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Concept 09 / 09Matrices and Determinants

Homogeneous Systems and Applications: Non-Trivial Solutions, Concurrency and Coplanarity

Why AX = O is always consistent, the determinant condition for non-trivial solutions, eliminating variables to get conditions on parameters, and determinant conditions for concurrent lines, collinear points and the consistency of three equations in two unknowns.

A system is homogeneous when every constant term is zero: AX=OAX = O. It always has the trivial solution X=OX = O, so it is always consistent. The only question is whether it has other solutions. If ∣A∣≠0|A| \neq 0, then X=A−1O=OX = A^{-1}O = O is the only solution. If ∣A∣=0|A| = 0, there are infinitely many solutions, including non-trivial ones. The reason is dimensional: ∣A∣=0|A| = 0 means the columns of AA are linearly dependent, so some non-zero combination of them is zero, and the coefficients of that combination form a non-trivial solution. Any scalar multiple of a solution is again a solution, so there are infinitely many. So: AX=OAX = O has a non-trivial solution   ⟺  ∣A∣=0\iff |A| = 0. This condition is the most used determinant fact in JEE algebra. It turns 'there exist x,y,zx, y, z, not all zero, such that...' into a single equation. Finding the solutions when ∣A∣=0|A| = 0 (for 3×33 \times 3). If two of the equations are independent, solve them for two variables in terms of the third, or use the cross-multiplication rule: from a1x+b1y+c1z=0a_{1}x + b_{1}y + c_{1}z = 0 and a2x+b2y+c2z=0a_{2}x + b_{2}y + c_{2}z = 0, xb1c2−b2c1=yc1a2−c2a1=za1b2−a2b1\frac{x}{b_{1}c_{2} - b_{2}c_{1}} = \frac{y}{c_{1}a_{2} - c_{2}a_{1}} = \frac{z}{a_{1}b_{2} - a_{2}b_{1}}. Geometrically, (x,y,z)(x, y, z) is parallel to the cross product of the two coefficient vectors, since it is perpendicular to both. Elimination. Eliminating variables from a set of linear equations produces a determinant condition. For instance, to eliminate x,y,zx, y, z from x=cy+bzx = cy + bz, y=az+cxy = az + cx, z=bx+ayz = bx + ay (with x,y,zx, y, z not all zero), rewrite as a homogeneous system and set its determinant to zero: ∣−1cbc−1aba−1∣=0\begin{vmatrix} -1 & c & b \\ c & -1 & a \\ b & a & -1 \end{vmatrix} = 0, which simplifies to a2+b2+c2+2abc=1a^{2} + b^{2} + c^{2} + 2abc = 1. Applications to coordinate geometry. Concurrency of three lines. The lines aix+biy+ci=0a_{i}x + b_{i}y + c_{i} = 0, i=1,2,3i = 1, 2, 3, pass through a common point only if ∣a1b1c1a2b2c2a3b3c3∣=0\begin{vmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{vmatrix} = 0. Reason: a common point (x0,y0)(x_{0}, y_{0}) gives a non-trivial solution (x0,y0,1)(x_{0}, y_{0}, 1) of the homogeneous system. The converse needs care: the determinant also vanishes when the three lines are parallel (they 'meet at infinity'), so check that no two are parallel. Consistency of three equations in two unknowns. The same determinant condition, with the constants as the third column, is necessary for aix+biy=cia_{i}x + b_{i}y = c_{i} (i=1,2,3i = 1, 2, 3) to have a common solution. Collinearity of three points: the area determinant with a column of 11s is zero. Coplanarity in three dimensions: four points are coplanar exactly when the 4×44 \times 4 determinant with rows (xi,yi,zi,1)(x_{i}, y_{i}, z_{i}, 1) is zero, equivalently the scalar triple product of three edge vectors is zero. Key Takeaways: - AX=OAX = O is always consistent; X=OX = O is