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JEE Mathematics 2023 Question with Solution

Let A=[aij]2×2A = [a_{ij}]_{2\times 2}, where aij≠0a_{ij} \ne 0 for all i,ji, j, and A2=IA^2 = I. Let aa be the sum of all diagonal elements of AA and b=∣A∣b = |A|. Then 3a2+4b23a^2 + 4b^2 is equal to:

  • A

    1414

  • B

    44

  • C

    33

  • D

    77

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A=[aij]2×2A = [a_{ij}]_{2\times 2} with aij≠0a_{ij} \ne 0 for all entries, and A2=IA^2 = I.

Find: The value of 3a2+4b23a^2 + 4b^2, where aa is the sum of diagonal elements of AA and b=∣A∣b = |A|.

Let

A=[pqrs]A = \begin{bmatrix} p & q \\ r & s \end{bmatrix}

Then

A2=[p2+qrpq+qspr+rsrq+s2]A^2 = \begin{bmatrix} p^2 + qr & pq + qs \\ pr + rs & rq + s^2 \end{bmatrix}

Since A2=IA^2 = I, we get

p2+qr=1pq+qs=0pr+rs=0rq+s2=1\begin{aligned} p^2 + qr &= 1 \\ pq + qs &= 0 \\ pr + rs &= 0 \\ rq + s^2 &= 1 \end{aligned}

From the middle two equations,

q(p+s)=0,r(p+s)=0q(p+s) = 0, \qquad r(p+s) = 0

Since all entries are non-zero, q≠0q \ne 0 and r≠0r \ne 0. Therefore,

p+s=0p+s = 0

So the sum of diagonal elements is

a=p+s=0a = p+s = 0

Also, taking determinant on both sides of A2=IA^2 = I,

∣A2∣=∣I∣  ⟹  ∣A∣2=1|A^2| = |I| \implies |A|^2 = 1

Hence

b2=1b^2 = 1

Now compute

3a2+4b2=3(0)2+4(1)=43a^2 + 4b^2 = 3(0)^2 + 4(1) = 4

Therefore, the value of 3a2+4b23a^2 + 4b^2 is 44. Hence, the correct option is B.

Common mistakes

  • Using p2=s2p^2 = s^2 to conclude only p=sp=s is incorrect. Here the needed conclusion is p+s=0p+s=0 because q≠0q \ne 0 and r≠0r \ne 0 force p+s=0p+s=0 from the off-diagonal equations.

  • Forgetting that all entries are non-zero leads to missing the key step. Since qq and rr are non-zero, q(p+s)=0q(p+s)=0 and r(p+s)=0r(p+s)=0 imply p+s=0p+s=0, not q=0q=0 or r=0r=0.

  • Confusing bb with ∣b∣|b| is a common error. From ∣A∣2=1|A|^2 = 1, we only need b2=1b^2=1; there is no need to decide whether b=1b=1 or b=−1b=-1 separately.

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