Q.3. Let A be a matrix such that , and . If , then is equal to :

Q: What is the correct answer for JEE Main Mathematics Q3?

The correct answer is D. Sol. Let $A = \begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 \\ \beta_1 & \beta_2 & \beta_3 \\ \gamma_1 & \gamma_2 & \gamma_3 \end{pmatrix}$Now $A^T \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} \alpha_1 & \beta_1 & \gamma_1 \\ \alpha_2 & \beta_2 & \gamma_2 \\ \alpha_3 & \beta_3 & \gamma_3 \end{pmatrix} \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2\beta_1 + \gamma_1 \

Q: Q.3. Let A be a matrix such that , and . If , then is equal to :

The correct answer is D. Sol. Let $A = \begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 \\ \beta_1 & \beta_2 & \beta_3 \\ \gamma_1 & \gamma_2 & \gamma_3 \end{pmatrix}$Now $A^T \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} \alpha_1 & \beta_1 & \gamma_1 \\ \alpha_2 & \beta_2 & \gamma_2 \\ \alpha_3 & \beta_3 & \gamma_3 \end{pmatrix} \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2\beta_1 + \gamma_1 \

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

A. 16 · B. 25 · C. 49 · D. 64

Question

Q.3. Let A be a  matrix such that ,  and . If , then  is equal to :

