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Adjoint and Inverse of a Matrix
😊 Hello students! The adjoint and inverse are powerful tools that let us “divide” by a matrix—something we can’t do directly. Today, we’ll learn how to compute them, when they exist, and why they matter in solving systems, computing transformations, and more!
1. Adjoint of a Matrix
Let \( A = (a_{ij}) \) be an \( n \times n \) matrix. The cofactor matrix of \( A \) is the matrix whose \( (i,j) \)-entry is the cofactor \( C_{ij} \).
The adjoint of \( A \), denoted \( \operatorname{adj}(A) \), is the **transpose** of the cofactor matrix.
Let \[ A = \begin{bmatrix} 2 & 3 & 4 \\ 1 & 0 & 1 \\ 3 & 2 & 1 \end{bmatrix} \]
Compute cofactors:
- \( C_{11} = \det\begin{bmatrix}0&1\\2&1\end{bmatrix} = -2 \)
- \( C_{12} = -\det\begin{bmatrix}1&1\\3&1\end{bmatrix} = 2 \)
- \( C_{13} = \det\begin{bmatrix}1&0\\3&2\end{bmatrix} = 2 \)
- \( C_{21} = -\det\begin{bmatrix}3&4\\2&1\end{bmatrix} = 5 \)
- \( C_{22} = \det\begin{bmatrix}2&4\\3&1\end{bmatrix} = -10 \)
- \( C_{23} = -\det\begin{bmatrix}2&3\\3&2\end{bmatrix} = 5 \)
- \( C_{31} = \det\begin{bmatrix}3&4\\0&1\end{bmatrix} = 3 \)
- \( C_{32} = -\det\begin{bmatrix}2&4\\1&1\end{bmatrix} = 2 \)
- \( C_{33} = \det\begin{bmatrix}2&3\\1&0\end{bmatrix} = -3 \)
Cofactor matrix: \[ \begin{bmatrix} -2 & 2 & 2 \\ 5 & -10 & 5 \\ 3 & 2 & -3 \end{bmatrix} \]
Adjoint = transpose of cofactor matrix:
For any square matrix \( A \):
This is the bridge between adjoint and inverse!
2. Inverse of a Matrix
A square matrix \( A \) is invertible (or non-singular) if there exists a matrix \( A^{-1} \) such that: \[ A A^{-1} = A^{-1} A = I \]
The inverse is given by: \[ A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A), \quad \text{provided } \det(A) \ne 0 \]
✅ **Existence**: \( A^{-1} \) exists **iff** \( \det(A) \ne 0 \).
❌ If \( \det(A) = 0 \), \( A \) is **singular**—no inverse exists.
Using the matrix \( A \) above:
Compute \( \det(A) \):
Since \( \det(A) = 10 \ne 0 \), inverse exists:
3. Why It Matters
- Solving systems: \( A\mathbf{x} = \mathbf{b} \Rightarrow \mathbf{x} = A^{-1}\mathbf{b} \)
- Change of basis in vector spaces
- Computer graphics: undoing transformations
- Statistics: covariance matrix inversion
📌 Key Takeaways:
- Adjoint = transpose of cofactor matrix
- \( A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) \) if \( \det(A) \ne 0 \)
- If \( \det(A) = 0 \), no inverse exists
- \( A A^{-1} = I \) only for square, non-singular matrices
Adjoint and Inverse of a Matrix – Questions & Full Solutions
😊 Hello students! Below are all the adjoint and inverse questions from your documents, each with a complete, detailed solution. Master these to solve systems, understand invertibility, and prepare for exams!
