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?
Finding the Adjoint of a Matrix: A Step-by-Step Guide
The question asks us to find the adjoint of the given matrix B.
The given matrix is:
\(B = \left[ {\begin{array}{*{20}{c}} 3&2&0\\ 2&4&0\\ 1&1&0 \end{array}} \right]\)
The adjoint of a square matrix is the transpose of its cofactor matrix. To find the adjoint of matrix B, we first need to calculate the cofactor matrix of B.
Steps to Calculate the Adjoint Matrix
Find the minor for each element of the matrix.
Find the cofactor for each element using the minors.
Form the cofactor matrix.
Transpose the cofactor matrix to get the adjoint matrix.
Calculating Minors of Matrix B
The minor \(M_{ij}\) of an element \(b_{ij}\) is the determinant of the submatrix obtained by deleting the i-th row and j-th column of the original matrix B.
\(M_{11}\): Delete row 1 and column 1. Submatrix is \(\left[ {\begin{array}{*{20}{c}} 4&0\\ 1&0 \end{array}} \right]\). Minor \(M_{11} = \det\left[ {\begin{array}{*{20}{c}} 4&0\\ 1&0 \end{array}} \right] = (4 \times 0) - (0 \times 1) = 0 - 0 = 0\).
\(M_{12}\): Delete row 1 and column 2. Submatrix is \(\left[ {\begin{array}{*{20}{c}} 2&0\\ 1&0 \end{array}} \right]\). Minor \(M_{12} = \det\left[ {\begin{array}{*{20}{c}} 2&0\\ 1&0 \end{array}} \right] = (2 \times 0) - (0 \times 1) = 0 - 0 = 0\).
\(M_{13}\): Delete row 1 and column 3. Submatrix is \(\left[ {\begin{array}{*{20}{c}} 2&4\\ 1&1 \end{array}} \right]\). Minor \(M_{13} = \det\left[ {\begin{array}{*{20}{c}} 2&4\\ 1&1 \end{array}} \right] = (2 \times 1) - (4 \times 1) = 2 - 4 = -2\).
\(M_{21}\): Delete row 2 and column 1. Submatrix is \(\left[ {\begin{array}{*{20}{c}} 2&0\\ 1&0 \end{array}} \right]\). Minor \(M_{21} = \det\left[ {\begin{array}{*{20}{c}} 2&0\\ 1&0 \end{array}} \right] = (2 \times 0) - (0 \times 1) = 0 - 0 = 0\).
\(M_{22}\): Delete row 2 and column 2. Submatrix is \(\left[ {\begin{array}{*{20}{c}} 3&0\\ 1&0 \end{array}} \right]\). Minor \(M_{22} = \det\left[ {\begin{array}{*{20}{c}} 3&0\\ 1&0 \end{array}} \right] = (3 \times 0) - (0 \times 1) = 0 - 0 = 0\).
\(M_{23}\): Delete row 2 and column 3. Submatrix is \(\left[ {\begin{array}{*{20}{c}} 3&2\\ 1&1 \end{array}} \right]\). Minor \(M_{23} = \det\left[ {\begin{array}{*{20}{c}} 3&2\\ 1&1 \end{array}} \right] = (3 \times 1) - (2 \times 1) = 3 - 2 = 1\).
\(M_{31}\): Delete row 3 and column 1. Submatrix is \(\left[ {\begin{array}{*{20}{c}} 2&0\\ 4&0 \end{array}} \right]\). Minor \(M_{31} = \det\left[ {\begin{array}{*{20}{c}} 2&0\\ 4&0 \end{array}} \right] = (2 \times 0) - (0 \times 4) = 0 - 0 = 0\).
\(M_{32}\): Delete row 3 and column 2. Submatrix is \(\left[ {\begin{array}{*{20}{c}} 3&0\\ 2&0 \end{array}} \right]\). Minor \(M_{32} = \det\left[ {\begin{array}{*{20}{c}} 3&0\\ 2&0 \end{array}} \right] = (3 \times 0) - (0 \times 2) = 0 - 0 = 0\).
\(M_{33}\): Delete row 3 and column 3. Submatrix is \(\left[ {\begin{array}{*{20}{c}} 3&2\\ 2&4 \end{array}} \right]\). Minor \(M_{33} = \det\left[ {\begin{array}{*{20}{c}} 3&2\\ 2&4 \end{array}} \right] = (3 \times 4) - (2 \times 2) = 12 - 4 = 8\).
