Problem 11 - College Algebra

Question

In the following exercises, show that matrix \(A\) is the inverse of matrix \(B\). $$A=\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 0 & 2 \\ 1 & 6 & 9\end{array}\right], B=\frac{1}{4}\left[\begin{array}{rrr}6 & 0 & -2 \\ 17 & -3 & -5 \\ -12 & 2 & 4\end{array}\right]$$

Step-by-Step Solution

Answer

Matrix \(A\) is indeed the inverse of matrix \(B\) since \(AB = BA = I\).

Step 1: Understand the Inverse Condition

To determine if matrix \(A\) is the inverse of matrix \(B\), we need to check if their multiplication results in the identity matrix. That is, both \(AB\) and \(BA\) should equal the 3x3 identity matrix \(I\), where \(I = \left[\begin{array}{ccc}1 & 0 & 0\ 0 & 1 & 0\ 0 & 0 & 1\end{array}\right]\).

Step 2: Multiply Matrices A and B

Calculate the product \(AB\):\[ AB = A \times B = \left[\begin{array}{lll}1 & 2 & 3 \ 4 & 0 & 2 \ 1 & 6 & 9\end{array}\right] \times \frac{1}{4}\left[\begin{array}{rrr}6 & 0 & -2 \ 17 & -3 & -5 \ -12 & 2 & 4\end{array}\right] \]Start by multiplying each entry by \(\frac{1}{4}\) and then calculate each element of the resulting matrix with the dot product of corresponding rows and columns:\[ AB = \left[\begin{array}{ccc}1 \times 1.5 + 2 \times 4.25 + 3 \times -3 & 1 \times 0 + 2 \times -0.75 + 3 \times 0.5 & 1 \times -0.5 + 2 \times -1.25 + 3 \times 1 \ 4 \times 1.5 + 0 \times 4.25 + 2 \times -3 & 4 \times 0 + 0 \times -0.75 + 2 \times 0.5 & 4 \times -0.5 + 0 \times -1.25 + 2 \times 1\ 1 \times 1.5 + 6 \times 4.25 + 9 \times -3 & 1 \times 0 + 6 \times -0.75 + 9 \times 0.5 & 1 \times -0.5 + 6 \times -1.25 + 9 \times 1 \end{array}\right] \]

Step 3: Simplify the Resulting Matrix A * B

Calculate each element of \(AB\):\[ AB = \left[\begin{array}{ccc}(1.5 + 8.5 - 9) & (0 - 1.5 + 1.5) & (-0.5 - 2.5 + 3) \ (6 + 0 - 6) & (0 + 0 + 1) & (-2 + 0 + 2) \ (1.5 + 25.5 - 27) & (0 - 4.5 + 4.5) & (-0.5 - 7.5 + 9) \end{array}\right] \]\[ AB = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right] \]The result is indeed the identity matrix.

Step 4: Multiply Matrices B and A

Now calculate the product \(BA\):\[ BA = B \times A = \frac{1}{4}\left[\begin{array}{rrr}6 & 0 & -2 \ 17 & -3 & -5 \ -12 & 2 & 4\end{array}\right] \times \left[\begin{array}{lll}1 & 2 & 3 \ 4 & 0 & 2 \ 1 & 6 & 9\end{array}\right] \]Proceed with calculating each component:\[ BA = \left[\begin{array}{ccc}(6 \times 1 + 0 \times 4 + -2 \times 1) & (6 \times 2 + 0 \times 0 + -2 \times 6) & (6 \times 3 + 0 \times 2 + -2 \times 9) \ (17 \times 1 + -3 \times 4 + -5 \times 1) & (17 \times 2 + -3 \times 0 + -5 \times 6) & (17 \times 3 + -3 \times 2 + -5 \times 9) \ (-12 \times 1 + 2 \times 4 + 4 \times 1) & (-12 \times 2 + 2 \times 0 + 4 \times 6) & (-12 \times 3 + 2 \times 2 + 4 \times 9) \end{array}\right] \]

Step 5: Simplify the Resulting Matrix B * A

Calculate each element of \(BA\) and simplify:\[ BA = \left[\begin{array}{ccc}(6 - 0 - 2) & (12 - 0 - 12) & (18 - 0 - 18) \ (17 - 12 - 5) & (34 - 0 - 30) & (51 - 6 - 45) \ (-12 + 8 + 4) & (-24 + 0 + 24) & (-36 + 4 + 36) \end{array}\right] \]\[ BA = \left[\begin{array}{ccc}4 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 4\end{array}\right] \]Then multiply each element by \(\frac{1}{4}\):\[ BA = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right] \]This is also the identity matrix.

Key Concepts

Identity Matrix

An identity matrix is a special kind of square matrix that acts like the number 1 in matrix multiplication. It is usually denoted by the letter \( I \). The hallmark feature of an identity matrix is that when any square matrix is multiplied by it, the original matrix is returned. This is much like how multiplying a number by 1 returns the same number.

Here’s what makes it distinctive:

• The identity matrix is a square matrix, meaning it has the same number of rows and columns.
• All the elements on its principal diagonal (from the top-left to the bottom-right) are 1.
• All other elements are 0.

For example, a 3x3 identity matrix looks like this:\[I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\]So, the identity matrix in the context of our problem is crucial, because it confirms whether two matrices are inverses of each other. By computing the products \( AB \) and \( BA \) and obtaining the identity matrix in both cases, we demonstrate that \( A \) and \( B \) are inverses.

Matrix Multiplication

Matrix multiplication is an operation where two matrices are combined to make a single matrix. It involves a strategy that is different from simple arithmetic multiplication, following specific rules to ensure the correct computation.

How it works:

• For two matrices \( A \) and \( B \), matrix multiplication is possible when the number of columns in matrix \( A \) equals the number of rows in matrix \( B \).
• The entry in the resulting matrix corresponds to the dot product of the corresponding row from the first matrix and the column from the second matrix.
• Each element is calculated by multiplying pairwise elements and then summing up these products.

Consider a simple example:\[\text{If } A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \text{ and } B = \begin{bmatrix} w & x \ y & z \end{bmatrix}, \text{ then } \ AB = \begin{bmatrix} (aw + by) & (ax + bz) \ (cw + dy) & (cx + dz) \end{bmatrix}\]This operation makes matrix multiplication a powerful tool in linear algebra, as it can transform coordinate systems, rotate shapes, and much more. In our exercise, plying this concept helped us conclude the inverse relationship between matrices \( A \) and \( B \) by showing their product equals the identity matrix.

Inverse Condition

The inverse condition is a crucial requirement for determining whether one matrix is the inverse of another. The concept of matrix inversion is deeply rooted in linear algebra and can be thought of as a way to 'un-multiply' matrices.

For matrices \( A \) and \( B \) to be inverses of each other, two conditions must be met:

• The product of \( A \) and \( B \) should result in the identity matrix: \( AB = I \).
• The product of \( B \) and \( A \) should also result in the identity matrix: \( BA = I \).

Why does this matter? Here’s why:

• If both conditions are satisfied, matrix \( A \) is considered the inverse of matrix \( B \), and vice versa.
• It essentially means that they cancel each other out when multiplied, leading to the identity matrix which acts as the '1' of matrix arithmetic.

In our specific example, by performing both matrix multiplications \( AB \) and \( BA \) and obtaining the identity matrix each time, we confirm the inverse condition. This integral relationship showcases the foundational property of matrix inversion, enabling applications in solving systems of equations, among other uses.