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:
2 and 3 only
The question asks us to examine the properties of \(A^n\), where \(A\) is a given matrix and \(n\) is a positive integer. The matrix \(A\) is given as:
\(A=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{array}\right]\)
This is a special type of matrix known as a diagonal matrix because all the non-diagonal elements are zero.
When we raise a diagonal matrix to a positive integer power \(n\), we simply raise each diagonal element to that power. For the given matrix \(A\), \(A^n\) is:
\(A^n = \left[\begin{array}{lll} 2^n & 0 & 0 \\ 0 & 3^n & 0 \\ 0 & 0 & 4^n \end{array}\right]\)
Now let's evaluate each statement provided in the question.
A matrix is singular if its determinant is equal to zero. Let's find the determinant of \(A^n\).
The determinant of a diagonal matrix is the product of its diagonal elements.
\(\det(A^n) = 2^n \times 3^n \times 4^n\)
We can also express this as:
\(\det(A^n) = (2 \times 3 \times 4)^n = 24^n\)
For any positive integer \(n\), \(24^n\) will always be a positive number, and thus it will never be equal to zero.
Since the determinant of \(A^n\) is never zero, \(A^n\) is always non-singular (or invertible) for any positive integer \(n\).
Therefore, Statement 1 is incorrect.
A diagonal matrix is a square matrix where all elements outside the main diagonal are zero. The elements on the main diagonal can be any value.
We found that \(A^n = \left[\begin{array}{lll} 2^n & 0 & 0 \\ 0 & 3^n & 0 \\ 0 & 0 & 4^n \end{array}\right]\).
In this matrix, the elements at positions \((1,2), (1,3), (2,1), (2,3), (3,1), (3,2)\) are all zero. The elements on the main diagonal are \(2^n, 3^n, 4^n\).
By the definition of a diagonal matrix, \(A^n\) is indeed a diagonal matrix for any positive integer \(n\).
Therefore, Statement 2 is correct.
A square matrix \(M\) is called symmetric if it is equal to its transpose, i.e., \(M = M^T\). The transpose of a matrix is obtained by swapping its rows and columns.
Let's find the transpose of \(A^n\):
\((A^n)^T = \left[\begin{array}{lll} 2^n & 0 & 0 \\ 0 & 3^n & 0 \\ 0 & 0 & 4^n \end{array}\right]^T\)
Swapping rows and columns:
\((A^n)^T = \left[\begin{array}{lll} 2^n & 0 & 0 \\ 0 & 3^n & 0 \\ 0 & 0 & 4^n \end{array}\right]\)
We can see that \((A^n)^T\) is exactly the same as \(A^n\).
Since \(A^n = (A^n)^T\), \(A^n\) is always a symmetric matrix for any positive integer \(n\).
Therefore, Statement 3 is correct.
The correct statements are 2 and 3.
Based on the analysis of each statement, the statements that are correct are 2 and 3.
| Statement | Analysis Result |
|---|---|
| 1. \(A^n\) is always singular | Incorrect (Determinant is \(24^n \ne 0\)) |
| 2. \(A^n\) is always a diagonal matrix | Correct (Structure remains diagonal) |
| 3. \(A^n\) is always a symmetric matrix | Correct (Transpose is the same) |
| Matrix Type / Property | Definition | Check for \(A^n\) |
|---|---|---|
| Singular Matrix | Determinant is 0 | \(\det(A^n) = 24^n \ne 0\). Not singular. |
| Diagonal Matrix | Non-diagonal elements are zero | \(A^n = \left[\begin{array}{lll} 2^n & 0 & 0 \\ 0 & 3^n & 0 \\ 0 & 0 & 4^n \end{array}\right]\). Is diagonal. |
| Symmetric Matrix | Matrix is equal to its transpose (\(M = M^T\)) | \((A^n)^T = A^n\). Is symmetric. |
Diagonal Matrices:
Symmetric Matrices:
Understanding these basic matrix properties helps in solving problems involving matrix operations and powers.
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
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 A is a 2 × 3 matrix and AB is a 2 × 5 matrix, then B must be a
The adjoint of matrix \(\left[ {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right]\)is
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
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
\(A=\begin{bmatrix} -1 & 2 \\\ 3 & 4 \end{bmatrix} \ \text{and} \;\rm B =\ \begin{bmatrix} 3 & -2 \\\ 1 & 5 \end{bmatrix} \ ?\)