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Worked Examples · Example 45

Q.Use row reduction method to solve the given system of equations: 3x+2y−z=13x + 2y - z = 1, x+2y−2z=0x + 2y - 2z = 0 and 2x+y−3z=−12x + y - 3z = -1.

Chandigarh CbseNCERTSubjective· 5mImportance★★★★★
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Row-reduction gives x=211, y=611, z=711x=\tfrac{2}{11},\ y=\tfrac{6}{11},\ z=\tfrac{7}{11}.

Form the augmented matrix [A ∣ B][A\,|\,B] and reduce to row-echelon form using elementary row operations, then back-substitute.

  1. Augmented matrix of 3x+2y−z=1, x+2y−2z=0, 2x+y−3z=−13x+2y-z=1,\ x+2y-2z=0,\ 2x+y-3z=-1; swap to put a leading 11 on top (R1↔R2R_1\leftrightarrow R_2):

[12−2032−1121−3−1].\left[\begin{array}{ccc|c} 1 & 2 & -2 & 0 \\ 3 & 2 & -1 & 1 \\ 2 & 1 & -3 & -1 \end{array}\right].

  1. R2→R2−3R1, R3→R3−2R1R_2\to R_2-3R_1,\ R_3\to R_3-2R_1:

[12−200−4510−31−1].\left[\begin{array}{ccc|c} 1 & 2 & -2 & 0 \\ 0 & -4 & 5 & 1 \\ 0 & -3 & 1 & -1 \end{array}\right].

  1. R3→4R3−3R2R_3\to 4R_3-3R_2: 4(0,−3,1 ∣−1)−3(0,−4,5 ∣1)=(0,0,−11 ∣−7)4(0,-3,1\,|-1)-3(0,-4,5\,|1)=(0,0,-11\,|-7):

[12−200−45100−11−7].\left[\begin{array}{ccc|c} 1 & 2 & -2 & 0 \\ 0 & -4 & 5 & 1 \\ 0 & 0 & -11 & -7 \end{array}\right].

  1. Row 3: −11z=−7⇒z=711-11z=-7 \Rightarrow z=\dfrac{7}{11}.
  2. Row 2: −4y+5z=1⇒−4y=1−5⋅711=1−3511=−2411⇒y=611-4y+5z=1 \Rightarrow -4y=1-5\cdot\dfrac{7}{11}=1-\dfrac{35}{11}=-\dfrac{24}{11}\Rightarrow y=\dfrac{6}{11}.
  3. Row 1: x+2y−2z=0⇒x=2z−2y=1411−1211=211x+2y-2z=0 \Rightarrow x=2z-2y=\dfrac{14}{11}-\dfrac{12}{11}=\dfrac{2}{11}.
  4. Check in 3x+2y−z3x+2y-z: 611+1211−711=1111=1\dfrac{6}{11}+\dfrac{12}{11}-\dfrac{7}{11}=\dfrac{11}{11}=1.
✓Final answer

x=211, y=611, z=711.x=\dfrac{2}{11},\ y=\dfrac{6}{11},\ z=\dfrac{7}{11}.

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