Let A be an invertible n x n matrix, and let B be an n x p matrix. Explain why (A^-1)(B) can be computed by row reduction:
[A B] ~...~ [I X] X=(A^-1)(B)
a trust fund has $30,000 that must be invested in two different types of bonds. the first bond pays 5% interest per year, & the second bond pays 7% interest per year. using matrix multiplication, determine how to divide $ 30,000 among the two types of bonds. if the trust fund must obtain an annual total interest of: (a) $ 1800?
To understand why (A^(-1))(B) can be computed by row reduction, let's break down the steps involved:
1. Start by considering the augmented matrix [A | B], where A is an invertible n x n matrix and B is an n x p matrix.
2. Our goal is to reduce the left side of the augmented matrix [A | B] to the identity matrix I. This transformation is achieved through a series of elementary row operations.
3. By performing the same row operations on both sides of the augmented matrix, we maintain the equality and end up with [I | X], where X is a p x p matrix to be determined.
4. Now, we have the equation AX = B, where A is an invertible matrix and X is an unknown matrix.
5. To solve for X, we multiply both sides of the equation by A^(-1) (the inverse of matrix A). This results in A^(-1)(AX) = A^(-1)B.
6. Since A^(-1) is the inverse of A, the product (AX) simplifies to X. Therefore, we have A^(-1)X = A^(-1)B.
7. By substituting the values of X and B back into the augmented matrix [I | X], we can conclude that X = (A^(-1))(B).
By following these steps, we can see that row reducing the augmented matrix [A | B] leads us to the desired expression (A^(-1))(B).
To understand why (A^-1)(B) can be computed by row reduction, let's break down the process step by step:
1. Firstly, we start with the augmented matrix [A B], where A is an invertible n x n matrix, and B is an n x p matrix.
2. Our goal is to transform the augmented matrix [A B] into the form [I X], where I represents the identity matrix of order n and X represents the resulting matrix (A^-1)(B).
3. By performing row reduction on [A B], we can apply a series of elementary row operations to manipulate the augmented matrix and eventually obtain [I X].
4. The elementary row operations include:
a. Swapping rows: You can interchange the position of two rows in the matrix; for example, R1 <-> R2.
b. Scaling rows: You can multiply a row by a nonzero scalar; for example, 2R1.
c. Adding multiples of rows: You can multiply a row by a scalar and add it to another row; for example, 3R1 + R2.
5. Using these elementary row operations, we can carefully manipulate the augmented matrix [A B] until the leftmost n x n submatrix becomes the identity matrix I.
6. As we apply the same row operations to both A and B, the matrix A transforms into the identity matrix I, and the matrix B transforms into the desired matrix X.
7. Therefore, at the end of row reduction, the resulting matrix is in the form [I X]. The matrix X represents the result of multiplying A^-1 (the inverse of A) by B, which is precisely what we wanted to compute.
By using row reduction, we are effectively applying the same sequence of steps to both matrices A and B, and the final result X corresponds to (A^-1)(B).