Calculating Cofactors of Matrix B
The cofactor \(C_{ij}\) of an element \(b_{ij}\) is given by \(C_{ij} = (-1)^{i+j} M_{ij}\).
\(C_{11} = (-1)^{1+1} M_{11} = (+1) \times 0 = 0\).
\(C_{12} = (-1)^{1+2} M_{12} = (-1) \times 0 = 0\).
\(C_{13} = (-1)^{1+3} M_{13} = (+1) \times (-2) = -2\).
\(C_{21} = (-1)^{2+1} M_{21} = (-1) \times 0 = 0\).
\(C_{22} = (-1)^{2+2} M_{22} = (+1) \times 0 = 0\).
\(C_{23} = (-1)^{2+3} M_{23} = (-1) \times 1 = -1\).
\(C_{31} = (-1)^{3+1} M_{31} = (+1) \times 0 = 0\).
\(C_{32} = (-1)^{3+2} M_{32} = (-1) \times 0 = 0\).
\(C_{33} = (-1)^{3+3} M_{33} = (+1) \times 8 = 8\).
Forming the Cofactor Matrix
The cofactor matrix, denoted by C, is formed by placing the calculated cofactors in their corresponding positions:
| Col 1 | Col 2 | Col 3 | |
|---|---|---|---|
| Row 1 | \(C_{11} = 0\) | \(C_{12} = 0\) | \(C_{13} = -2\) |
| Row 2 | \(C_{21} = 0\) | \(C_{22} = 0\) | \(C_{23} = -1\) |
| Row 3 | \(C_{31} = 0\) | \(C_{32} = 0\) | \(C_{33} = 8\) |
So the cofactor matrix C is:
\(C = \left[ {\begin{array}{*{20}{c}} 0&0&-2\\ 0&0&-1\\ 0&0&8 \end{array}} \right]\)
Calculating the Adjoint Matrix
The adjoint of matrix B, written as adj(B), is the transpose of the cofactor matrix C (\(C^T\)). Transposing a matrix means swapping its rows and columns.
The first row of C becomes the first column of adj(B), the second row of C becomes the second column of adj(B), and the third row of C becomes the third column of adj(B).
adj(B) = \(C^T = \left[ {\begin{array}{*{20}{c}} 0&0&0\\ 0&0&0\\ -2&-1&8 \end{array}} \right]\)
Comparing with Options
Let's compare our calculated adjoint matrix with the given options:
Option 1: \(\left[ {\begin{array}{*{20}{c}} 0&0&0\\ 0&0&0\\ { - 2}&{ - 1}&8 \end{array}} \right]\)
Option 2: \(\left[ {\begin{array}{*{20}{c}} 0&0&{ - 2}\\ 0&0&{ - 1}\\ 0&0&8 \end{array}} \right]\)
Option 3: \(\left[ {\begin{array}{*{20}{c}} 0&0&2\\ 0&0&1\\ 0&0&0 \end{array}} \right]\)
Option 4: It does not exist
Our calculated adjoint matrix matches Option 1.
Revision Table: Matrix Adjoint Calculation
| Concept | Definition/Method | Application in this Problem |
|---|---|---|
| Minor \(M_{ij}\) | Determinant of submatrix after removing row i and col j. | Calculated \(M_{11}, \ldots, M_{33}\) from matrix B. |
| Cofactor \(C_{ij}\) | \(C_{ij} = (-1)^{i+j} M_{ij}\) | Calculated \(C_{11}, \ldots, C_{33}\) using the minors. |
| Cofactor Matrix | Matrix [\(C_{ij}\)] | Formed matrix C from calculated cofactors. |
| Adjoint Matrix adj(B) | Transpose of the Cofactor Matrix (\(C^T\)) | Transposed matrix C to get adj(B). |
Additional Information: Properties of Adjoint Matrix
The adjoint matrix is defined for any square matrix.
A property relating a matrix, its adjoint, and its determinant is \(A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = \det(A) \cdot I\), where I is the identity matrix.
A matrix A is invertible if and only if \(\det(A) \neq 0\). If A is invertible, its inverse is given by \(A^{-1} = \frac{1}{\det(A)} \text{adj}(A)\).
In this specific problem, the determinant of matrix B is \(\det(B) = 3(0-0) - 2(0-0) + 0(2-4) = 0\). Since \(\det(B) = 0\), the matrix B is singular and does not have an inverse. However, the adjoint of a singular matrix still exists and can be calculated as shown above.
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