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Worked Examples · Example 37

Q.Solve the following system of equation finding the inverse of coefficient matrix: x−y=5x - y = 5; 2x+3y=−12x + 3y = -1.

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Solving AX=BAX=B via X=A−1BX=A^{-1}B gives x=145, y=−115x=\tfrac{14}{5},\ y=-\tfrac{11}{5}.

For AX=BAX=B with det⁡A≠0\det A\neq0: X=A−1BX=A^{-1}B, and for a 2×22\times2 matrix A−1=1det⁡A[d−b−ca]A^{-1}=\dfrac{1}{\det A}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.

  1. Matrix form of x−y=5, 2x+3y=−1x-y=5,\ 2x+3y=-1:

[1−123][xy]=[5−1].\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ -1 \end{bmatrix}.

  1. det⁡A=(1)(3)−(−1)(2)=3+2=5≠0\det A = (1)(3)-(-1)(2) = 3+2 = 5 \neq 0, so a unique solution exists.

  2. Inverse:

A−1=15[31−21].A^{-1} = \frac{1}{5}\begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix}.

  1. Compute X=A−1BX=A^{-1}B:

X=15[31−21][5−1]=15[15−1−10−1]=15[14−11].X = \frac{1}{5}\begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 5 \\ -1 \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 15-1 \\ -10-1 \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 14 \\ -11 \end{bmatrix}.

  1. Hence x=145, y=−115x=\dfrac{14}{5},\ y=-\dfrac{11}{5}.

  2. Check: x−y=145+115=255=5x-y = \tfrac{14}{5}+\tfrac{11}{5}=\tfrac{25}{5}=5 ✓; 2x+3y=285−335=−12x+3y=\tfrac{28}{5}-\tfrac{33}{5}=-1 ✓.

✓Final answer

x=145,y=−115x = \dfrac{14}{5},\quad y = -\dfrac{11}{5}

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