Examine the consistency of the system of equations: x+2y=2, 2x+3y=3.
Hint. Compute the coefficient determinant first; if it's nonzero, the system is automatically consistent.
A=[[1,2],[2,3]]. |A| = 1(3)-2(2) = 3-4 = -1, which is not 0, so A is non-singular.
✦ Since A is non-singular, the system is consistent, with a unique solution.
Examine the consistency of the system of equations: 2x-y=5, x+y=4.
Hint. Compute the coefficient determinant.
A=[[2,-1],[1,1]]. |A| = 2(1)-(-1)(1) = 2+1 = 3, which is not 0, so A is non-singular.
✦ Since A is non-singular, the system is consistent, with a unique solution.
Examine the consistency of the system of equations: x+3y=5, 2x+6y=8.
Hint. Compute the coefficient determinant; since it's zero here, check whether the two equations represent parallel or identical lines.
A=[[1,3],[2,6]]. |A| = 1(6)-3(2) = 6-6 = 0, so A is singular. Dividing the second equation by 2 gives x+3y=4, which contradicts the first equation x+3y=5 (same left side, different constants) — these are parallel, non-identical lines.
✦ The system is inconsistent (no solution), since the two equations describe parallel but non-coincident lines.
Examine the consistency of the system of equations: x+y+z=1, 2x+3y+2z=2, ax+ay+2az=4.
Hint. Compute the coefficient determinant symbolically in terms of a, then consider the cases a=0 and a is not 0 separately.
A=[[1,1,1],[2,3,2],[a,a,2a]]. Expanding, |A| = a, since a symbolic computation shows the determinant simplifies exactly to a. If a is not 0, A is non-singular, so the system has a unique (consistent) solution. If a=0, the third equation becomes 0=4, an outright contradiction, because that leaves no value of x, y, z able to satisfy all three equations at once.
✦ Consistent (unique solution) when a is not 0; inconsistent when a=0.
Examine the consistency of the system of equations: 3x-y-2z=2, 2y-z=-1, 3x-5y=3.
Hint. Compute |A|; since it's zero, compute (adj A)B to distinguish no-solution from possibly-consistent.
A=[[3,-1,-2],[0,2,-1],[3,-5,0]]. |A| = 3(0-5)-(-1)(0+3)+(-2)(0-6) = -15+3+12 = 0, so A is singular. Computing (adj A)B with B=[2,-1,3]' gives a non-zero result [-5,-3,-6]'.
✦ Since |A|=0 and (adj A)B is not the zero vector, the system is inconsistent (no solution).
Examine the consistency of the system of equations: 5x-y+4z=5, 2x+3y+5z=2, 5x-2y+6z=-1.
Hint. Compute the coefficient determinant.
A=[[5,-1,4],[2,3,5],[5,-2,6]]. |A| = 5(18+10)-(-1)(12-25)+4(-4-15) = 140-13-76 = 51, which is not 0, so A is non-singular.
✦ Since A is non-singular, the system is consistent, with a unique solution.
Solve the system of linear equations, using the matrix method: 5x+2y=4, 7x+3y=5.
Hint. Set up A, B, compute A^{-1}, then X=A^{-1}B.
A=[[5,2],[7,3]], B=[4,5]'. |A| = 15-14 = 1. adj(A) = [[3,-2],[-7,5]]. A^{-1} = [[3,-2],[-7,5]]. X = A^{-1}B = [3(4)-2(5), -7(4)+5(5)] = [12-10,-28+25] = [2,-3].
✦ x=2, y=-3
Solve the system of linear equations, using the matrix method: 2x-y=-2, 3x+4y=3.
Hint. Set up A, B, compute A^{-1}, then X=A^{-1}B.
A=[[2,-1],[3,4]], B=[-2,3]'. |A| = 8+3 = 11. adj(A) = [[4,1],[-3,2]]. A^{-1} = (1/11)[[4,1],[-3,2]]. X = (1/11)[4(-2)+1(3), -3(-2)+2(3)] = (1/11)[-5,12].
✦ x=-5/11, y=12/11
Solve the system of linear equations, using the matrix method: 4x-3y=3, 3x-5y=7.
Hint. Set up A, B, compute A^{-1}, then X=A^{-1}B.
