Use Cramer's rule to solve 2x+3y=5 and x-2y=1

easy CBSE JEE-MAIN 4 min read

Question

Solve the system of linear equations using Cramer’s rule:

2x+3y=52x + 3y = 5 x2y=1x - 2y = 1

Solution — Step by Step

For a system a1x+b1y=c1a_1x + b_1y = c_1 and a2x+b2y=c2a_2x + b_2y = c_2:

The coefficient determinant (D) and modified determinants (DxD_x, DyD_y) are:

D=a1b1a2b2,Dx=c1b1c2b2,Dy=a1c1a2c2D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}, \quad D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix}, \quad D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}

For our equations (a1=2,b1=3,c1=5a_1 = 2, b_1 = 3, c_1 = 5; a2=1,b2=2,c2=1a_2 = 1, b_2 = -2, c_2 = 1):

D=2312=(2)(2)(3)(1)=43=7D = \begin{vmatrix} 2 & 3 \\ 1 & -2 \end{vmatrix} = (2)(-2) - (3)(1) = -4 - 3 = -7

Since D0D \neq 0, the system has a unique solution.

Replace the first column (coefficients of xx) with the constants:

Dx=5312=(5)(2)(3)(1)=103=13D_x = \begin{vmatrix} 5 & 3 \\ 1 & -2 \end{vmatrix} = (5)(-2) - (3)(1) = -10 - 3 = -13

Replace the second column (coefficients of yy) with the constants:

Dy=2511=(2)(1)(5)(1)=25=3D_y = \begin{vmatrix} 2 & 5 \\ 1 & 1 \end{vmatrix} = (2)(1) - (5)(1) = 2 - 5 = -3
x=DxD=137=137x = \frac{D_x}{D} = \frac{-13}{-7} = \frac{13}{7} y=DyD=37=37y = \frac{D_y}{D} = \frac{-3}{-7} = \frac{3}{7}

Verification: 2x+3y=2(13/7)+3(3/7)=26/7+9/7=35/7=52x + 3y = 2(13/7) + 3(3/7) = 26/7 + 9/7 = 35/7 = 5

x2y=13/76/7=7/7=1x - 2y = 13/7 - 6/7 = 7/7 = 1

Why This Works

Cramer’s rule provides a direct formula for the solution of a linear system using determinants. It comes from the theory of matrices: if AX=BAX = B, then X=A1BX = A^{-1}B, which can be expressed using the adjugate matrix. For a 2×2 system, this simplifies to the determinant formula.

The key insight: DD is the determinant of the coefficient matrix. If D0D \neq 0, the matrix is invertible and the system has a unique solution. The values DxD_x and DyD_y are obtained by replacing the column of the variable being solved with the constants column.

Geometrically, each equation represents a straight line, and the solution (x,y)=(13/7,3/7)(x, y) = (13/7, 3/7) is the unique point where the two lines intersect.

Alternative Method

Substitution method (for comparison):

From the second equation: x=1+2yx = 1 + 2y

Substituting into the first: 2(1+2y)+3y=52(1 + 2y) + 3y = 5

2+4y+3y=57y=3y=3/72 + 4y + 3y = 5 \Rightarrow 7y = 3 \Rightarrow y = 3/7

Then x=1+2(3/7)=1+6/7=13/7x = 1 + 2(3/7) = 1 + 6/7 = 13/7

Same answer. Cramer’s rule is more systematic for larger systems (3×3 and beyond) and avoids the algebraic manipulation of substitution.

For CBSE Class 12 determinants chapter, Cramer’s rule questions are standard 4-mark problems. The systematic approach (find D, Dx, Dy, then compute ratios) is safer than substitution for board exams because each step is clearly examinable. Always write out each determinant expansion in full — the calculation for D, Dx, and Dy each earns partial marks. If D = 0, state that the system either has no solution or infinitely many solutions (you need to investigate further).

Common Mistake

The most common error in Cramer’s rule is the column replacement. For DxD_x, replace the x-column (first column) with constants. For DyD_y, replace the y-column (second column) with constants. Students often replace the wrong column, swapping DxD_x and DyD_y. A good check: if you compute both xx and yy and verify them in the original equations, a column-swap error will show up immediately. Always verify your answer in both equations.

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