Chapter 2: Problem 11
If \(A\left[\begin{array}{rr}0 & 1 \\ 2 & -1\end{array}\right]=\left[\begin{array}{rr}2 & 1 \\ -1 & 0\end{array}\right]\) where \(A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\), then \(A\) is (a) \(\left[\begin{array}{ll}2 & 1 \\ 0 & 0\end{array}\right]\) (b) \(\left[\begin{array}{rr}0 & 1 \\ 2 & -1\end{array}\right]\) (c) \(\left[\begin{array}{rr}2 & 1 \\ -1 & 0\end{array}\right]\) (d) \(\left[\begin{array}{cc}2 & 1 \\ -1 / 2 & -1 / 2\end{array}\right]\)
Short Answer
Step by step solution
Setup the Matrix Equation
Multiply Matrices on Left-Hand Side
Equate Result to Given Matrix
Solve for Variables
Assemble Matrix A
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Multiplication
- The element in row 1, column 1 of the product is computed as: \( a \times 0 + b \times 2 = 2b \).
- The element in row 1, column 2 is: \( a \times 1 + b \times (-1) = a - b \).
- The element in row 2, column 1 is: \( c \times 0 + d \times 2 = 2d \).
- The element in row 2, column 2 is: \( c \times 1 + d \times (-1) = c - d \).
Matrix Equation
- Equation 1: \( 2b = 2 \)
- Equation 2: \( a - b = 1 \)
- Equation 3: \( 2d = -1 \)
- Equation 4: \( c - d = 0 \)
Linear Systems
- \( 2b = 2 \)
- \( a - b = 1 \)
- \( 2d = -1 \)
- \( c - d = 0 \)
Matrices and Determinants
- Finding the matrix's invertibility — if the determinant is non-zero, the matrix is invertible.
- Solving equations uniquely — a non-zero determinant implies unique solutions.
- Understanding geometric transformations — determinants give insight into scaling actions.