Problem 14 - College Algebra

Question

In Exercises \(13-18,\) use the fact that if \(A=\left[\begin{array}{ll}a & b \\\ c & d\end{array}\right]\), then \(A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]\) to find the inverse of cach matrix, if possible. Check that \(A A^{-1}=I_{2}\) and \(A^{-1} A=I_{2}\) $$A=\left[\begin{array}{rr}0 & 3 \\\4 & -2\end{array}\right]$$

Step-by-Step Solution

Answer

The inverse of the given matrix A is \[A^{-1}=\left[\begin{array}{cc}1/6 & 1/4 \ 1/3 & 0\end{array}\right]\].

Step 1: Calculate the determinant

The formula for the determinant of a 2x2 matrix \[A=\left[\begin{array}{cc}a & b \c & d\end{array}\right]\] is \(ad - bc\). Substituting the values from the given matrix, we get \(0*(-2) - 3*4 = -12\)

Step 2: calculate the inverse if determinant is Non-Zero

Since the determinant isn't zero, the inverse exists. We follow the formula given, which involves swapping a and d, changing the signs of b and c, and dividing all elements by the determinant. This results in \[A^{-1}=\frac{1}{-12}\left[\begin{array}{cc}-2 & -3 \ -4 & 0\end{array}\right] = \left[\begin{array}{cc}1/6 & 1/4 \ 1/3 & 0\end{array}\right]\]

Step 3: Check the result

To confirm if the calculated inverse is correct, we multiply the given matrix \(A\) by its inverse and from both sides, it should yield the identity matrix \(I\). The identity matrix is \[I =\left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}\right]\].

Key Concepts

Matrix Determinant

When dealing with a 2x2 matrix, finding the determinant is an essential step. The determinant helps us understand whether a matrix has an inverse. For a matrix \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\], the determinant is calculated using the formula \(ad - bc\). Each element plays a crucial role:

• \(a\) and \(d\) are the diagonal elements, and their product is part of the determinant.
• \(b\) and \(c\) are the other elements and their product is subtracted from \(ad\).

In our exercise, the matrix given was \[\begin{bmatrix} 0 & 3 \ 4 & -2 \end{bmatrix}\]. Plugging the values into \(ad - bc\), we compute \(0(-2) - 3(4) = -12\). Since this is not zero, the matrix is invertible. If the determinant were zero, the matrix would not have an inverse.

Identity Matrix

The identity matrix is like the "1" in matrix arithmetic. It's a special kind of matrix that doesn't change anything when used in multiplication. For a 2x2 matrix, the identity matrix is:\[I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]
When you multiply any matrix by the identity matrix (as long as they are compatible in size), you'll get back the original matrix. This property is vital in verifying whether you've correctly found the inverse of a matrix.

• If \(A\) is your original matrix and \(A^{-1}\) is the inverse, then multiplying them should yield the identity matrix, i.e., \(AA^{-1} = I\) and \(A^{-1}A = I\).

This consistency check is crucial in confirming that the computed inverse is accurate.

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra, especially for verifying the inverse of a matrix. To multiply matrices, remember a few key rules:

• Each element in the resulting matrix is a sum of products. You multiply the elements of the rows of the first matrix by the corresponding elements of the columns of the second matrix.
• The size of the first matrix's row must match the size of the second matrix's column for multiplication to work.

In the exercise, after finding the inverse, the calculation was verified through multiplication. By multiplying the original matrix\[A = \begin{bmatrix} 0 & 3 \ 4 & -2 \end{bmatrix}\]with its computed inverse\[A^{-1} = \begin{bmatrix} 1/6 & 1/4 \ 1/3 & 0 \end{bmatrix}\]from both sides (i.e., \(AA^{-1}\) and \(A^{-1}A\)), they confirmed the result as:\[I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]This shows that the product yields the identity matrix, ensuring the inverse has been correctly found.