Problem 31
Question
The matrices \(A, B, C, D, E, F, G\) and \(H\) are defined as follows. $$A=\left[\begin{array}{rr}2 & -5 \\\0 & 7\end{array}\right] \quad B=\left[\begin{array}{rrr}3 &\frac{1}{2} & 5 \\\1 & -1 & 3\end{array}\right] \quad C=\left[\begin{array}{rrr}2 & -\frac{5}{2} &0 \\\0 & 2 & -3\end{array}\right]$$ $$D=\left[\begin{array}{lll}7 & 3\end{array}\right] \quad E=\left[\begin{array}{l}1 \\\2 \\\0\end{array}\right] \quad F=\left[\begin{array}{lll}1 & 0 & 0 \\\0 & 1 & 0 \\\0 & 0 & 1\end{array}\right]$$ $$G=\left[\begin{array}{rrr}5 & -3 & 10 \\\6 & 1 & 0 \\\\-5 & 2 & 2\end{array}\right] \quadH=\left[\begin{array}{rr}3 & 1 \\\2 & -1\end{array}\right]$$ Carry out the indicated algebraic operation, or explain why it cannot be performed. (a) \(G F\) (b) \(G E\)
Step-by-Step Solution
VerifiedKey Concepts
Matrix Dimensions
It's important to remember the order of dimensions, as it affects compatibility in operations like addition and multiplication. Comparing dimensions helps determine if matrices can be combined. Often, simply knowing the dimensions is key to solving matrix-based problems.
Identity Matrix
For example, the identity matrix \( F \) in the exercise is a \( 3 \times 3 \) matrix:
- \( F = \begin{bmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \)
Matrix Compatibility
Assessing compatibility is the first step before multiplication. In the exercise, both \( G \) and \( F \) are \( 3 \times 3 \) matrices, making them compatible. Similarly, matrix \( G \) and vector (or single-column matrix) \( E \), with dimensions \( 3 \times 3 \) and \( 3 \times 1 \) respectively, are also compatible.
Matrix Calculations
In multiplication, you multiply rows of the first matrix by columns of the second, summing the results to form each element of the product matrix. To illustrate, in the exercise, the product \( GE \) was calculated as follows:
- For each element in \( GE \), multiply corresponding elements from row of \( G \) and column of \( E \), then sum them up.
- Example from first row: \(5 \times 1 + (-3) \times 2 + 10 \times 0 = -1\).