Discover how a single MATLAB command can save you hours of frustrating math work!
Why Solving linear systems (A\b) in MATLAB? - Purpose & Use Cases
Imagine you have a set of equations to find unknown values, like figuring out how many apples and oranges you bought given the total cost and quantity. Doing this by hand means juggling numbers and rewriting equations repeatedly.
Manually solving these equations is slow and easy to mess up, especially when there are many variables. It's like trying to solve a puzzle with many pieces without a clear guide, leading to mistakes and frustration.
Using MATLAB's A\b operator quickly and accurately solves these equations for you. It's like having a smart calculator that instantly finds the right answers without the hassle.
x = inv(A)*b; % manually calculating inverse and multiplyingx = A\b; % direct and efficient solutionThis lets you solve complex systems of equations instantly, freeing you to focus on understanding results instead of tedious calculations.
Engineers use this to find forces in structures or currents in circuits, where many variables interact and manual solving would be impractical.
Manual solving is slow and error-prone.
A\b provides a fast, reliable solution.
It enables solving complex problems easily.
Practice
x = A\b do?Solution
Step 1: Understand the expression A\b
The operator \ in MATLAB is used to solve linear systems of the form Ax = b.Step 2: Identify the purpose of A\b
It finds the vector x that satisfies the equation Ax = b without explicitly calculating the inverse of A.Final Answer:
Solves the linear system Ax = b for x -> Option AQuick Check:
A\b solves Ax = b [OK]
- Confusing A\b with matrix multiplication
- Thinking it calculates inverse of A
- Using b\A instead of A\b
Solution
Step 1: Recall MATLAB syntax for solving linear systems
The correct syntax uses the backslash operator: x = A\b;Step 2: Check other options for correctness
x = b/A; uses b/A which is incorrect; x = inv(A)\b; uses inv(A)\b which solves inv(A)x = b (wrong system); x = A*b; multiplies A and b which is not solving the system.Final Answer:
x = A\b; -> Option AQuick Check:
Use backslash operator for solving Ax = b [OK]
- Using b/A instead of A\b
- Using inv(A)\b which solves wrong system
- Confusing multiplication with solving
A = [2 1; 3 4] and b = [5; 10], what is the result of x = A\b?Solution
Step 1: Write the system of equations
2x1 + x2 = 5
3x1 + 4x2 = 10.Step 2: Solve using elimination
Multiply first equation by 4: 8x1 + 4x2 = 20.
Subtract second: 5x1 = 10 -> x1 = 2.
Substitute: x2 = 5 - 2*2 = 1.Step 3: Verify solution
A*[2; 1] = [2*2 + 1*1; 3*2 + 4*1] = [5; 10] = b.Final Answer:
[2; 1] -> Option DQuick Check:
A\b = [2; 1] [OK]
- Miscalculating substitution (e.g., picking [1; 2])
- Confusing with matrix multiplication A*b
- Selecting x = b
A = [1 2; 3 4]; b = [5; 6; 7]; x = A\b;
Solution
Step 1: Check dimensions of A and b
Matrix A is 2x2, vector b is 3x1. For Ax = b, number of rows in A must equal length of b.Step 2: Identify dimension mismatch
Since A has 2 rows but b has 3 elements, dimensions are incompatible, causing an error when solving.Final Answer:
Matrix A and vector b have incompatible dimensions -> Option BQuick Check:
Rows of A must match length of b [OK]
- Ignoring dimension mismatch errors
- Thinking backslash works for any sizes
- Assuming A must be square always
A = [1 2 3; 4 5 6] and vector b = [7; 8]. What does x = A\b compute in MATLAB?Solution
Step 1: Recognize matrix shape and operation
A is 2x3 (more columns than rows), so system is underdetermined.Step 2: Understand MATLAB's backslash behavior for non-square A
MATLAB computes the least-squares solution x that minimizes ||Ax - b|| (minimum norm among solutions).Final Answer:
A least-squares solution minimizing ||Ax - b|| -> Option CQuick Check:
Backslash solves least-squares for non-square A [OK]
- Expecting error for non-square A
- Thinking it calculates inverse always
- Assuming exact solution always exists
Ready to prove it?
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