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Question 31 of 34

Q.Express the following equations in matrix form and solve them by the method of reduction:
x−y+z=1x - y + z = 1, 2x−y=12x - y = 1, 3x+3y−4z=23x + 3y - 4z = 2
Solution:
The given equations can be written in the matrix form as:
[1−112−1033−4][xyz]=[112]\begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}
By R2→R2−2R1R_2 \rightarrow R_2 - 2R_1,
[1−11□□□33−4][xyz]=[1−12]\begin{bmatrix} 1 & -1 & 1 \\ \square & \square & \square \\ 3 & 3 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}
By R3→R3−3R1R_3 \rightarrow R_3 - 3R_1
[1−1101−2□□□][xyz]=[1−1−1]\begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -2 \\ \square & \square & \square \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}
By R3→R3−6R2R_3 \rightarrow R_3 - 6R_2
[1−1101−2005][xyz]=[1−1□]\begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ \square \end{bmatrix}
We write equations as
x−y+z=1x - y + z = 1 ...(I)
y−2z=−1y - 2z = -1 ...(II)
5z=55z = 5 ...(III)
Solving equations (I), (II) and (III)
We get x=□,y=□,z=□x = \square, y = \square, z = \square

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2026Subjective· 4mImportance★★★★★
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Express the system as AX=BAX = B, apply the row operations R2→R2−2R1R_2 \to R_2 - 2R_1, R3→R3−3R1R_3 \to R_3 - 3R_1, then R3→R3−6R2R_3 \to R_3 - 6R_2 to reach an upper-triangular form, and back-substitute to get x=y=z=1x = y = z = 1.

The system is

x−y+z=1,2x−y=1,3x+3y−4z=2x - y + z = 1,\quad 2x - y = 1,\quad 3x + 3y - 4z = 2

In matrix form AX=BAX = B:

[1−112−1033−4][xyz]=[112]\begin{bmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}

Apply R2→R2−2R1R_2 \to R_2 - 2R_1: the new second row is (0,  1,  −2)(0,\; 1,\; -2) with RHS 1−2(1)=−11 - 2(1) = -1.

[1−1101−233−4][xyz]=[1−12]\begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -2 \\ 3 & 3 & -4 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}

Apply R3→R3−3R1R_3 \to R_3 - 3R_1: the new third row is (0,  6,  −7)(0,\; 6,\; -7) with RHS 2−3(1)=−12 - 3(1) = -1.

[1−1101−206−7][xyz]=[1−1−1]\begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -2 \\ 0 & 6 & -7 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}

Apply R3→R3−6R2R_3 \to R_3 - 6R_2: the new third row is (0,  0,  −7+12)=(0,0,5)(0,\; 0,\; -7 + 12) = (0,0,5) with RHS −1−6(−1)=5-1 - 6(-1) = 5. …

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