FirstInTest · Accepted Answer

Sol. Let $A = \begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 \ \beta_1 & \beta_2 & \beta_3 \ \gamma_1 & \gamma_2 & \gamma_3 \end{pmatrix}$Now $A^T \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} = \begin{pmatrix} \alpha_1 & \beta_1 & \gamma_1 \ \alpha_2 & \beta_2 & \gamma_2 \ \alpha_3 & \beta_3 & \gamma_3 \end{pmatrix} \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} = \begin{pmatrix} 2\beta_1 + \gamma_1 \ 2\beta_2 + \gamma_2 \ 2\beta_3 + \gamma_3 \end{pmatrix} = \begin{pmatrix} 5 \ 0 \ 2 \end{pmatrix}$ ...(1)Now $A^T \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} \alpha_1 & \beta_1 & \gamma_1 \ \alpha_2 & \beta_2 & \gamma_2 \ \alpha_3 & \beta_3 & \gamma_3 \end{pmatrix} \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} \beta_1 + \gamma_1 \ \beta_2 + \gamma_2 \ \beta_3 + \gamma_3 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix}$ ...(2)Now $A^T \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} \alpha_1 & \beta_1 & \gamma_1 \ \alpha_2 & \beta_2 & \gamma_2 \ \alpha_3 & \beta_3 & \gamma_3 \end{pmatrix} \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} \gamma_1 \ \gamma_2 \ \gamma_3 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}$From (1), $2\beta_1 + 3 = 5 \Rightarrow \beta_1 = 1$From (2), $\beta_1 + 3 = 3 \Rightarrow \beta_1 = 0$This is a contradiction. Let's re-evaluate the problem statement.The given equations are: $A^T \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} = \begin{pmatrix} 5 \ 0 \ 2 \end{pmatrix}$, $A^T \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix}$, $A^T \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}$.Let $X_1 = \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix}$, $Y_1 = \begin{pmatrix} 5 \ 0 \ 2 \end{pmatrix}$. Let $X_2 = \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix}$, $Y_2 = \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix}$. Let $X_3 = \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix}$, $Y_3 = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}$.So $A^T X_1 = Y_1$, $A^T X_2 = Y_2$, $A^T X_3 = Y_3$.Let $X = \begin{pmatrix} 0 & 0 & 0 \ 2 & 1 & 0 \ 1 & 1 & 1 \end{pmatrix}$ and $Y = \begin{pmatrix} 5 & 3 & 3 \ 0 & 1 & 4 \ 2 & 1 & 1 \end{pmatrix}$.Then $A^T X = Y$.So $A^T = Y X^{-1}$.First, find $X^{-1}$.$\det(X) = 0(1-0) - 0(2-0) + 0(2-1) = 0$.This means $X$ is singular, and $X^{-1}$ does not exist.The problem statement must be interpreted differently. The OCR might have misread the matrix structure.Let's assume the matrices are given as columns of $A^T$.The problem states: $A^T \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} = \begin{pmatrix} 5 \ 0 \ 2 \end{pmatrix}$, $A^T \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix}$, $A^T \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}$.This implies that if $A^T = \begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix}$, then:$\begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix} \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} = \begin{pmatrix} 2b+c \ 2e+f \ 2h+i \end{pmatrix} = \begin{pmatrix} 5 \ 0 \ 2 \end{pmatrix}$$\begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix} \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} b+c \ e+f \ h+i \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix}$$\begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix} \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} c \ f \ i \end{pmatrix} = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}$From the third equation: $c=3, f=4, i=1$.Substitute into the second equation: $b+3=3 \Rightarrow b=0$. $e+4=1 \Rightarrow e=-3$. $h+1=1 \Rightarrow h=0$.Substitute into the first equation: $2(0)+3=5$ (consistent). $2(-3)+4=0 \Rightarrow -6+4=0 \Rightarrow -2=0$ (contradiction).There is an issue with the problem statement as interpreted from the OCR. Let's look at the solution provided in the document.The solution starts with:Let $A = \begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 \ \beta_1 & \beta_2 & \beta_3 \ \gamma_1 & \gamma_2 & \gamma_3 \end{pmatrix}$Now $A \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} = \begin{pmatrix} 5 \ 0 \ 2 \end{pmatrix}$, $A \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix}$, $A \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}$ (The solution uses A, not A^T)Let $C_1 = \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix}$, $C_2 = \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix}$, $C_3 = \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix}$.Let $R_1 = \begin{pmatrix} 5 \ 0 \ 2 \end{pmatrix}$, $R_2 = \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix}$, $R_3 = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}$.Then $A C_1 = R_1$, $A C_2 = R_2$, $A C_3 = R_3$.Let $C = \begin{pmatrix} 0 & 0 & 0 \ 2 & 1 & 0 \ 1 & 1 & 1 \end{pmatrix}$ and $R = \begin{pmatrix} 5 & 3 & 3 \ 0 & 1 & 4 \ 2 & 1 & 1 \end{pmatrix}$.Then $A C = R$.So $A = R C^{-1}$.However, $\det(C) = 0$. This is still an issue.The OCR for the question itself shows $A^T$ and then the matrices are written horizontally.Let's re-read the question carefully from the