Q.Examine the consistency of the following system of equations: 3x−y−2z=2 2y−z=−1 3x−5y=3
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Solving a System of Equations by the Matrix Method
A system of linear equations can be written as a single matrix equation and solved in one clean step using the inverse of a matrix. This is the Class-12 "matrix method" for simultaneous equations.
Writing the system as AX=B
Take the system
a1x+b1y+c1z=d1,a2x+b2y+c2z=d2,a3x+b3y+c3z=d3.
Collect the coefficients, the unknowns, and the constants into matrices:
A=a1a2a3b1b2b3c1c2c3,X=xyz,B=d1d2d3.
Then the whole system is just
AX=B.
Solving when A is invertible
If det(A)=0, then A−1 exists, and multiplying both sides on the left by A−1 gives
X=A−1B,where A−1=det(A)1adj(A).
So you compute det(A), then adj(A), form A−1, and multiply by B. The single column X=A−1B hands you x, y, z at once, and because A−1 is unique, the solution is unique.
Multiply in the correct order: X=A−1B, not BA−1. Matrix multiplication is not commutative, and BA−1 is not even defined here.
When det(A)=0
If det(A)=0, A−1 does not exist and the inverse method fails. The system is then either inconsistent (no solution) or has infinitely many solutions. Decide which by computing (adjA)B:
- (adjA)B=O → no solution (inconsistent).
- (adjA)B=O → infinitely many solutions (consistent, dependent). …
Write the system 3x−y−2z=2, 2y−z=−1, 3x−5y=3 as AX=B with A=303−12−5−2−10.
Determinant: detA=3(2⋅0−(−1)(−5))−(−1)(0−(−1)(3))+(−2)(0−2⋅3)=3(−5)+1(3)−2(−6)=−15+3+12=0. …
The coefficient determinant is 0, and the equations force 4y−2z=−1 and 4y−2z=−2 at once — a contradiction, so the system has no solution.
Write the equations in standard form, filling in the missing terms:
⎩⎨⎧3x−y−2z=20x+2y−z=−13x−5y+0z=3
1. Determinant test
A=303−12−5−2−10,detA=32−5−10−(−1)03−10+(−2)032−5.
=3(0−5)+1(0+3)−2(0−6)=−15+3+12=0.
A zero determinant means A−1 does not exist, so the system has either no solution or infinitely many — we must test which.
2. Check consistency by elimination
Subtract eq3 from eq1 to remove x:
(3x−y−2z)−(3x−5y)=2−3⇒4y−2z=−1. …
Method: When the Determinant is Zero — Distinguishing 'No Solution' from 'Infinitely Many'
A zero determinant only tells you A−1 does not exist; it never by itself tells you whether the system is inconsistent or has infinitely many solutions. This method shows how to decide which.
Steps
Step 1: Write the system as AX=B and compute det(A)
Expand a 3×3 determinant along any row (the first row is usually simplest), using the alternating cofactor sign pattern +,−,+:
det(A)=a11a22a32a23a33−a12a21a31a23a33+a13a21a31a22a32
Step 2: If det(A)=0, do NOT conclude 'inconsistent' yet
A zero determinant only rules out a unique solution — the correct next move is to test the equations directly for a contradiction.
Step 3: Combine two of the original equations to try to isolate a common expression …
Common Mistakes
Mistake 1: Concluding 'no solution' the instant det(A)=0, without checking further
Why it's wrong: det(A)=0 only rules out a UNIQUE solution — it is consistent with either 'no solution' or 'infinitely many.' Here it happens to be no solution, but that had to be confirmed separately by finding a genuine contradiction (4y−2z=−1 from one combination versus 4y−2z=−2 from another), not assumed from the zero determinant alone. Correct approach: treat det(A)=0 as a trigger to test the equations directly, never as the final answer by itself.
Mistake 2: A sign error in the 3×3 cofactor expansion …
- KEAM 2026Set eng-2026-04194 marksMCQQ.If (x3−1)1−1110011−1231=0, then the values of x are (A) -2 (B) 3−1 (C) -3 (D) 32 (E) 3−2
›Reveal solutionSolution
Do the matrix multiplication in stages (right pair first), reduce the row-vector × column-vector to a scalar, set it to 0, and solve for x.
