If A is an identity matrix of order 3, then its inverse (A -1 )
is equal to A
This question asks about the inverse of an identity matrix of order 3. Let's first understand what an identity matrix is and what an inverse of a matrix means.
An identity matrix, usually denoted by \( I \), is a square matrix where all the elements on the main diagonal are 1, and all other elements are 0. The order of the identity matrix is the same as the number of rows or columns. For example, an identity matrix of order 3 is a 3x3 matrix:
\( I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \)
When you multiply any square matrix \( A \) by an identity matrix of the same order, the result is the matrix \( A \) itself. That is, \( A \times I = I \times A = A \).
For a square matrix \( A \), its inverse, denoted by \( A^{-1} \), is a matrix such that when \( A \) is multiplied by \( A^{-1} \) (in either order), the result is the identity matrix \( I \). So, if \( A^{-1} \) exists, we have:
\( A \times A^{-1} = A^{-1} \times A = I \)
Not all square matrices have an inverse. A matrix that has an inverse is called an invertible or non-singular matrix.
We are looking for the inverse of the identity matrix \( A = I_3 \). Let's call its inverse \( A^{-1} \). By the definition of the inverse, we must have:
\( A \times A^{-1} = I \)
Substituting \( A = I_3 \), we get:
\( I_3 \times A^{-1} = I_3 \)
From the property of the identity matrix, we know that \( I_3 \times X = X \) for any matrix \( X \) that can be multiplied by \( I_3 \). Comparing this property with the equation \( I_3 \times A^{-1} = I_3 \), it follows that \( A^{-1} \) must be equal to \( I_3 \).
Also, consider the second part of the definition: \( A^{-1} \times A = I \).
Substituting \( A = I_3 \), we get:
\( A^{-1} \times I_3 = I_3 \)
Again, using the property that \( X \times I_3 = X \) for any matrix \( X \), it follows that \( A^{-1} \) must be equal to \( I_3 \).
Therefore, the inverse of an identity matrix of any order is the identity matrix of the same order itself. For an identity matrix \( A \) of order 3, its inverse \( A^{-1} \) is equal to \( A \).
Let's examine the given options in light of our finding:
| Option | Analysis |
|---|---|
| is equal to null matrix | The null matrix (a matrix with all zeros) multiplied by any non-null matrix gives the null matrix, not the identity matrix. So, this is incorrect. |
| is equal to A | We found that the inverse of the identity matrix \( A \) is \( A \) itself. This option matches our conclusion. |
| is equal to 3A | If \( A = I_3 \), then \( 3A = 3I_3 = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} \). \( A \times (3A) = I_3 \times 3I_3 = 3I_3 \). This is not equal to \( I_3 \). So, this is incorrect. |
| does not exist | The identity matrix is always invertible because its determinant is 1 (non-zero). Therefore, its inverse exists. So, this is incorrect. |
Based on the analysis, the inverse of an identity matrix of order 3 is equal to the matrix itself, which is \( A \).
| Concept | Description | Property related to Inverse |
|---|---|---|
| Identity Matrix (\( I \)) | Square matrix with 1s on the main diagonal, 0s elsewhere. | \( A \times I = I \times A = A \) for any compatible matrix \( A \). |
| Inverse Matrix (\( A^{-1} \)) | For a square matrix \( A \), \( A^{-1} \) satisfies \( A \times A^{-1} = A^{-1} \times A = I \). | Exists only for non-singular matrices (determinant is non-zero). |
| Inverse of Identity Matrix | The matrix \( X \) such that \( I \times X = X \times I = I \). | The identity matrix itself satisfies this: \( I \times I = I \times I = I \). Thus, \( I^{-1} = I \). |
Here are some more key points about matrix inverses:
A matrix has an inverse if and only if its determinant is non-zero. This is a crucial condition for a matrix to be invertible.
The inverse of an inverse is the original matrix: \( (A^{-1})^{-1} = A \).
The inverse of a product of invertible matrices is the product of their inverses in reverse order: \( (AB)^{-1} = B^{-1} A^{-1} \).
Identity matrices are always invertible because their determinant is always 1.
The concept of matrix inverse is fundamental in solving systems of linear equations and in various areas of mathematics, physics, and engineering.
What should be the value of x so that the matrix \(\left( {\begin{array}{*{20}{c}} 2&4\\ { - 8}&{\rm{x}} \end{array}} \right)\) does not have an inverse?
What is the adjoint of the matrix \(\left( {\begin{array}{*{20}{c}} {\cos \left( { - \theta } \right)}&{ - \sin \left( { - \theta } \right)}\\ { - \sin \left( { - \theta } \right)}&{\cos \left( { - \theta } \right)} \end{array}} \right)\) ?
What is the inverse of the matrix?
\(A = \left( {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta }&0\\ { - \sin \theta }&{\cos \theta }&0\\ 0&0&1 \end{array}} \right)\)
If A and B are two invertible square matrices of same order, then what is (AB) -1 equal to?
If \(B = \left[ {\begin{array}{*{20}{c}} 3&2&0\\ 2&4&0\\ 1&1&0 \end{array}} \right]\) , then what is adjoint of B equal to?
For a square matrix A, which of the following properties hold?
1) (A -1 )-1 = A
2) \(\det \left( {{A^{ - 1}}} \right) = \frac{1}{{detA}}\)
3) (λA) -1 = λA -1 where λ is a scalar
Select the correct answer using the code given below:The adjoint of the matrix \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&0&2\\ 2&1&0\\ 0&3&1 \end{array}} \right]\) is
If A is a square matrix, then what is adj (A -1 ) – (adj A) -1 equal to?
The matrix \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&3&2\\ 1&{{\rm{x}} - 1}&1\\ 2&7&{{\rm{x}} - 3} \end{array}} \right]\)
Will have inverse for every real number x except forIf $A = \begin{bmatrix} 2 & 7 \\ 1 & 5 \end{bmatrix}$, then what is $A + 3A^{-1}$ equal to, where $A$ is a matrix of order 2?
What should be the value of x so that the matrix \(\left( {\begin{array}{*{20}{c}} 2&4\\ { - 8}&{\rm{x}} \end{array}} \right)\) does not have an inverse?
What is the adjoint of the matrix \(\left( {\begin{array}{*{20}{c}} {\cos \left( { - \theta } \right)}&{ - \sin \left( { - \theta } \right)}\\ { - \sin \left( { - \theta } \right)}&{\cos \left( { - \theta } \right)} \end{array}} \right)\) ?
The inverse of the matrix A = \(\left( {\begin{array}{} 1&1&3\\ 1&3&{ - 3}\\ { - 2}&{ - 4}&{ - 4} \end{array}} \right)\) is:
What is the inverse of the matrix?
\(A = \left( {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta }&0\\ { - \sin \theta }&{\cos \theta }&0\\ 0&0&1 \end{array}} \right)\)
If A and B are two invertible square matrices of same order, then what is (AB) -1 equal to?