The adjoint of matrix \(\left[ {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right]\)is
Adjoint Matrix Definition
The adjoint of a square matrix is a crucial concept in linear algebra. It plays a significant role in various matrix operations, particularly in finding the inverse of a matrix. For a square matrix \(A\), its adjoint, denoted as \(adj(A)\), is defined as the transpose of its cofactor matrix.
While the general procedure for finding the adjoint involves calculating all cofactors and then transposing the resulting cofactor matrix, for a 2x2 matrix, there is a very straightforward and direct formula.
Adjoint of a 2x2 Matrix Formula
Consider a generic 2x2 matrix \(A\) expressed as:
$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$
To find the adjoint of this 2x2 matrix, \(adj(A)\), you perform two simple operations:
- Swap the elements on the main diagonal: The elements \(a\) and \(d\) swap their positions.
- Change the signs of the elements on the off-diagonal: The elements \(b\) and \(c\) remain in their positions, but their signs are reversed (i.e., \(b\) becomes \(-b\) and \(c\) becomes \(-c\)).
Applying these rules, the adjoint matrix for a 2x2 matrix \(A\) is given by the formula:
$$ adj(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} $$
Calculating Adjoint for the Given Matrix
The question asks us to find the adjoint of the matrix:
$$ \begin{bmatrix} a&b\\ c&d \end{bmatrix} $$
Let's refer to this given matrix as \(M\):
$$ M = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$
Now, we will apply the direct formula for the adjoint of a 2x2 matrix:
- The elements on the main diagonal are \(a\) and \(d\). Swapping these elements gives \(d\) in the top-left position and \(a\) in the bottom-right position.
- The elements on the off-diagonal are \(b\) and \(c\). Changing their signs gives \(-b\) in the top-right position and \(-c\) in the bottom-left position.
Therefore, the adjoint of matrix \(M\) is calculated as:
$$ adj(M) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} $$
Comparing with Options
Let's compare our calculated adjoint matrix with the provided options to identify the correct one:
- Option 1: $ \begin{bmatrix} a&b\\ c&d \end{bmatrix} $ - This is the original matrix itself, not its adjoint.
- Option 2: $ \begin{bmatrix} -a&{-b}\\ -c&{-d} \end{bmatrix} $ - This matrix results from negating all elements of the original matrix, which is not the definition of an adjoint.
- Option 3: $ \begin{bmatrix} d&{-b}\\ -c&a \end{bmatrix} $ - This matrix exactly matches our calculated adjoint matrix. The main diagonal elements (\(a\) and \(d\)) are swapped, and the off-diagonal elements (\(b\) and \(c\)) have their signs changed.
- Option 4: $ \begin{bmatrix} d&b\\ c&a \end{bmatrix} $ - In this matrix, only the main diagonal elements are swapped, but the signs of the off-diagonal elements are not changed. This is incorrect for the adjoint.
Thus, based on the definition and calculation, Option 3 is the correct representation of the adjoint of the given matrix.
Important Questions from Operations on Matrices
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
\(A=\begin{bmatrix} -1 & 2 \\\ 3 & 4 \end{bmatrix} \ \text{and} \;\rm B =\ \begin{bmatrix} 3 & -2 \\\ 1 & 5 \end{bmatrix} \ ?\)
The solution of the matrix equation \(\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&3\\ 1&1&1\\ 1&{ - 1}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 9\\ 6\\ 2 \end{array}} \right]\) is:
