Q.5. Let and , . If , then the value of is:

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

The correct answer is C. Sol. Given $A = \begin{pmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{pmatrix}$. We need to find $A^{-1}$. We can write $A = I + P$, where $P = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{pmatrix}$. Calculate powers of $P$: $P^2 = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{pmatrix} = \begin{pmatrix} 0 &

Q: Q.5. Let and , . If , then the value of is:

The correct answer is C. Sol. Given $A = \begin{pmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 9 & 3 & 1 \end{pmatrix}$. We need to find $A^{-1}$. We can write $A = I + P$, where $P = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{pmatrix}$. Calculate powers of $P$: $P^2 = \begin{pmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{pmatrix} = \begin{pmatrix} 0 &

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

A. 99 · B. 199 · C. 149 · D. 159

Question

Q.5. Let  and , . If , then the value of  is:

FirstInTest · Accepted Answer

Sol. Given $A = \begin{pmatrix} 1 & 0 & 0 \ 3 & 1 & 0 \ 9 & 3 & 1 \end{pmatrix}$. We need to find $A^{-1}$. We can write $A = I + P$, where $P = \begin{pmatrix} 0 & 0 & 0 \ 3 & 0 & 0 \ 9 & 3 & 0 \end{pmatrix}$. Calculate powers of $P$: $P^2 = \begin{pmatrix} 0 & 0 & 0 \ 3 & 0 & 0 \ 9 & 3 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \ 3 & 0 & 0 \ 9 & 3 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 9 & 0 & 0 \end{pmatrix}$. $P^3 = \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 9 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \ 3 & 0 & 0 \ 9 & 3 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix} = 0$. Since $P^3 = 0$, $A^{-1} = (I+P)^{-1} = I - P + P^2 - P^3 + \dots = I - P + P^2$. $A^{-1} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} - \begin{pmatrix} 0 & 0 & 0 \ 3 & 0 & 0 \ 9 & 3 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 9 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \ -3 & 1 & 0 \ 0 & -3 & 1 \end{pmatrix}$. So $B = A^{-1} = \begin{pmatrix} 1 & 0 & 0 \ -3 & 1 & 0 \ 0 & -3 & 1 \end{pmatrix}$. From this, $b_{31} = 0$, $b_{21} = -3$, $b_{32} = -3$. The value of $\frac{b_{31}-b_{21}}{b_{32}} = \frac{0 - (-3)}{-3} = \frac{3}{-3} = -1$. The solution provided in the image has $A^{-1} = I + 99P + C_2P^2$. This is incorrect. The solution seems to be for a different problem or a typo. Let's re-calculate $A^{-1}$ using the adjoint method. $\det(A) = 1(1-0) - 0 + 0 = 1$. Adjoint of $A$: $C_{11}=1, C_{12}=-3, C_{13}=0$. $C_{21}=0, C_{22}=1, C_{23}=-3$. $C_{31}=0, C_{32}=0, C_{33}=1$. Adjoint matrix is $\begin{pmatrix} 1 & 0 & 0 \ -3 & 1 & 0 \ 0 & -3 & 1 \end{pmatrix}$. So $A^{-1} = \frac{1}{\det(A)} 	ext{adj}(A) = \begin{pmatrix} 1 & 0 & 0 \ -3 & 1 & 0 \ 0 & -3 & 1 \end{pmatrix}$. This matches my previous calculation. So $b_{31}=0$, $b_{21}=-3$, $b_{32}=-3$. $\frac{b_{31}-b_{21}}{b_{32}} = \frac{0 - (-3)}{-3} = -1$. The options are positive integers. There must be a mistake in my understanding or the question/solution. Let's re-read the question. The question says $B = A^{-1}$. The solution says $A^{-1} = I + 99P + C_2P^2$. This is not $A^{-1}$ for the given $A$. The solution's calculation of $A^{-1}$ is wrong. Let's assume the question is correct and my $A^{-1}$ is correct. Then the answer is -1, which is not among the options. Let's check the solution's calculation for $b_{31}, b_{21}, b_{32}$. The solution has $b_{31} = 44550$, $b_{21} = 99 	imes 3 = 297$, $b_{32} = 99 	imes 3 = 297$. This implies $A^{-1}$ is a very different matrix. This suggests the solution is for a different problem. Let's assume the question is as written and my calculation of $A^{-1}$ is correct. Then the answer is -1. Since -1 is not an option, there might be a typo in the question or options. However, I must provide an answer from the options. Let's re-examine the solution's values: $b_{31} = 44550$, $b_{21} = 297$, $b_{32} = 297$. Then $\frac{b_{31}-b_{21}}{b_{32}} = \frac{44550-297}{297} = \frac{44253}{297} = 149$. This matches option (3). This means the matrix $A$ or the definition of $B$ in the solution is different from the question. I will use the question as written and my derived $A^{-1}$. Since my answer is not in the options, I will mark confidence lower and state the discrepancy. However, the instruction is to provide the correct answer if found in the document. The document's solution leads to 149. I will follow the document's solution for the answer, assuming there's an unstated modification to the problem or a typo in the printed question. The solution's $A^{-1}$ is not derived from the given $A$. The solution's $A^{-1}$ seems to be $I + 99P + 99P^2$ or something similar, where $P = \begin{pmatrix} 0 & 0 & 0 \ 3 & 0 & 0 \ 9 & 3 & 0 \end{pmatrix}$. If $A^{-1} = I + 99P + 99P^2$, then $A^{-1} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} + 99\begin{pmatrix} 0 & 0 & 0 \ 3 & 0 & 0 \ 9 & 3 & 0 \end{pmatrix} + 99\begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 9 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \ 297 & 1 & 0 \ 891+891 & 297 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \ 297 & 1 & 0 \ 1782 & 297 & 1 \end{pmatrix}$. In this case, $b_{31}=1782$, $b_{21}=297$, $b_{32}=297$. Then $\frac{b_{31}-b_{21}}{b_{32}} = \frac{1782-297}{297} = \frac{1485}{297} = 5$. This is also not 149. The solution's values $b_{31} = 44550$, $b_{21} = 99 	imes 3$, $b_{32} = 99 	imes 3$ are very specific. It seems the solution is for a different problem where $A^{-1}$ is given by a specific formula or is a different matrix. Given the discrepancy, I will extract the answer as per the solution provided in the document, which is 149. This implies the question in the document is different from what is written, or the solution is for a different problem. I will set confidence to 0.7 due to this discrepancy.

Answer

99

Answer

199

Answer

149

Answer

159

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