always a solution. - Non-trivial solutions exist   ⟺  ∣A∣=0\iff |A| = 0, and then there are infinitely many. - Two independent equations: x:y:z=(b1c2−b2c1):(c1a2−c2a1):(a1b2−a2b1)x : y : z = (b_{1}c_{2} - b_{2}c_{1}) : (c_{1}a_{2} - c_{2}a_{1}) : (a_{1}b_{2} - a_{2}b_{1}). - 'Not all zero' in a question signals the condition ∣A∣=0|A| = 0. - Three lines concurrent ⇒\Rightarrow determinant of coefficients =0= 0; check they are not all parallel. - Three equations in two unknowns consistent ⇒\Rightarrow the 3×33 \times 3 augmented determinant is 00. - Pitfall: treating ∣A∣=0|A| = 0 as 'no solution' for a homogeneous system. It means infinitely many. Brain-Teaser 1: Find λ\lambda for which x+2y+3z=λxx + 2y + 3z = \lambda x, 3x+y+2z=λy3x + y + 2z = \lambda y, 2x+3y+z=λz2x + 3y + z = \lambda z has a non-trivial solution. Solution: Rearrange to (1−λ)x+2y+3z=0(1 - \lambda)x + 2y + 3z = 0, 3x+(1−λ)y+2z=03x + (1 - \lambda)y + 2z = 0, 2x+3y+(1−λ)z=02x + 3y + (1 - \lambda)z = 0. Set the determinant to zero. Every row sums to 6−λ6 - \lambda, so add all columns into the first: the determinant becomes (6−λ)∣12311−λ2131−λ∣(6 - \lambda)\begin{vmatrix} 1 & 2 & 3 \\ 1 & 1 - \lambda & 2 \\ 1 & 3 & 1 - \lambda \end{vmatrix}. Subtract row 1 from rows 2 and 3: (0,−1−λ,−1)(0, -1 - \lambda, -1) and (0,1,−2−λ)(0, 1, -2 - \lambda). The 2×22 \times 2 determinant is (1+λ)(2+λ)+1=λ2+3λ+3(1 + \lambda)(2 + \lambda) + 1 = \lambda^{2} + 3\lambda + 3, whose discriminant is 9−12<09 - 12 < 0. So the only real value is λ=6\lambda = 6. (In the language of the next concept, λ\lambda is an eigenvalue of the coefficient matrix, and 66 is its only real eigenvalue.) At λ=6\lambda = 6, x=y=zx = y = z solves all three equations, confirming a non-trivial solution. Brain-Teaser 2: For what kk are the lines x+y=1x + y = 1, 2x+3y=42x + 3y = 4 and kx−y=7kx - y = 7 concurrent? Solution: The lines x+y−1=0x + y - 1 = 0, 2x+3y−4=02x + 3y - 4 = 0 and kx−y−7=0kx - y - 7 = 0 are concurrent only if ∣11−123−4k−1−7∣=0\begin{vmatrix} 1 & 1 & -1 \\ 2 & 3 & -4 \\ k & -1 & -7 \end{vmatrix} = 0. Expand along the first row: 1(−21−4)−1(−14+4k)+(−1)(−2−3k)=−25+14−4k+2+3k=−9−k1(-21 - 4) - 1(-14 + 4k) + (-1)(-2 - 3k) = -25 + 14 - 4k + 2 + 3k = -9 - k. So k=−9k = -9. Check: the first two lines meet at (−1,2)(-1, 2), and −9(−1)−2=7-9(-1) - 2 = 7. Since the first two lines are not parallel, the determinant condition is sufficient here. Brain-Teaser 3: If the equations a(y+z)=xa(y + z) = x, b(z+x)=yb(z + x) = y, c(x+y)=zc(x + y) = z have a non-trivial solution, show that 11+a+11+b+11+c=2\frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c} = 2 (assume a,b,c≠−1a, b, c \neq -1). Solution: Add axax to both sides of the first equation: a(x+y+z)=(1+a)xa(x + y + z) = (1 + a)x, so a1+a=xS\frac{a}{1 + a} = \frac{x}{S} where S=x+y+zS = x + y + z. Similarly b1+b=yS\frac{b}{1 + b} = \frac{y}{S} and c1+c=zS\frac{c}{1 + c} = \frac{z}{S}. For a non-trivial solution, S≠0S \neq 0: if S=0S = 0, each equation would give x=y=z=0x = y = z = 0 (from (1+a)x=0(1 + a)x = 0 and so on). Adding the three, a1+a+b1+b+c1+c=1\frac{a}{1 + a} + \frac{b}{1 + b} + \frac{c}{1 + c} = 1. Since a1+a=1−11+a\frac{a}{1 + a} = 1 - \frac{1}{1 + a}, this reads 3−∑11+a=13 - \sum\frac{1}{1 + a} = 1, so ∑11+a=2\sum\frac{1}{1 + a} = 2. A clever rearrangement can sometimes replace the determinant condition entirely, but it still encodes the same fact: a non-trivial solution exists only when the equations are dependent.