Engineering Mathematics
Linear Combinations & Systems of Linear Equations
Example 2
Showing a vector is a linear combination of two others
Problem
Consider the vectors \(u = (1,\,2,\,-1)\) and \(v = (6,\,4,\,2)\) in \(\mathbb{R}^3\). Show that \(w = (9,\,2,\,7)\) is a linear combination of \(u\) and \(v\).
Given
\(u = (1,\,2,\,-1)\)
\(v = (6,\,4,\,2)\)
\(w = (9,\,2,\,7)\)
Suppose \(w\) is a linear combination of \(u\) and \(v\). Then there exist scalars \(\alpha\) and \(\beta\) such that \[\alpha u + \beta v = w.\] Substituting the component form: \[\alpha(1,\,2,\,-1) + \beta(6,\,4,\,2) = (9,\,2,\,7).\] Equating each component gives the following system of three equations in two unknowns.
System of equations
| \(\alpha + 6\beta\) | \(= 9\) | — (1) |
| \(2\alpha + 4\beta\) | \(= 2\) | — (2) |
| \(-\alpha + 2\beta\) | \(= 7\) | — (3) |
Solution steps
Step 1 — Eliminate \(\alpha\) using equations (1) and (2)
Multiply equation (1) by \(2\) and subtract equation (2):
\[
2(\alpha + 6\beta) - (2\alpha + 4\beta) = 2(9) - 2
\]
\[
2\alpha + 12\beta - 2\alpha - 4\beta = 16
\]
\[
8\beta = 16 \;\Rightarrow\; \beta = 2
\]
Result: β = 2
Step 2 — Find \(\alpha\) by substituting \(\beta = 2\) into equation (1)
\[
\alpha + 6(2) = 9
\]
\[
\alpha + 12 = 9 \;\Rightarrow\; \alpha = -3
\]
Result: α = −3
Step 3 — Verify the solution satisfies equation (3)
Substitute \(\alpha = -3\) and \(\beta = 2\) into equation (3):
\[
-(-3) + 2(2) = 3 + 4 = 7 \checkmark
\]
Equation (3) is satisfied. The system is consistent.
Full verification
\[
\alpha u + \beta v = -3(1,\,2,\,-1) + 2(6,\,4,\,2)
\]
\[
= (-3,\,-6,\,3) + (12,\,8,\,4)
\]
\[
= (9,\,2,\,7) = w \checkmark
\]
\(w = (9,\,2,\,7)\) is a linear combination of \(u\) and \(v\).
Specifically, \(w = -3\,u + 2\,v\).
Specifically, \(w = -3\,u + 2\,v\).
Tutorial Video: Linear Combination & System of Equations
No comments:
Post a Comment