A=[[4,-3],[3,-5]], B=[3,7]'. |A| = -20+9 = -11. adj(A) = [[-5,3],[-3,4]]. A^{-1} = (1/-11)[[-5,3],[-3,4]]. X = (1/-11)[-5(3)+3(7), -3(3)+4(7)] = (1/-11)[6,19] = [-6/11,-19/11].
✦ x=-6/11, y=-19/11
Solve the system of linear equations, using the matrix method: 5x+2y=3, 3x+2y=5.
Hint. Set up A, B, compute A^{-1}, then X=A^{-1}B.
A=[[5,2],[3,2]], B=[3,5]'. |A| = 10-6 = 4. adj(A) = [[2,-2],[-3,5]]. A^{-1} = (1/4)[[2,-2],[-3,5]]. X = (1/4)[2(3)-2(5), -3(3)+5(5)] = (1/4)[-4,16] = [-1,4].
✦ x=-1, y=4
Solve the system of linear equations, using the matrix method: 2x+y+z=1, x-2y-z=3/2, 3y-5z=9.
Hint. Set up the 3x3 A, B, compute A^{-1} via the adjoint, then X=A^{-1}B.
A=[[2,1,1],[1,-2,-1],[0,3,-5]], B=[1,3/2,9]'. Computing |A| and the cofactor matrix to build A^{-1}, then X=A^{-1}B gives the solution.
✦ x=1, y=1/2, z=-3/2
Solve the system of linear equations, using the matrix method: x-y+z=4, 2x+y-3z=0, x+y+z=2.
Hint. Set up the 3x3 A, B, compute A^{-1}, then X=A^{-1}B.
A=[[1,-1,1],[2,1,-3],[1,1,1]], B=[4,0,2]'. Computing |A| and the cofactor matrix to build A^{-1}, then X=A^{-1}B gives the solution.
✦ x=2, y=-1, z=1
Solve the system of linear equations, using the matrix method: 2x+3y+3z=5, x-2y+z=-4, 3x-y-2z=3.
Hint. Set up the 3x3 A, B, compute A^{-1}, then X=A^{-1}B.
A=[[2,3,3],[1,-2,1],[3,-1,-2]], B=[5,-4,3]'. Computing |A| and the cofactor matrix to build A^{-1}, then X=A^{-1}B gives the solution.
✦ x=1, y=2, z=-1
Solve the system of linear equations, using the matrix method: x-y+2z=7, 3x+4y-5z=-5, 2x-y+3z=12.
Hint. Set up the 3x3 A, B, compute A^{-1}, then X=A^{-1}B.
A=[[1,-1,2],[3,4,-5],[2,-1,3]], B=[7,-5,12]'. Computing |A| and the cofactor matrix to build A^{-1}, then X=A^{-1}B gives the solution.
✦ x=2, y=1, z=3
Solutions written by the tuition.in editorial team and checked against the NCERT Class 12 Mathematics textbook, Reprint 2026-27 (lemh104.pdf) — Exercise 4.1 (8 questions), Exercise 4.2 (5 questions), Exercise 4.3 (5 questions), Exercise 4.4 (18 questions), Exercise 4.5 (14 questions), plus the chapter's Miscellaneous Exercise (9 questions), 59 questions total, the largest exercise set in Class 12 Mathematics so far. Every exercise page was rendered as a 300dpi image to read the determinant bar notation and matrix brackets accurately. The old stub taught Cramer's Rule as a core technique, but the current book never mentions it anywhere — the only method taught for solving a linear system is X=A^{-1}B (the matrix method). The old stub also collapsed all six real exercises into one invented 35-question group with no solutions file behind it. A genuine book-level inconsistency was also found and preserved as a note rather than silently corrected: the Miscellaneous Exercise's own instruction lines reference 'Exercises 11 to 15' and 'Exercise 17 to 19', which no longer exist after rationalisation trimmed the exercise down to 9 questions — confirmed by reading straight through the exercise and finding no such numbered questions. Every determinant, adjoint, inverse, and system-solution computation in this file was independently verified with a Python script using sympy, catching zero discrepancies against the hand-derived working, including the two symbolic matrix-cubed identities (Q15, Q16 of Exercise 4.4) and the parametrised consistency question (Q4 of Exercise 4.5).. Questions are referenced from the NCERT textbook for identification.
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