Q.Use row reduction method to solve the given system of equations: 3x+2y−z=1, x+2y−2z=0 and 2x+y−3z=−1.
Concept understanding — Solving Linear Systems
Solving Linear Systems: From Intuition to Precision
Imagine you're at a market. You buy 2 apples and 3 bananas for ₹17. Your friend buys 1 apple and 2 bananas for ₹10. How much does one apple cost? One banana?
You already know how to solve this with a single variable — but here, two things are unknown at once. The price of an apple and the price of a banana are linked together by two different conditions. That's a linear system: a set of equations that must all be true at the same time.
The Intuition: Two Lines, One Meeting Point
Each equation in a system of two variables (say x and y) can be drawn as a straight line on a graph. The equation 2x+3y=17 is one line; x+2y=10 is another.
Solving the system means finding the (x,y) pair that lies on both lines at once. That's the point where the two lines cross — their intersection. If you think about it, that's the only place where both conditions are satisfied simultaneously.
Visualise it: two straight lines on a plane. They either cross at exactly one point (one solution), run parallel and never meet (no solution), or lie exactly on top of each other (infinitely many solutions). The first case is what you'll see most often.
The Precise Statement
A linear system (in two variables) is a collection of equations of the form:
{a1x+b1y=c1a2x+b2y=c2
where a1,b1,c1,a2,b2,c2 are known numbers (constants), and x and y are the unknowns we need to find. The word "linear" means each variable appears only to the first power — no x2, no xy, no x1.
A solution is an ordered pair (x,y) that makes both equations true when substituted in.
a1x+b1y=c1anda2x+b2y=c2
The Three Possibilities
| Case | Graph | Number of Solutions |
|---|---|---|
| Lines intersect at one point | Two crossing lines | Exactly one |
| Lines are parallel (same slope, different intercept) | Two parallel lines | None |
| Lines are identical (same slope and intercept) | One line on top of another | Infinitely many |
How to Solve: Two Core Methods
1. Substitution Method
Solve one equation for one variable, then plug that into the other equation.
From the apple-banana problem:
{2x+3y=17x+2y=10
From the second equation: x=10−2y. Substitute into the first:
2(10−2y)+3y=17
20−4y+3y=17
20−y=17
y=3
Then x=10−2(3)=4. So an apple costs ₹4, a banana costs ₹3.
2. Elimination Method
Add or subtract the equations to cancel one variable.
Multiply the second equation by 2: 2x+4y=20. Subtract the first equation:
(2x+4y)−(2x+3y)=20−17
y=3
Same result, different path.
A common mistake: when substituting, forget to distribute the coefficient correctly. In 2(10−2y), the 2 multiplies both the 10 and the −2y. Write it out step by step until it becomes automatic.
Why This Matters
Linear systems are the foundation for solving problems with multiple constraints — in physics (balancing forces), economics (supply and demand), chemistry (balancing equations), and beyond. Once you master two variables, the same ideas extend to three, ten, or a thousand variables. The methods change, but the core question remains: find the values that satisfy all conditions at once.
Row-reducing the augmented matrix of the three equations to echelon form, then back-substituting, gives the values of x, y and z.
x=112,y=116,z=117.
Row-reduction gives x=112, y=116, z=117.
Form the augmented matrix [A∣B] and reduce to row-echelon form using elementary row operations, then back-substitute.
- Augmented matrix of 3x+2y−z=1, x+2y−2z=0, 2x+y−3z=−1; swap to put a leading 1 on top (R1↔R2):
132221−2−1−301−1.
- R2→R2−3R1, R3→R3−2R1:
1002−4−3−25101−1.
- R3→4R3−3R2: 4(0,−3,1∣−1)−3(0,−4,5∣1)=(0,0,−11∣−7):
1002−40−25−1101−7.
- Row 3: −11z=−7⇒z=117.
- Row 2: −4y+5z=1⇒−4y=1−5⋅117=1−1135=−1124⇒y=116.
- Row 1: x+2y−2z=0⇒x=2z−2y=1114−1112=112.
- Check in 3x+2y−z: 116+1112−117=1111=1.
x=112, y=116, z=117.
- CBSE 2025Set 465/S/WXYZ/41 markMCQQ.The system of linear equations 2x+ky=7 3x+2y=7 will be consistent, if : (A) k=34 (B) k=34 (C) k=43 (D) k=43
›Reveal solutionSolution
The system fails (parallel lines) only at k=34; hence it is consistent for every k=34.
For a1x+b1y=c1, a2x+b2y=c2: a unique solution exists iff a2a1=b2b1; the system is inconsistent iff a2a1=b2b1=c2c1.
- Identify coefficients: a1=2, b1=k, c1=7 and a2=3, b2=2, c2=7.
- Unique solution (consistent) when 32=2k, i.e. k=34.
- Test the boundary k=34: then 32=2k but c2c1=77=1=32, so the lines are parallel and distinct — inconsistent.
- Therefore the system is consistent for all k=34.
✓Final answer(B) k=34
- CBSE 2023Set 465/EF1GH/41 markMCQQ.If [x+y2x−yx+216]=[8153y+1], then the values of x and y are :(a) x=3,y=5(b) x=5,y=3(c) x=2,y=7(d) x=7,y=2
›Reveal solutionSolution
Equal matrices have equal corresponding entries; solving the simplest entries gives x=3, y=5.
Two matrices are equal iff they have the same order and every corresponding entry is equal: aij=bij.
- From the (1,2) entries: x+2=5⇒x=3.
- From the (1,1) entries: x+y=8⇒3+y=8⇒y=5.
- Check (2,1): 2x−y=2(3)−5=1 ✓.
- Check (2,2): 3y+1=3(5)+1=16 ✓. All four entries agree, so x=3, y=5.
✓Final answer(a) x=3, y=5
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