of the following statements are correct?
1. An will always be singular for any positive integer n.
2. An will always be a diagonal matrix for any positive integer n.
3. An will always be a symmetric matrix for any positive integer n.
Select the correct answer using the code given below:
If \(A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\), then what is 23A3 - 19A2 - 4A equal to ?
If A is an orthogonal matrix of order 3 and \({\rm{B}} = \left[ {\begin{array}{*{20}{c}} 1&2&3\\ { - 3}&0&2\\ 2&5&0 \end{array}} \right]\) , then which of the following is/are correct?
1. |AB| = ± 47
2. AB = BA
Select the correct answer using the code given below:If \({\rm{E}}\left( {\rm{\theta }} \right) = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{\sin {\rm{\theta }}}\\ { - \sin {\rm{\theta \;}}}&{\cos {\rm{\theta }}} \end{array}} \right]\) then E(α) E(β) is equal to