image: "Let A be a 3x3 matrix such that $A^T \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} = \begin{pmatrix} 5 \ 0 \ 2 \end{pmatrix}$, $A^T \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix}$ and $A^T \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}$"The solution provided in the image seems to be using $A$ instead of $A^T$.Let's follow the solution's steps as closely as possible, assuming the solution is correct and the question text has a typo ($A^T$ vs $A$).The solution writes: "Now $A = \begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 \ \beta_1 & \beta_2 & \beta_3 \ \gamma_1 & \gamma_2 & \gamma_3 \end{pmatrix}$"Then it shows: $A \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} = \begin{pmatrix} 2\alpha_2 + \alpha_3 \ 2\beta_2 + \beta_3 \ 2\gamma_2 + \gamma_3 \end{pmatrix} = \begin{pmatrix} 5 \ 0 \ 2 \end{pmatrix}$ (1)And $A \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} \alpha_2 + \alpha_3 \ \beta_2 + \beta_3 \ \gamma_2 + \gamma_3 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix}$ (2)And $A \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} \alpha_3 \ \beta_3 \ \gamma_3 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}$ (3)From (3): $\alpha_3 = 3, \beta_3 = 4, \gamma_3 = 1$.Substitute into (2):$\alpha_2 + 3 = 3 \Rightarrow \alpha_2 = 0$$\beta_2 + 4 = 1 \Rightarrow \beta_2 = -3$$\gamma_2 + 1 = 1 \Rightarrow \gamma_2 = 0$Substitute into (1) to check consistency:$\alpha_2 = 0, \alpha_3 = 3 \Rightarrow 2(0)+3 = 3 
e 5$.This means the solution provided in the image is also inconsistent with itself or the problem statement.The way the matrices are written in the question, it's more likely that the columns of $A^T$ are given.Let $A^T = B$. Then $B \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} = \begin{pmatrix} 5 \ 0 \ 2 \end{pmatrix}$, $B \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix}$, $B \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}$.Let $B = \begin{pmatrix} b_{11} & b_{12} & b_{13} \ b_{21} & b_{22} & b_{23} \ b_{31} & b_{32} & b_{33} \end{pmatrix}$.From $B \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}$, we get $b_{13}=3, b_{23}=4, b_{33}=1$.From $B \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix}$, we get $b_{12}+b_{13}=3 \Rightarrow b_{12}+3=3 \Rightarrow b_{12}=0$.$b_{22}+b_{23}=1 \Rightarrow b_{22}+4=1 \Rightarrow b_{22}=-3$.$b_{32}+b_{33}=1 \Rightarrow b_{32}+1=1 \Rightarrow b_{32}=0$.From $B \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} = \begin{pmatrix} 5 \ 0 \ 2 \end{pmatrix}$, we get $2b_{12}+b_{13}=5 \Rightarrow 2(0)+3=5 \Rightarrow 3=5$ (Contradiction).This means the problem statement as written (or OCR'd) is fundamentally flawed, or my interpretation of the matrix multiplication is wrong.However, the solution in the document proceeds by defining $A$ as: $A = \begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 \ \beta_1 & \beta_2 & \beta_3 \ \gamma_1 & \gamma_2 & \gamma_3 \end{pmatrix}$.Then it shows: $A \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} = \begin{pmatrix} 5 \ 0 \ 2 \end{pmatrix}$, $A \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix}$, $A \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}$.This is what I tried above and found inconsistent.Let's re-examine the image's solution for Q3. It seems to be defining $A$ and then using the given vectors to find the elements of $A$.The solution's first step is: "Let $A = \begin{pmatrix} \alpha_1 & \alpha_2 & \alpha_3 \ \beta_1 & \beta_2 & \beta_3 \ \gamma_1 & \gamma_2 & \gamma_3 \end{pmatrix}$"Then it shows: $A \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} = \begin{pmatrix} 2\alpha_2 + \alpha_3 \ 2\beta_2 + \beta_3 \ 2\gamma_2 + \gamma_3 \end{pmatrix} = \begin{pmatrix} 5 \ 0 \ 2 \end{pmatrix}$ (1)This is consistent with $A$ multiplying a column vector.The next step in the solution is: $A \begin{pmatrix} 0 \ 1 \ 1 \end{pmatrix} = \begin{pmatrix} \alpha_2 + \alpha_3 \ \beta_2 + \beta_3 \ \gamma_2 + \gamma_3 \end{pmatrix} = \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix}$ (2)This is also consistent.The next step in the solution is: $A \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} = \begin{pmatrix} \alpha_3 \ \beta_3 \ \gamma_3 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}$ (3)This is also consistent.Now, let's use these equations to find the elements of A.From (3): $\alpha_3 = 3, \beta_3 = 4, \gamma_3 = 1$.Substitute into (2):$\alpha_2 + 3 = 3 \Rightarrow \alpha_2 = 0$$\beta_2 + 4 = 1 \Rightarrow \beta_2 = -3$$\gamma_2 + 1 = 1 \Rightarrow \gamma_2 = 0$Now, check with (1):$2\alpha_2 + \alpha_3 = 2(0) + 3 = 3$. But (1) says $2\alpha_2 + \alpha_3 = 5$. So $3=5$, which is a contradiction.This means the problem as presented in the image (both question and solution) has an error or I am misinterpreting the notation.Given the instruction

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16

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25

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49

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64

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