Problem 7 - Precalculus with Limits

Question

In Exercises 5-12, show that \(B\) is the inverse of \(A\). \(A = \left[ \begin{array}{r} 1 && 2 \\ 3 && 4 \end{array} \right]\), \(B = \left[ \begin{array}{r} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{array} \right]\).

Step-by-Step Solution

Answer

The matrix B is proven to be the inverse of the matrix A as the multiplication of A and B - as well as B and A - results in the identity matrix.

Step 1: Verify the Inverse Definition

The inverse of a matrix A is another matrix B, if the product of A and B is the identity matrix. The identity matrix being \[ I = \left[ \begin{array}{r} 1 && 0 \ 0 && 1 \end{array} \right]\]

Step 2: Multiplication of A and B

First, multiply matrices A and B in the order AB which equates to: \[ (AB) = \left[ \begin{array}{r} 1 && 2 \ 3 && 4 \end{array} \right] \times \left[ \begin{array}{r} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{array} \right] = \left[ \begin{array}{r} 1*(-2) + 2*(1.5) && 1*1 + 2*(-0.5) \ 3*(-2) + 4*(1.5) && 3*1 + 4*(-0.5) \end{array} \right] = \left[ \begin{array}{r} 1 && 0 \ 0 && 1 \end{array} \right] = I \]

Step 3: Multiplication of B and A

Now, we multiply them in the order BA: \[ (BA) = \left[ \begin{array}{r} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{array} \right] \times \left[ \begin{array}{r} 1 && 2 \ 3 && 4 \end{array} \right] = \left[ \begin{array}{r} -2*1 + 1*3 && -2*2 + 1*4 \ 1.5*1 + (-0.5)*3 && 1.5*2 + (-0.5)*4 \end{array} \right] = \left[ \begin{array}{r} 1 && 0 \ 0 && 1 \end{array} \right] = I \]

Step 4: Conclusion

Since matrix BA and AB both resulted in the identity matrix, we can conclude that B is indeed the inverse of A

Key Concepts

Identity Matrix

The identity matrix is a square matrix with ones on the diagonal and zeros elsewhere. It's denoted by the letter 'I' and plays a crucial role in matrix algebra, analogous to the number 1 in regular multiplication. Just as multiplying any number by 1 leaves that number unchanged, multiplying any matrix by the identity matrix preserves the original matrix.

For a 2x2 matrix, the identity matrix is:
\[ I = \left[ \begin{array}{cc} 1 & 0 \ 0 & 1 \end{array} \right] \]
In the exercise, we see matrix A being multiplied by matrix B and the result is the identity matrix. This outcome certifies the core property of the inverse - it 'resets' the matrix multiplication back to the identity, affirming B as the inverse of A.

Matrix Multiplication

Matrix multiplication is a fundamental operation where two matrices are multiplied together to produce a new matrix. For this to work, the number of columns in the first matrix must match the number of rows in the second matrix.

Here's how two matrices, say A and B, are multiplied to form an AB matrix:
\[ (AB)_{ij} = \sum_{k=1}^{n} A_{ik} \times B_{kj} \]
This equation means that each element of the resulting matrix, AB, is the sum of products of corresponding elements from the rows of A and the columns of B. The exercise describes how by multiplying given matrix A with B, and vice versa, the result is an identity matrix, proving that B is an inverse of A according to the definition of matrix inverses.

Inverse of a Matrix

The inverse of a matrix A, denoted as A^-1, is a special matrix that, when multiplied with A, results in the identity matrix I. Not all matrices have inverses; only square matrices with certain properties are invertible.

To verify if B is the inverse of A, as the exercise requires, we multiplied A by B and then B by A. In both cases, the product was the identity matrix I, satisfying the condition for B to be the inverse of A. The steps clearly demonstrate that:
\[ A \times B = I \]\[ B \times A = I \]
Thus, satisfying one of the most important properties of matrix inverses: a matrix A multiplied with its inverse B can arrive at the identity matrix, regardless of the order of multiplication. This property is useful for solving systems of linear equations and in many applications within mathematics, physics, and engineering.