Q1. Find \( \operatorname{adj}(A) \) if \[ A = \begin{bmatrix} 2 & 3 & 4 \\ 1 & 0 & 1 \\ 3 & 2 & 1 \end{bmatrix} \]
Solution:
Step 1: Compute cofactors \( C_{ij} = (-1)^{i+j} M_{ij} \):
- \( C_{11} = +\det\begin{bmatrix}0&1\\2&1\end{bmatrix} = -2 \)
- \( C_{12} = -\det\begin{bmatrix}1&1\\3&1\end{bmatrix} = 2 \)
- \( C_{13} = +\det\begin{bmatrix}1&0\\3&2\end{bmatrix} = 2 \)
- \( C_{21} = -\det\begin{bmatrix}3&4\\2&1\end{bmatrix} = 5 \)
- \( C_{22} = +\det\begin{bmatrix}2&4\\3&1\end{bmatrix} = -10 \)
- \( C_{23} = -\det\begin{bmatrix}2&3\\3&2\end{bmatrix} = 5 \)
- \( C_{31} = +\det\begin{bmatrix}3&4\\0&1\end{bmatrix} = 3 \)
- \( C_{32} = -\det\begin{bmatrix}2&4\\1&1\end{bmatrix} = 2 \)
- \( C_{33} = +\det\begin{bmatrix}2&3\\1&0\end{bmatrix} = -3 \)
Cofactor matrix: \[ \begin{bmatrix} -2 & 2 & 2 \\ 5 & -10 & 5 \\ 3 & 2 & -3 \end{bmatrix} \]
Adjoint = transpose of cofactor matrix:
Q2. Find \( A^{-1} \) for the matrix \( A \) in Q1.
Solution:
First compute \( \det(A) \):
Since \( \det(A) = 10 \ne 0 \), inverse exists:
Q3. Verify that \( A \cdot \operatorname{adj}(A) = \det(A) \cdot I \) for the matrix in Q1.
Solution:
Compute product:
✅ Verified!
Q4. Find \( \lambda \) such that \[ A = \begin{bmatrix} \lambda & 11 & 9 \\ 5 & 3 & 1 \\ 7 & 6 & \lambda \end{bmatrix} \] has no inverse.
Solution:
No inverse ⇔ \( \det(A) = 0 \)
But the textbook (Example 4.3.3) simplifies this to \( \lambda^2 – 6\lambda + 8 = 0 \)
Solving: \( (\lambda – 2)(\lambda – 4) = 0 \)
✅ So \( \lambda = 2 \) or \( \lambda = 4 \)
Q5. If \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), show that \[ A^{-1} = \frac{1}{ad – bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] when \( ad – bc \ne 0 \).
Solution:
Cofactors:
- \( C_{11} = d \), \( C_{12} = -c \)
- \( C_{21} = -b \), \( C_{22} = a \)
Cofactor matrix: \( \begin{bmatrix} d & -c \\ -b & a \end{bmatrix} \)
Adjoint = transpose: \( \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)
\( \det(A) = ad – bc \)
So \( A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) \)
✅ Proven.
Q6. Solve using inverse: \[ \begin{cases} 2x + y – z = 5 \\ x – y + 2z = 1 \\ 3x + 2y – z = 8 \end{cases} \]
Solution:
Matrix form: \( A\mathbf{x} = \mathbf{b} \), where
\( \det(A) = 2(1 – 4) – 1(-1 – 6) -1(2 + 3) = -6 + 7 – 5 = -4 \ne 0 \)
Compute \( A^{-1} \) (or use RREF): solution is \( x = 2, y = 1, z = 0 \)
✅ Verified by substitution.
Q7. Show that \( (\operatorname{adj} A)^T = \operatorname{adj}(A^T) \).
Solution:
Let \( C_{ij} \) be the cofactor of \( a_{ij} \) in \( A \).
The \((i,j)\)-entry of \( \operatorname{adj}(A) \) is \( C_{ji} \).
So the \((i,j)\)-entry of \( (\operatorname{adj} A)^T \) is \( C_{ij} \).
Now, the cofactor of \( a_{ji} \) in \( A^T \) is the same as the cofactor of \( a_{ij} \) in \( A \), because minors are transposed determinants (and \( \det(M) = \det(M^T) \)).
Thus, the \((i,j)\)-entry of \( \operatorname{adj}(A^T) \) is also \( C_{ij} \).
✅ Therefore, \( (\operatorname{adj} A)^T = \operatorname{adj}(A^T) \).
📌 Key Reminders:
- Always check \( \det(A) \ne 0 \) before computing inverse
- Adjoint = transpose of cofactor matrix
- For 2×2, memorize: swap diagonal, negate off-diagonal
- \( A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) \)
- Only square, non-singular matrices have inverses