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
Finding the Adjoint of a Matrix
The question asks us to find the adjoint of the given matrix \({\rm{A}}\).
The given matrix is:
\({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&0&2\\ 2&1&0\\ 0&3&1 \end{array}} \right]\)
The adjoint of a matrix \({\rm{A}}\) is defined as the transpose of the cofactor matrix of \({\rm{A}}\). Let \({\rm{C}}\) be the cofactor matrix of \({\rm{A}}\). The element \(C_{ij}\) of the cofactor matrix is given by \(C_{ij} = (-1)^{i+j} M_{ij}\), where \(M_{ij}\) is the minor of the element \(a_{ij}\) (the determinant of the submatrix obtained by deleting the i-th row and j-th column).
We need to calculate the cofactor for each element of the matrix \({\rm{A}}\).
- Cofactor \(C_{11}\): \(i=1, j=1\) \(C_{11} = (-1)^{1+1} \det \left[ {\begin{array}{*{20}{c}} 1&0\\ 3&1 \end{array}} \right] = 1 \times ((1)(1) - (0)(3)) = 1 \times (1 - 0) = 1\)
- Cofactor \(C_{12}\): \(i=1, j=2\) \(C_{12} = (-1)^{1+2} \det \left[ {\begin{array}{*{20}{c}} 2&0\\ 0&1 \end{array}} \right] = -1 \times ((2)(1) - (0)(0)) = -1 \times (2 - 0) = -2\)
- Cofactor \(C_{13}\): \(i=1, j=3\) \(C_{13} = (-1)^{1+3} \det \left[ {\begin{array}{*{20}{c}} 2&1\\ 0&3 \end{array}} \right] = 1 \times ((2)(3) - (1)(0)) = 1 \times (6 - 0) = 6\)
- Cofactor \(C_{21}\): \(i=2, j=1\) \(C_{21} = (-1)^{2+1} \det \left[ {\begin{array}{*{20}{c}} 0&2\\ 3&1 \end{array}} \right] = -1 \times ((0)(1) - (2)(3)) = -1 \times (0 - 6) = -1 \times (-6) = 6\)
- Cofactor \(C_{22}\): \(i=2, j=2\) \(C_{22} = (-1)^{2+2} \det \left[ {\begin{array}{*{20}{c}} 1&2\\ 0&1 \end{array}} \right] = 1 \times ((1)(1) - (2)(0)) = 1 \times (1 - 0) = 1\)
- Cofactor \(C_{23}\): \(i=2, j=3\) \(C_{23} = (-1)^{2+3} \det \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&3 \end{array}} \right] = -1 \times ((1)(3) - (0)(0)) = -1 \times (3 - 0) = -3\)
- Cofactor \(C_{31}\): \(i=3, j=1\) \(C_{31} = (-1)^{3+1} \det \left[ {\begin{array}{*{20}{c}} 0&2\\ 1&0 \end{array}} \right] = 1 \times ((0)(0) - (2)(1)) = 1 \times (0 - 2) = -2\)
- Cofactor \(C_{32}\): \(i=3, j=2\) \(C_{32} = (-1)^{3+2} \det \left[ {\begin{array}{*{20}{c}} 1&2\\ 2&0 \end{array}} \right] = -1 \times ((1)(0) - (2)(2)) = -1 \times (0 - 4) = -1 \times (-4) = 4\)
- Cofactor \(C_{33}\): \(i=3, j=3\) \(C_{33} = (-1)^{3+3} \det \left[ {\begin{array}{*{20}{c}} 1&0\\ 2&1 \end{array}} \right] = 1 \times ((1)(1) - (0)(2)) = 1 \times (1 - 0) = 1\)
Now, we can form the cofactor matrix \({\rm{C}}\) using these calculated cofactors:
\({\rm{C}} = \left[ {\begin{array}{*{20}{c}} {C_{11}}&{C_{12}}&{C_{13}}\\ {C_{21}}&{C_{22}}&{C_{23}}\\ {C_{31}}&{C_{32}}&{C_{33}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&6\\ 6&1&{ - 3}\\ { - 2}&4&1 \end{array}} \right]\)
The adjoint of matrix \({\rm{A}}\), denoted as \(\text{adj}({\rm{A}})\), is the transpose of the cofactor matrix \({\rm{C}}\). The transpose of a matrix is obtained by swapping its rows and columns.
