Problem 9 - Differential Equations and Linear Algebra

Question

Find \(\operatorname{det}(A) .\) If \(A\) is invertible, use the adjoint method to find \(A^{-1}\). $$A=\left[\begin{array}{ll}3 & 5 \\\2 & 7\end{array}\right]$$

Step-by-Step Solution

Answer

The determinant of matrix A is 11, and since it's non-zero, the matrix is invertible. The inverse of matrix A using the adjoint method is: \[ A^{-1} =\left[\begin{array}{cc}\frac{7}{11} & -\frac{5}{11} \\ -\frac{2}{11} & \frac{3}{11}\end{array}\right] \]

Step 1: To find the determinant of a \(2\times2\) matrix, we use the formula: \[ \operatorname{det}(A) = a_{11}a_{22} - a_{12}a_{21} \] where \(a_{ij}\) represents the matrix elements. In our case, \(a_{11}=3\), \(a_{12}=5\), \(a_{21}=2\), and \(a_{22}=7\). Plugging these values into the formula, we have: \[ \operatorname{det}(A) = (3)(7) - (5)(2) \] #Step 2: Compute the determinant of A#

Now we compute the determinant using the values calculated in the previous step: \[ \operatorname{det}(A) = 21 - 10 = 11 \] Since the determinant is non-zero, this means that the matrix A is invertible. #Step 3: Find the adjoint matrix of A#

Step 2: The adjoint matrix of A, denoted as \(\operatorname{adj}(A)\), is given by: \[ \operatorname{adj}(A) = \left[\begin{array}{cc}a_{22} & -a_{12} \\ -a_{21} & a_{11}\end{array}\right] \] Plugging the values of \(a_{ij}\) into the formula, we have: \[ \operatorname{adj}(A) = \left[\begin{array}{cc}7 & -5 \\ -2 & 3\end{array}\right] \] #Step 4: Compute the inverse of A#

Finally, we can find the inverse of A by dividing the adjoint matrix by the determinant of A: \[ A^{-1} = \frac{1}{\operatorname{det}(A)}\operatorname{adj}(A) \] Substituting the values calculated in the previous steps, we get: \[ A^{-1} = \frac{1}{11}\left[\begin{array}{cc}7 & -5 \\ -2 & 3\end{array}\right] \] Therefore, the inverse of matrix A is \[ A^{-1} =\left[\begin{array}{cc}\frac{7}{11} & -\frac{5}{11} \\ -\frac{2}{11} & \frac{3}{11}\end{array}\right] \]

Key Concepts

Adjoint Matrix

The adjoint matrix plays a crucial role in finding the inverse of a square matrix. Imagine you have a matrix and you want to make another matrix that helps you find its inverse. That's where the adjoint matrix comes in. To construct the adjoint of a 2x2 matrix, we simply switch the positions of the elements on the main diagonal and change the signs of the other two elements. This gives us a new matrix.Let's see how it works with matrix A:- Original: \( A = \begin{bmatrix} 3 & 5 \ 2 & 7 \end{bmatrix} \)- Adjoint: Swap diagonals and change signs of off-diagonals: \( \text{adj}(A) = \begin{bmatrix} 7 & -5 \ -2 & 3 \end{bmatrix} \)Once we have the adjoint matrix, it's ready to use in calculating the inverse of the original matrix, provided it is invertible.

Inverse Matrix

Finding the inverse of a matrix, specifically a 2x2 matrix, follows a simple procedure if the matrix is invertible. The inverse matrix, denoted as \( A^{-1} \), is the matrix you multiply by the original to get the identity matrix. For a matrix \( A \), the identity matrix \( I \) looks like \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \).To find \( A^{-1} \), we use the adjoint of \( A \) and the determinant of \( A \):\[A^{-1} = \frac{1}{\operatorname{det}(A)} \cdot \operatorname{adj}(A)\]In our example, substituting the values:- \( \operatorname{det}(A) = 11 \)- \( \text{adj}(A) = \begin{bmatrix} 7 & -5 \ -2 & 3 \end{bmatrix} \)So the inverse is:\[A^{-1} = \frac{1}{11} \begin{bmatrix} 7 & -5 \ -2 & 3 \end{bmatrix} = \begin{bmatrix} \frac{7}{11} & -\frac{5}{11} \ -\frac{2}{11} & \frac{3}{11} \end{bmatrix}\]

Invertible Matrix

An invertible matrix is a matrix that has an inverse. Not all matrices have this property. The key to determining if a matrix is invertible is its determinant. For a 2x2 matrix, if the determinant is not equal to zero, then the matrix is invertible. Let's delve into it:- The determinant of matrix \( A \) is given by: \[ \operatorname{det}(A) = a_{11}a_{22} - a_{12}a_{21} \]- If \( \operatorname{det}(A) = 0 \), the matrix is not invertible, hence no inverse exists.In our exercise, we calculated:- \( \operatorname{det}(A) = 11 \), which is not zero. - This means matrix \( A \) is indeed invertible.An invertible matrix allows us to solve systems of equations, and it's fundamental in many areas of mathematics, including linear algebra and calculus.