Problem 88
Question
Let \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right], B=\left[\begin{array}{ll}b_{11} & b_{12} \\ b_{21} & b_{22}\end{array}\right],\) and \(C=\left[\begin{array}{ll}c_{11} & c_{12} \\\ c_{21} & c_{22}\end{array}\right]\) where all the elements are real numbers. Use these matrices to show that each statement is true for \(2 \times 2\) matrices. Let \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right], B=\left[\begin{array}{ll}b_{11} & b_{12} \\ b_{21} & b_{22}\end{array}\right],\) and \(C=\left[\begin{array}{ll}c_{11} & c_{12} \\\ c_{21} & c_{22}\end{array}\right]\) where all the elements are real numbers. Use these matrices to show that each statement is true for \(2 \times 2\) matrices. \((c+d) A=c A+d A,\) for any real numbers \(c\) and \(d\)
Step-by-Step Solution
Verified\(cA + dA = \begin{bmatrix} ca_{11}+da_{11} & ca_{12}+da_{12} \\ ca_{21}+da_{21} & ca_{22}+da_{22} \end{bmatrix}\)
By the distributive property of real numbers: \((c+d)a_{ij} = ca_{ij} + da_{ij}\) for all \(i, j\).
Therefore \((c+d)A = cA + dA\), proving that scalar multiplication distributes over scalar addition for \(2 \times 2\) matrices. \(\square\)
Key Concepts
Scalar Multiplication
For example, consider a matrix \( A \) with elements \( a_{11}, a_{12}, a_{21}, \) and \( a_{22} \). When we multiply this matrix by a scalar \( c \), each element in the matrix is multiplied by \( c \). The resulting matrix \( cA \) becomes:
- \( cA = \left[ \begin{array}{ll} c \cdot a_{11} & c \cdot a_{12} \ c \cdot a_{21} & c \cdot a_{22} \end{array} \right] \)
- \( dA = \left[ \begin{array}{ll} d \cdot a_{11} & d \cdot a_{12} \ d \cdot a_{21} & d \cdot a_{22} \end{array} \right] \)
2x2 Matrices
- \( \left[ \begin{array}{cc} a_{11} & a_{12} \ a_{21} & a_{22} \end{array} \right] \)
2x2 matrices are especially common in introductory linear algebra courses due to their simplicity and ease of computation. They serve as building blocks for larger and more complex matrices.
Distributive Property
Start by calculating \( (c + d)A \):
- \( (c+d)A = \left[ \begin{array}{cc} (c+d) \cdot a_{11} & (c+d) \cdot a_{12} \ (c+d) \cdot a_{21} & (c+d) \cdot a_{22} \end{array} \right] \)
- \( cA = \left[ \begin{array}{cc} c \cdot a_{11} & c \cdot a_{12} \ c \cdot a_{21} & c \cdot a_{22} \end{array} \right] \)
- \( dA = \left[ \begin{array}{cc} d \cdot a_{11} & d \cdot a_{12} \ d \cdot a_{21} & d \cdot a_{22} \end{array} \right] \)
- \( cA + dA = \left[ \begin{array}{cc} c \cdot a_{11} + d \cdot a_{11} & c \cdot a_{12} + d \cdot a_{12} \ c \cdot a_{21} + d \cdot a_{21} & c \cdot a_{22} + d \cdot a_{22} \end{array} \right] \)