Concept. The product of a 1×3 row, a 3×3 matrix, and a 3×1 column is a 1×1 scalar. Multiplication is associative, so evaluate the 3×3 times the column first.
Step 1 — multiply the matrix by the column (2,3,1)T.
1−1110011−1231=1⋅2+1⋅3+1⋅1−1⋅2+0⋅3+1⋅11⋅2+0⋅3−1⋅1=6−11. …
- KEAM 2024Set eng-2024-06064 marksMCQQ.Let A=(3−1−231−1) and B=1α−1. If AB=(−26), then the value of α is equal to (A) -1 (B) 1 (C) -2 (D) 2 (E) 0
›Reveal solutionSolution
Multiply A (2×3) by B (3×1) and match to (−26).
First entry of AB: 3(1)+(−2)(α)+1(−1)=2−2α. Setting 2−2α=−2 gives α=2. …
- KEAM 2024Set eng-2024-06054 marksMCQQ.Let A=(0−112) and B=(1−11−1). If XA=B, then X is (A) (−31−11) (B) (−331−1) (C) (3−3−11) (D) (1001) (E) (−100−1)
›Reveal solutionSolution
From XA=B, X=BA−1; computing gives (3−3−11).
detA=0⋅2−1⋅(−1)=1, so A−1=(21−10). …
- KEAM 2025Set eng-2025-04274 marksMCQQ.Let A=(0324), I=(1001). If (I+A)(42−3−1)=(822−5x), then the value of x is equal to (A) 14 (B) -14 (C) 12 (D) -12 (E) 15
›Reveal solutionSolution
Compute (I+A)(42−3−1); the bottom-right entry gives x=−14.
First,
I+A=(1001)+(0324)=(1325).
Multiply: …
- KEAM 2026Set eng-2026-04184 marksMCQQ.Let A=[3−25−3]. If BA2=A, where B is a 2×2 matrix, then B= (A) [−32−53] (B) [32−5−3] (C) [35−2−3] (D) [3−52−3] (E) [3−25−3]
›Reveal solutionSolution
B=AA−2=A−1, and A−1=[−32−53].
From BA2=A, multiply on the right by A−2:
B=AA−2=A−1.
For A=[3−25−3], detA=(3)(−3)−(5)(−2)=−9+10=1, so …
- KEAM 2025Set eng-2025-04294 marksMCQQ.If A=[X101] and B=[16501] and if A2=B then the value X is equal to (A) 2 (B) 3 (C) 4 (D) 5 (E) 6
›Reveal solutionSolution
Squaring A gives entries X2 and X+1; matching to B gives X2=16 and X+1=5, both satisfied by X=4.
Compute A2 with A=[X101]:
A2=[X101][X101]=[X2X+101]. …
- KEAM 2026Set eng-2026-04194 marksMCQQ.If A=[0−110] and (αI+βA)2=A where I is 2×2 unit matrix, then α2−β2= (A) 2 (B) -2 (C) -1 (D) 1 (E) 0
›Reveal solutionSolution
Squaring αI+βA and matching with A makes the diagonal entries α2−β2 equal 0.
Note A2=(0−110)2=−I. Then
(αI+βA)2=α2I+2αβA+β2A2=(α2−β2)I+2αβA.
In matrix form this is (α2−β2−2αβ2αβα2−β2). …
- KEAM 2026Set eng-2026-04224 marksMCQQ.If A is a non-singular matrix of order n satisfying the matrix equation I+A+A2+A3+…+A10=O, where I and O are, respectively, unit and null matrices of order n, then A10= (A) A−1 (B) I (C) A (D) I+A (E) O
›Reveal solutionSolution
Multiply the series by A, subtract, to collapse it to A11=I.
Given I+A+A2+⋯+A10=O.
Multiply by A: A+A2+⋯+A11=O.
Subtract the original: A11−I=O⇒A11=I. …
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