\(\text{adj}({\rm{A}}) = {\rm{C}}^T = \left[ {\begin{array}{*{20}{c}} 1&6&{ - 2}\\ { - 2}&1&4\\ 6&{ - 3}&1 \end{array}} \right]\)
Comparing this result with the given options, we see that it matches option 2.
| Cofactor | Calculation | Value |
|---|---|---|
| \(C_{11}\) | \((-1)^{1+1} \det \left[ {\begin{array}{*{20}{c}} 1&0\\ 3&1 \end{array}} \right]\) | 1 |
| \(C_{12}\) | \((-1)^{1+2} \det \left[ {\begin{array}{*{20}{c}} 2&0\\ 0&1 \end{array}} \right]\) | -2 |
| \(C_{13}\) | \((-1)^{1+3} \det \left[ {\begin{array}{*{20}{c}} 2&1\\ 0&3 \end{array}} \right]\) | 6 |
| \(C_{21}\) | \((-1)^{2+1} \det \left[ {\begin{array}{*{20}{c}} 0&2\\ 3&1 \end{array}} \right]\) | 6 |
| \(C_{22}\) | \((-1)^{2+2} \det \left[ {\begin{array}{*{20}{c}} 1&2\\ 0&1 \end{array}} \right]\) | 1 |
| \(C_{23}\) | \((-1)^{2+3} \det \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&3 \end{array}} \right]\) | -3 |
| \(C_{31}\) | \((-1)^{3+1} \det \left[ {\begin{array}{*{20}{c}} 0&2\\ 1&0 \end{array}} \right]\) | -2 |
| \(C_{32}\) | \((-1)^{3+2} \det \left[ {\begin{array}{*{20}{c}} 1&2\\ 2&0 \end{array}} \right]\) | 4 |
| \(C_{33}\) | \((-1)^{3+3} \det \left[ {\begin{array}{*{20}{c}} 1&0\\ 2&1 \end{array}} \right]\) | 1 |
Revision Table: Matrix Adjoint Calculation
| Step | Description |
|---|---|
| 1 | Calculate the minor \(M_{ij}\) for each element \(a_{ij}\). |
| 2 | Calculate the cofactor \(C_{ij} = (-1)^{i+j} M_{ij}\) for each element. |
| 3 | Form the cofactor matrix \({\rm{C}}\) using \(C_{ij}\) as elements. |
| 4 | Find the transpose of the cofactor matrix \({\rm{C}}^T\). This is the adjoint matrix \(\text{adj}({\rm{A}})\). |
Additional Information: Properties of Matrix Adjoint
The adjoint of a matrix has several important properties:
- For a square matrix \({\rm{A}}\) of order \(n\), \({\rm{A}} (\text{adj}({\rm{A}})) = (\text{adj}({\rm{A}})) {\rm{A}} = \det({\rm{A}}) I_n\), where \(I_n\) is the identity matrix of order \(n\).
- A matrix \({\rm{A}}\) is invertible if and only if \(\det({\rm{A}}) \neq 0\).
- If \({\rm{A}}\) is invertible, then \({\rm{A}}^{-1} = \frac{1}{\det({\rm{A}})} \text{adj}({\rm{A}})\).
- The adjoint of the transpose is the transpose of the adjoint: \(\text{adj}({\rm{A}}^T) = (\text{adj}({\rm{A}}))^T\).
- The adjoint of the product of two matrices \({\rm{A}}\) and \({\rm{B}}\) (if they are square and of the same order) is given by \(\text{adj}({\rm{AB}}) = \text{adj}({\rm{B}}) \text{adj}({\rm{A}})\).
These properties are fundamental in linear algebra and are often used to find the inverse of a matrix or solve systems of linear equations.
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