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Exercise D · Q3

Q.Solve the following system of equations by

(i) Matrix method
(ii) Row reduction method:
(a) 2x−3y=−42x - 3y = -4, 3x+5y=133x + 5y = 13
(b) x+y=1x + y = 1, 5x−7y=295x - 7y = 29
(c) 5x−4y=95x - 4y = 9, 3x+7y=−43x + 7y = -4
(d) x−y+2z=1x - y + 2z = 1, 2y−3z=12y - 3z = 1, 3x−2y+4z=23x - 2y + 4z = 2
(e) 2x−3y+5z=12x - 3y + 5z = 1, 3x+2y−4z=−53x + 2y - 4z = -5, x+y−2z=−3x + y - 2z = -3.
Andaman Nicobar CbseNCERTSubjective· 5mImportance★★★★★
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Each system is written as AX=BAX=B and solved by X=A−1BX=A^{-1}B; the same solutions follow by Gaussian row reduction of [A ∣ B][A\,|\,B].

Matrix method: AX=B⇒X=A−1BAX=B\Rightarrow X=A^{-1}B (when det⁡A≠0\det A\neq0). Row-reduction: reduce [A ∣ B][A\,|\,B] to row-echelon form and back-substitute.

(a) 2x−3y=−4,  3x+5y=132x-3y=-4,\;3x+5y=13

  1. A=[2−335],  det⁡A=10+9=19.A=\begin{bmatrix}2&-3\\3&5\end{bmatrix},\;\det A=10+9=19.
  2. A−1=119[53−32].A^{-1}=\dfrac{1}{19}\begin{bmatrix}5&3\\-3&2\end{bmatrix}.
  3. X=119[53−32][−413]=119[−20+3912+26]=119[1938]=[12].X=\dfrac{1}{19}\begin{bmatrix}5&3\\-3&2\end{bmatrix}\begin{bmatrix}-4\\13\end{bmatrix}=\dfrac{1}{19}\begin{bmatrix}-20+39\\12+26\end{bmatrix}=\dfrac{1}{19}\begin{bmatrix}19\\38\end{bmatrix}=\begin{bmatrix}1\\2\end{bmatrix}.
  4. x=1,  y=2.x=1,\;y=2.

(b) x+y=1,  5x−7y=29x+y=1,\;5x-7y=29

  1. det⁡A=(1)(−7)−(1)(5)=−12.\det A=(1)(-7)-(1)(5)=-12.
  2. R2→R2−5R1:  −12y=24⇒y=−2;R_2\to R_2-5R_1:\;-12y=24\Rightarrow y=-2; then x=1−(−2)=3.x=1-(-2)=3.
  3. x=3,  y=−2.x=3,\;y=-2.

(c) 5x−4y=9,  3x+7y=−45x-4y=9,\;3x+7y=-4

  1. det⁡A=35+12=47.\det A=35+12=47.
  2. x=9⋅7−(−4)(−4)47=63−1647=1,    y=5(−4)−3(9)47=−4747=−1.x=\dfrac{9\cdot7-(-4)(-4)}{47}=\dfrac{63-16}{47}=1,\;\; y=\dfrac{5(-4)-3(9)}{47}=\dfrac{-47}{47}=-1.
  3. x=1,  y=−1.x=1,\;y=-1.

(d) x−y+2z=1,  2y−3z=1,  3x−2y+4z=2x-y+2z=1,\;2y-3z=1,\;3x-2y+4z=2

  1. A=[1−1202−33−24],  det⁡A=1(8−6)+1(0+9)+2(0−6)=2+9−12=−1.A=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix},\;\det A=1(8-6)+1(0+9)+2(0-6)=2+9-12=-1.
  2. Row reduce: R3→R3−3R1⇒(0, 1, −2 ∣ −1)R_3\to R_3-3R_1\Rightarrow(0,\,1,\,-2\,|\,-1). With R2:(0,2,−3 ∣ 1)R_2:(0,2,-3\,|\,1), do R3→2R3−R2⇒(0,0,−1 ∣ −3)⇒z=3.R_3\to 2R_3-R_2\Rightarrow(0,0,-1\,|\,-3)\Rightarrow z=3.
  3. Back-substitute: 2y−3(3)=1⇒y=5;2y-3(3)=1\Rightarrow y=5; then x−5+6=1⇒x=0.x-5+6=1\Rightarrow x=0.
  4. x=0,  y=5,  z=3.x=0,\;y=5,\;z=3. Check: 3(0)−2(5)+4(3)=2.3(0)-2(5)+4(3)=2. ✓

(e) 2x−3y+5z=1,  3x+2y−4z=−5,  x+y−2z=−32x-3y+5z=1,\;3x+2y-4z=-5,\;x+y-2z=-3 …

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