Problem 4 - Precalculus: Mathematical for Calculus

Question

Calculate the products \(A B\) and \(B A\) to verify that \(B\) is the inverse of \(A\). $$A=\left[\begin{array}{rrr} 3 & 2 & 4 \\ 1 & 1 & -6 \\ 2 & 1 & 12 \end{array}\right], \quad B=\left[\begin{array}{rrr} 9 & -10 & -8 \\ -12 & 14 & 11 \\ -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{array}\right]$$

Step-by-Step Solution

Answer

Matrix B is the inverse of matrix A.

Step 1: Multiply Matrices A and B

To find the product \(AB\), multiply each element in row \(i\) of matrix \(A\) by each element in column \(j\) of matrix \(B\), then take the sum of these products for each element in the resulting matrix. Compute each cell of the product matrix:- First row: - \( (3 \times 9) + (2 \times -12) + (4 \times -\frac{1}{2}) = 27 - 24 - 2 = 1 \) - \( (3 \times -10) + (2 \times 14) + (4 \times \frac{1}{2}) = -30 + 28 + 2 = 0 \) - \( (3 \times -8) + (2 \times 11) + (4 \times \frac{1}{2}) = -24 + 22 + 2 = 0 \)- Second row: - \( (1 \times 9) + (1 \times -12) + (-6 \times -\frac{1}{2}) = 9 - 12 + 3 = 0 \) - \( (1 \times -10) + (1 \times 14) + (-6 \times \frac{1}{2}) = -10 + 14 - 3 = 1 \) - \( (1 \times -8) + (1 \times 11) + (-6 \times \frac{1}{2}) = -8 + 11 - 3 = 0 \)- Third row: - \( (2 \times 9) + (1 \times -12) + (12 \times -\frac{1}{2}) = 18 - 12 - 6 = 0 \) - \( (2 \times -10) + (1 \times 14) + (12 \times \frac{1}{2}) = -20 + 14 + 6 = 0 \) - \( (2 \times -8) + (1 \times 11) + (12 \times \frac{1}{2}) = -16 + 11 + 6 = 1 \)Thus, \( AB = \left[\begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right] \)

Step 2: Multiply Matrices B and A

To find the product \(BA\), multiply each element in row \(i\) of matrix \(B\) by each element in column \(j\) of matrix \(A\), then take the sum of these products for each element in the resulting matrix. Compute each cell of the product matrix:- First row: - \( (9 \times 3) + (-10 \times 1) + (-8 \times 2) = 27 - 10 - 16 = 1 \) - \( (9 \times 2) + (-10 \times 1) + (-8 \times 1) = 18 - 10 - 8 = 0 \) - \( (9 \times 4) + (-10 \times -6) + (-8 \times 12) = 36 + 60 - 96 = 0 \)- Second row: - \( (-12 \times 3) + (14 \times 1) + (11 \times 2) = -36 + 14 + 22 = 0 \) - \( (-12 \times 2) + (14 \times 1) + (11 \times 1) = -24 + 14 + 11 = 1 \) - \( (-12 \times 4) + (14 \times -6) + (11 \times 12) = -48 - 84 + 132 = 0 \)- Third row: - \( (-\frac{1}{2} \times 3) + (\frac{1}{2} \times 1) + (\frac{1}{2} \times 2) = -\frac{3}{2} + \frac{1}{2} + 1 = 0 \) - \( (-\frac{1}{2} \times 2) + (\frac{1}{2} \times 1) + (\frac{1}{2} \times 1) = -1 + \frac{1}{2} + \frac{1}{2} = 0 \) - \( (-\frac{1}{2} \times 4) + (\frac{1}{2} \times -6) + (\frac{1}{2} \times 12) = -2 - 3 + 6 = 1 \)Thus, \( BA = \left[\begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right] \)

Step 3: Verify if B is the Inverse of A

If A and B matrices multiply to the identity matrix in both orders, namely \(AB = I\) and \(BA = I\), where \(I\) is the identity matrix, then \(B\) is the inverse of \(A\). Since we found both \(AB\) and \(BA\) to be the identity matrix \( \left[\begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right] \), we confirm that \(B\) is indeed the inverse of \(A\).

Key Concepts

Matrix Multiplication

Matrix multiplication is a method used in mathematics when you need to multiply two matrices together. It involves an array of numbers in rows and columns. You multiply two matrices by taking each row from the first matrix and each column from the second matrix, then performing a series of operations. Here’s how you do it:

• Take the first row of the first matrix.
• Multiply each element of this row by the corresponding element of the first column of the second matrix.
• Sum all the results to get a single number which becomes an element of the product matrix.
• Repeat this process for each row of the first matrix and each column of the second matrix.

Remember, the number of columns in the first matrix must equal the number of rows in the second matrix to be compatible for multiplication. The resulting matrix will have dimensions corresponding to the rows of the first matrix and the columns of the second matrix. In our example, multiplying matrix \(A\) and matrix \(B\) results in another matrix. Paying attention to detail during multiplication is key to making sure the outcome is accurate.

Inverse Matrix Verification

Verifying if a matrix is indeed the inverse of another is crucial in linear algebra. If one matrix is the inverse of another, their multiplication should result in the identity matrix. This is a clear indication that the inverse operation was successful. To verify this:

• Compute the product of the two matrices in both possible orders (i.e., First \(AB\), then \(BA\)).
• Check if the result of both products yields the identity matrix.

In the exercise, we multiplied matrices \(A\) and \(B\). Each of these operations produced an identity matrix. This identity matrix provides proof that \(B\) is indeed the inverse of \(A\). You can't judge an inverse solely by one direction of multiplication – both \(AB\) and \(BA\) need to yield the identity matrix. Otherwise, the results are not valid, and the matrices do not exhibit an inverse relationship.

Identity Matrix

In the realm of matrices, the identity matrix is a very special type. It has 1s on its diagonal (from top-left to bottom-right) and 0s elsewhere. For example, a 3x3 identity matrix looks like this:\[\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\]An identity matrix acts like the number 1 in regular multiplication. When you multiply any matrix by it, you get the original matrix back:

• It doesn't change the other matrix during multiplication.
• It is the only matrix that can do this unique task.

In matrix theory, it's essential to know about identity matrices as they offer a basis for verifying inverses and preserving the structural integrity of other matrices when used in operations. Anytime you successfully find that multiplying two matrices gives you an identity matrix, you've confirmed that they are inverses.