Q.4. Let and . If and , then among the two statements :
(S1):
and
(S2): ,
(1) only (S1) is correc

Q: What is the correct answer for JEE Main Mathematics Q4?

The correct answer is B. Sol. For matrix A: $A^2 - 4A + I = O$ By Cayley-Hamilton theorem, the characteristic equation is $\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$. So, $\text{tr}(A) = 4$ and $\det(A) = 1$. $\text{tr}(A) = 1 + \alpha = 4 \Rightarrow \alpha = 3$. $\det(A) = 1 \cdot \alpha - 2 \cdot 1 = \alpha - 2 = 1 \Rightarrow \alpha = 3$. (Consistent) So, $A = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}$. For mat

Q: Q.4. Let and . If and , then among the two statements : (S1): and (S2): , (1) only (S1) is correct (2) only (S2) is correct (3) both (S1) and (S2) are correct (4) both (S1) and (S2) are wrong

The correct answer is B. Sol. For matrix A: $A^2 - 4A + I = O$ By Cayley-Hamilton theorem, the characteristic equation is $\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$. So, $\text{tr}(A) = 4$ and $\det(A) = 1$. $\text{tr}(A) = 1 + \alpha = 4 \Rightarrow \alpha = 3$. $\det(A) = 1 \cdot \alpha - 2 \cdot 1 = \alpha - 2 = 1 \Rightarrow \alpha = 3$. (Consistent) So, $A = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}$. For mat

Q: What are the options for this JEE Main Mathematics question?

A. only (S1) is correct · B. only (S2) is correct · C. both (S1) and (S2) are correct · D. both (S1) and (S2) are wrong

Question

Q.4. Let  and . If  and , then among the two statements :
(S1): 
and
(S2): ,
(1) only (S1) is correct
(2) only (S2) is correct
(3) both (S1) and (S2) are correct
(4) both (S1) and (S2) are wrong

FirstInTest · Accepted Answer

Sol. For matrix A: $A^2 - 4A + I = O$
By Cayley-Hamilton theorem, the characteristic equation is $\lambda^2 - 	ext{tr}(A)\lambda + \det(A) = 0$.
So, $	ext{tr}(A) = 4$ and $\det(A) = 1$.
$	ext{tr}(A) = 1 + \alpha = 4 \Rightarrow \alpha = 3$.
$\det(A) = 1 \cdot \alpha - 2 \cdot 1 = \alpha - 2 = 1 \Rightarrow \alpha = 3$. (Consistent)
So, $A = \begin{pmatrix} 1 & 2 \ 1 & 3 \end{pmatrix}$.
For matrix B: $B^2 - 5B - 6I = O$
By Cayley-Hamilton theorem, $	ext{tr}(B) = 5$ and $\det(B) = -6$.
$	ext{tr}(B) = 3 + 2 = 5$. (Consistent)
$\det(B) = 3 \cdot 2 - 3 \cdot \beta = 6 - 3\beta = -6 \Rightarrow 3\beta = 12 \Rightarrow \beta = 4$.
So, $B = \begin{pmatrix} 3 & 3 \ 4 & 2 \end{pmatrix}$.

Statement (S1):
$B - A = \begin{pmatrix} 3 & 3 \ 4 & 2 \end{pmatrix} - \begin{pmatrix} 1 & 2 \ 1 & 3 \end{pmatrix} = \begin{pmatrix} 2 & 1 \ 3 & -1 \end{pmatrix}$
$B + A = \begin{pmatrix} 3 & 3 \ 4 & 2 \end{pmatrix} + \begin{pmatrix} 1 & 2 \ 1 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 5 \ 5 & 5 \end{pmatrix}$
$(B-A)(B+A) = \begin{pmatrix} 2 & 1 \ 3 & -1 \end{pmatrix} \begin{pmatrix} 4 & 5 \ 5 & 5 \end{pmatrix} = \begin{pmatrix} 2(4)+1(5) & 2(5)+1(5) \ 3(4)+(-1)(5) & 3(5)+(-1)(5) \end{pmatrix} = \begin{pmatrix} 8+5 & 10+5 \ 12-5 & 15-5 \end{pmatrix} = \begin{pmatrix} 13 & 15 \ 7 & 10 \end{pmatrix}$
Statement (S1) is correct.

Statement (S2):
$A + B = \begin{pmatrix} 4 & 5 \ 5 & 5 \end{pmatrix}$
$\det(A+B) = 4(5) - 5(5) = 20 - 25 = -5$.
For a $2 	imes 2$ matrix $M = \begin{pmatrix} a & b \ c & d \end{pmatrix}$, $	ext{adj}(M) = \begin{pmatrix} d & -b \ -c & a \end{pmatrix}$.
So, $	ext{adj}(A+B) = \begin{pmatrix} 5 & -5 \ -5 & 4 \end{pmatrix}$.
$\det(	ext{adj}(A+B)) = 5(4) - (-5)(-5) = 20 - 25 = -5$.
Statement (S2) is correct.

Both (S1) and (S2) are correct.

Answer

only (S1) is correct

Answer

only (S2) is correct

Answer

both (S1) and (S2) are correct

Answer

both (S1) and (S2) are wrong

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