Q.Let A=[23−14], B=[5724], C=[2358]. Find a matrix D such that CD−AB=O.
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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). …
Idea: CD−AB=O means CD=AB; since C is invertible, D=C−1(AB).
Step 1 — compute AB.
AB=[23−14][5724]=[10−715+284−46+16]=[343022].
Step 2 — invert C. detC=(2)(8)−(5)(3)=1, so
C−1=[8−3−52].
Step 3 — solve. …
From CD=AB with C invertible, D=C−1(AB)=[−19177−11044].
The equation CD−AB=O rearranges to CD=AB. Because C sits on the left of the unknown D, we undo it by left-multiplying both sides by C−1: C−1(CD)=C−1(AB), i.e. D=C−1(AB). So the plan is: compute AB, invert C, multiply.
Step 1 — compute AB
AB=[23−14][5724].
- (1,1): 2⋅5+(−1)⋅7=3
- (1,2): 2⋅2+(−1)⋅4=0
- (2,1): 3⋅5+4⋅7=43
- (2,2): 3⋅2+4⋅4=22
So AB=[343022].
Step 2 — invert C
detC=(2)(8)−(5)(3)=16−15=1=0,
so C is invertible. For a 2×2 matrix [acbd] the inverse is det1[d−c−ba]; here det=1, so
C−1=[8−3−52].
Step 3 — solve for D
D=C−1(AB)=[8−3−52][343022].
- (1,1): 8⋅3+(−5)⋅43=24−215=−191
- (1,2): 8⋅0+(−5)⋅22=−110
- (2,1): −3⋅3+2⋅43=−9+86=77 …
Method: Solving a matrix equation for an unknown matrix using an inverse
Use this when an equation like CD−AB=O must be solved for the matrix D and the coefficient matrix is square and invertible.
Steps
Step 1: Rearrange to isolate the unknown's term.
CD−AB=O⇒CD=AB.
Step 2: Left-multiply by the inverse of the coefficient on the correct side. …
Common Mistakes
Mistake 1: Multiplying by C−1 on the wrong side.
Why it's wrong: because C is on the left of D, you must left-multiply: D=C−1(AB), not (AB)C−1. Matrix multiplication is not commutative. Correct approach: apply the inverse on the same side as the coefficient.
Mistake 2: Errors in the 2×2 inverse formula. …
- 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 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-2025-04274 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 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 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. …
- 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 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 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]. …
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