Q.5. Let be a set of matrices. Then the number of matrices in S, for which the sum of the diagonal

Q: What is the correct answer for JEE Main Mathematics Q5?

The correct answer is D. Sol. The given matrix equation is $A^2 - 4A + 3I = 0$. This means that the eigenvalues of matrix A are the roots of the polynomial $x^2 - 4x + 3 = 0$, which are $(x-1)(x-3)=0$. So the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 3$. For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the characteristic equation is $x^2 - (a+d)x + (ad-bc) = 0$. Comparing this with $x^2

Q: Q.5. Let be a set of matrices. Then the number of matrices in S, for which the sum of the diagonal elements is equal to 4, is:

The correct answer is D. Sol. The given matrix equation is $A^2 - 4A + 3I = 0$. This means that the eigenvalues of matrix A are the roots of the polynomial $x^2 - 4x + 3 = 0$, which are $(x-1)(x-3)=0$. So the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 3$. For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the characteristic equation is $x^2 - (a+d)x + (ad-bc) = 0$. Comparing this with $x^2

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

A. 20 · B. 17 · C. 21 · D. 19

Question

Q.5. Let  be a set of  matrices. Then the number of matrices in S, for which the sum of the diagonal elements is equal to 4, is:

FirstInTest · Accepted Answer

Sol. The given matrix equation is $A^2 - 4A + 3I = 0$. This means that the eigenvalues of matrix A are the roots of the polynomial $x^2 - 4x + 3 = 0$, which are $(x-1)(x-3)=0$. So the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 3$. For a $2 	imes 2$ matrix $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$, the characteristic equation is $x^2 - (a+d)x + (ad-bc) = 0$. Comparing this with $x^2 - 4x + 3 = 0$, we get: 1. Sum of diagonal elements (trace) $a+d = 4$. 2. Determinant $ad-bc = 3$. We are given $a,b,c,d \in \{0,1,2,3,4\}$. We need to find the number of matrices satisfying these two conditions. Condition 1: $a+d=4$. Possible pairs $(a,d)$ are: $(0,4), (1,3), (2,2), (3,1), (4,0)$. Condition 2: $ad-bc=3$. Case I: $(a,d) = (0,4)$. Then $ad=0$. So $0-bc=3 \Rightarrow bc=-3$. Since $b,c \in \{0,1,2,3,4\}$, $bc$ cannot be negative. No solution. Case II: $(a,d) = (1,3)$. Then $ad=3$. So $3-bc=3 \Rightarrow bc=0$. Possible pairs $(b,c)$ are: $(0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (2,0), (3,0), (4,0)$. This gives $1+4+4 = 9$ ways. Case III: $(a,d) = (2,2)$. Then $ad=4$. So $4-bc=3 \Rightarrow bc=1$. Possible pairs $(b,c)$ are: $(1,1)$. This gives $1$ way. Case IV: $(a,d) = (3,1)$. Then $ad=3$. So $3-bc=3 \Rightarrow bc=0$. Same as Case II, gives $9$ ways. Case V: $(a,d) = (4,0)$. Then $ad=0$. So $0-bc=3 \Rightarrow bc=-3$. No solution. Total number of matrices = $0 + 9 + 1 + 9 + 0 = 19$.

Answer

20

Answer

17

Answer

21

Answer

19

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