Class CBSE Class 12 Mathematics Matrices and Determinants Q #1677
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
If $A^{2}=4A+3I$ and $A^{-1}=xA+yI$, then the value of $(x+y)$ is :
(A) -1
(B) 1
(C) $\frac{5}{3}$
(D) 7
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Correct Answer: A

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Detailed Solution

Step 1: Given Equation

We are given the matrix equation: $$A^{2} = 4A + 3I$$

Step 2: Multiply by Inverse

To find $A^{-1}$, multiply both sides of the equation by $A^{-1}$: $$A^{-1}(A^{2}) = A^{-1}(4A + 3I)$$ $$A = 4(A^{-1}A) + 3(A^{-1}I)$$ $$A = 4I + 3A^{-1}$$

Step 3: Isolate $A^{-1}$

Rearrange the equation to solve for $A^{-1}$: $$3A^{-1} = A - 4I$$ $$A^{-1} = \frac{1}{3}A - \frac{4}{3}I$$

Step 4: Compare and Solve

Comparing $A^{-1} = \frac{1}{3}A - \frac{4}{3}I$ with the given form $A^{-1} = xA + yI$, we get: $$x = \frac{1}{3}, \quad y = -\frac{4}{3}$$ Therefore, $x + y = \frac{1}{3} - \frac{4}{3} = -\frac{3}{3} = -1$

Final Answer: -1

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must apply the properties of matrix algebra and the definition of an inverse matrix to manipulate a given polynomial equation.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a step-by-step algorithmic approach to transform the matrix equation into the desired form.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. This tests the student's ability to handle matrix identities beyond simple arithmetic, which is a core competency in the Matrices and Determinants unit.

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