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Concept 05 / 09Matrices and Determinants

Systems of Linear Equations: Cramer's Rule, the Matrix Method and Consistency

Writing a linear system as AX = B, the unique solution X = A^-1 B when |A| is non-zero, Cramer's rule with its geometric picture, and a reliable procedure for deciding between no solution and infinitely many solutions when |A| = 0.

Figure 1
A system of nn linear equations in nn unknowns can be written in matrix form as AX=BAX = B, where AA is the coefficient matrix, XX the column of unknowns and BB the column of constants. The system is consistent if it has at least one solution and inconsistent if it has none. Geometrically, in two variables each equation is a line and a solution is a common point; in three variables each equation is a plane. Case ∣A∣≠0|A| \neq 0: there is exactly one solution, X=A−1BX = A^{-1}B. Multiply AX=BAX = B on the left by A−1A^{-1} to get it, and it clearly works. The same solution can be written with Cramer's rule: xi=ΔiΔx_{i} = \frac{\Delta_{i}}{\Delta}, where Δ=∣A∣\Delta = |A| and Δi\Delta_{i} is the determinant obtained by replacing column ii of AA by the column BB. For three variables, x=Δ1Δx = \frac{\Delta_{1}}{\Delta}, y=Δ2Δy = \frac{\Delta_{2}}{\Delta} and z=Δ3Δz = \frac{\Delta_{3}}{\Delta}. Why it works: substitute B=x1a1+x2a2+x3a3B = x_{1}a_{1} + x_{2}a_{2} + x_{3}a_{3} (a combination of the columns aja_{j} of AA) into column ii and split by linearity; every term except the one containing aia_{i} has two equal columns and vanishes, leaving Δi=xiΔ\Delta_{i} = x_{i}\Delta. Case ∣A∣=0|A| = 0: either no solution or infinitely many. The equations are then dependent in their left-hand sides, and everything depends on whether the right-hand sides respect the same dependence. Test with the adjoint: if (adj A)B≠O(\text{adj}\,A)B \neq O, the system is inconsistent. Reason: multiplying AX=BAX = B by adj A\text{adj}\,A gives ∣A∣X=(adj A)B|A|X = (\text{adj}\,A)B, and the left side is OO, so a non-zero right side is a contradiction. If (adj A)B=O(\text{adj}\,A)B = O: for 2×22 \times 2 systems (with A≠OA \neq O) this means infinitely many solutions. For 3×33 \times 3 systems it usually means infinitely many solutions, but not always. It is a necessary condition only, and there are examples (such as three parallel planes) where adj A=O\text{adj}\,A = O and the system is still inconsistent. Test with Cramer determinants: if Δ=0\Delta = 0 and at least one Δi≠0\Delta_{i} \neq 0, there is no solution. If Δ=Δ1=Δ2=Δ3=0\Delta = \Delta_{1} = \Delta_{2} = \Delta_{3} = 0, the system has either infinitely many solutions or none, and you must check directly. The reliable final check is elimination: reduce the equations and see whether you reach a contradiction such as 0=50 = 5 or an identity 0=00 = 0. The figure shows the two-variable picture. Two lines either cross at one point (∣A∣≠0|A| \neq 0), are parallel and distinct (no solution), or coincide (infinitely many). For two equations, parallel lines mean a1a2=b1b2\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}; they coincide if also a1a2=c1c2\frac{a_{1}}{a_{2}} = \frac{c_{1}}{c_{2}}. Parameter questions are the most common JEE format: 'find λ,μ\lambda, \mu so that the system has (i) a unique solution, (ii) no solution, (iii) infinitely many solutions'. The routine is: compute Δ\Delta as a function of the parameters; unique solution iff Δ≠0\Delta \neq 0; at the values making Δ=0\Delta = 0, substitute and use elimination (or the Δi\Delta_{i}) to separate the other two cases. Key Takeaways: - AX=BAX = B; ∣A∣≠0⇒|A| \neq 0 \Rightarrow unique solution X=A−1BX = A^{-1}B. - Cramer: xi=ΔiΔx_{i} = \frac{\Delta_{i}}{\Delta}, with column ii replaced by the constants. - ∣A∣=0|A| = 0 and (adj A)B≠O(\text{adj}\,A)B \neq O: no solution. - ∣A∣=0|A| = 0 and some Δi≠0\Delta_{i} \neq 0: no solution. - ∣A∣=0|A| = 0 and all Δi=0\Delta_{i} = 0: infinitely many or none; confirm by elimination. - In two variables: intersecting, parallel or coincident lines. - Unique-solution condition depends only on AA, never on BB. - Pitfall: declaring 'infinitely many solutions' for a 3×33 \times 3 system just because all Δi=0\Delta_{i} = 0. Brain-Teaser 1: For the system x+y+z=6x + y + z = 6, x+2y+3z=10x + 2y + 3z = 10, x+2y+λz=μx + 2y + \lambda z = \mu, find the conditions for a unique solution, no solution and infinitely many solutions. Solution: Δ=∣11112312λ∣\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix}. Subtract row 2 from row 3 to get (0,0,λ−3)(0, 0, \lambda - 3); expanding along it gives Δ=(λ−3)(2−1)=λ−3\Delta = (\lambda - 3)(2 - 1) = \lambda - 3. So there is a unique solution iff λ≠3\lambda \neq 3, for any μ\mu. If λ=3\lambda = 3, the third equation becomes x+2y+3z=μx + 2y + 3z = \mu, which has the same left side as the second equation. If μ=10\mu = 10 it duplicates the second equation, leaving two independent equations in three unknowns: infinitely many solutions. If μ≠10\mu \neq 10 the two equations contradict each other: no solution. Comparing equations directly in the singular case is quicker and safer than computing all the Δi\Delta_{i}. Brain-Teaser 2: Show that the system x+y+z=1x + y + z = 1, x+y+z=2x + y + z = 2, x+y+z=3x + y + z = 3 has Δ=Δ1=Δ2=Δ3=0\Delta = \Delta_{1} = \Delta_{2} = \Delta_{3} = 0 yet no solution. Solution: The coefficient matrix has three identical rows, so Δ=0\Delta = 0. Each Δi\Delta_{i} is formed by replacing one column with (1,2,3)(1, 2, 3); the remaining two columns are both (1,1,1)(1, 1, 1), so each Δi\Delta_{i} has two equal columns and is 00. Yet the equations say x+y+zx + y + z equals 11, 22 and 33 simultaneously, which is impossible. Geometrically these are three parallel planes. Here adj A=O\text{adj}\,A = O as well (every 2×22 \times 2 minor of a matrix of all ones is 00), so (adj A)B=O(\text{adj}\,A)B = O does not guarantee consistency either. This is the standard counterexample showing why the determinant tests are not the final word when Δ=0\Delta = 0. Brain-Teaser 3: For which kk do the lines kx+y=1kx + y = 1 and x+ky=1x + ky = 1 meet in exactly one point, never meet, or coincide? Solution: Δ=k2−1\Delta = k^{2} - 1, so a unique intersection needs k≠±1k \neq \pm 1. At k=1k = 1, both equations read x+y=1x + y = 1: the lines coincide, giving infinitely many common points. At k=−1k = -1, the equations are −x+y=1-x + y = 1 and x−y=1x - y = 1, that is, y=x+1y = x + 1 and y=x−1y = x - 1: parallel and distinct, so no solution. The unique solution for other kk is x=y=1k+1x = y = \frac{1}{k + 1}, which blows up exactly at k=−1k = -1 and becomes indeterminate as a limit at k=1k = 1. Watching how the Cramer formula breaks down at the singular values tells you which singular case you are in.

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