Visual SolutionPYQ 2024 · Apr 4 Shift 2Standard
Question
Let A=[1201]A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} and B=I+adj(A)+(adjA)2++(adjA)10B = I + \text{adj}(A) + (\text{adj}\,A)^2 + \ldots + (\text{adj}\,A)^{10}. Then the sum of all elements of the matrix BB is:
(A)124-124
(B)2222
(C)88-88
(D)110-110
Solution Path
adj(A)=[1201]\text{adj}(A) = \begin{bmatrix}1&-2\\0&1\end{bmatrix}. Power pattern: (adjA)k=[12k01](\text{adj}\,A)^k = \begin{bmatrix}1&-2k\\0&1\end{bmatrix}. Sum B=[11110011]B = \begin{bmatrix}11&-110\\0&11\end{bmatrix}. Element sum =88= -88.
01Question Setup
1/4
Find the sum of all elements of B=I+adj(A)+(adjA)2++(adjA)10B = I + \text{adj}(A) + (\text{adj}\,A)^2 + \cdots + (\text{adj}\,A)^{10} where A=[1201]A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}.
A=[1201]A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}
02Find Adjoint
2/4
For a 2×22 \times 2 matrix, swap the diagonal entries and negate the off-diagonal entries.
adj(A)=[1201]\text{adj}(A) = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}
03Power PatternKEY INSIGHT
3/4
By induction (or direct multiplication), (adjA)k=[12k01](\text{adj}\,A)^k = \begin{bmatrix} 1 & -2k \\ 0 & 1 \end{bmatrix}. The off-diagonal scales linearly.
(adjA)k=[12k01](\text{adj}\,A)^k = \begin{bmatrix} 1 & -2k \\ 0 & 1 \end{bmatrix}
04Final Answer
4/4
B=[112(0+1++10)011]=[11110011]B = \begin{bmatrix} 11 & -2(0+1+\cdots+10) \\ 0 & 11 \end{bmatrix} = \begin{bmatrix} 11 & -110 \\ 0 & 11 \end{bmatrix}. Sum =11110+0+11=88= 11 - 110 + 0 + 11 = -88.
88\boxed{-88}
Concepts from this question2 concepts unlocked

Cayley-Hamilton Theorem

STANDARD

Every square matrix satisfies its own characteristic equation: if p(x) = |A - xI|, then p(A) = 0.

A2tr(A)A+AI=0(for 2×2)A^2 - \text{tr}(A) \cdot A + |A| \cdot I = 0 \quad (\text{for } 2 \times 2)

Reduces higher powers of A to lower ones. Essential for expressing A⁻¹ in terms of A and I.

Matrix inverseHigher powers of AMatrix polynomial
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Determinant Product Rule

EASY

det(AB) = det(A) × det(B). Also det(Aᵀ) = det(A) and det(kA) = kⁿ det(A) for n×n matrix.

AB=AB,AT=A,kA=knA|AB| = |A| \cdot |B|, \quad |A^T| = |A|, \quad |kA| = k^n|A|

JEE loves chained matrix products like det(ABA^T). This rule avoids explicit multiplication entirely.

Product of matricesTranspose determinantScalar multiplication
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