Let A be a 3 \times 3 matrix of non-negative real elements such that A[\beginarrayl1 1 1\endarray]=3[\beginarrayl1 1 1\endarray] . Then the maximum value of \operatornamedet(A) is \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2024 · 4 Apr · Shift 1 · Q55 with Solution

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Correct answer: 27

We are given a $3\times 3$ matrix $A=(a_{ij})$ with all entries non-negative, and

$A\begin{bmatrix}1\\1\\1\end{bmatrix}=3\begin{bmatrix}1\\1\\1\end{bmatrix}.$

This means:

a_{11}+a_{12}+a_{13}\\ a_{21}+a_{22}+a_{23}\\ a_{31}+a_{32}+a_{33} \end{bmatrix} = \begin{bmatrix}3\\3\\3\end{bmatrix}.$$ So each row sum of $A$ is $3$. --- 2. **Use Hadamard's inequality** For any $3\times 3$ matrix, $$|\det A| \leq \|R_1\|\,\|R_2\|\,\|R_3\|,$$ where $R_1,R_2,R_3$ are the row vectors. So we want to maximize the product of row norms under the condition that each row has non-negative entries summing to $3$. --- 3. **Maximum possible norm of one row** Let a row be $(x,y,z)$ with $$x,y,z\ge 0, \qquad x+y+z=3.$$ Then $$x^2+y^2+z^2 \le (x+y+z)^2 = 9,$$ because $2xy+2yz+2zx\ge 0$. Hence $$\|(x,y,z)\|=\sqrt{x^2+y^2+z^2}\le 3.$$ Therefore each row norm is at most $3$. So by Hadamard, $$|\det A|\le 3\cdot 3\cdot 3=27.$$ Since all entries are non-negative and row sums are fixed, this shows $$\det A\le 27.$$ --- 4. **Check attainability** We now see whether $27$ can actually be achieved. Take $$A=\begin{bmatrix} 3&0&0\\ 0&3&0\\ 0&0&3 \end{bmatrix}=3I.$$ Then all entries are non-negative, and $$A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}3\\3\\3\end{bmatrix}=3\begin{bmatrix}1\\1\\1\end{bmatrix}.$$ Also, $$\det(A)=\det(3I)=3^3=27.$$ So the upper bound is attained. --- 5. **Conclusion** Hence the maximum value of $\det(A)$ is $$\boxed{27}.$$

View written solutionFree

Correct answer: 27

We are given a $3\times 3$ matrix $A=(a_{ij})$ with all entries non-negative, and

$A\begin{bmatrix}1\\1\\1\end{bmatrix}=3\begin{bmatrix}1\\1\\1\end{bmatrix}.$

This means:

a_{11}+a_{12}+a_{13}\\ a_{21}+a_{22}+a_{23}\\ a_{31}+a_{32}+a_{33} \end{bmatrix} = \begin{bmatrix}3\\3\\3\end{bmatrix}.$$ So each row sum of $A$ is $3$. --- 2. **Use Hadamard's inequality** For any $3\times 3$ matrix, $$|\det A| \leq \|R_1\|\,\|R_2\|\,\|R_3\|,$$ where $R_1,R_2,R_3$ are the row vectors. So we want to maximize the product of row norms under the condition that each row has non-negative entries summing to $3$. --- 3. **Maximum possible norm of one row** Let a row be $(x,y,z)$ with $$x,y,z\ge 0, \qquad x+y+z=3.$$ Then $$x^2+y^2+z^2 \le (x+y+z)^2 = 9,$$ because $2xy+2yz+2zx\ge 0$. Hence $$\|(x,y,z)\|=\sqrt{x^2+y^2+z^2}\le 3.$$ Therefore each row norm is at most $3$. So by Hadamard, $$|\det A|\le 3\cdot 3\cdot 3=27.$$ Since all entries are non-negative and row sums are fixed, this shows $$\det A\le 27.$$ --- 4. **Check attainability** We now see whether $27$ can actually be achieved. Take $$A=\begin{bmatrix} 3&0&0\\ 0&3&0\\ 0&0&3 \end{bmatrix}=3I.$$ Then all entries are non-negative, and $$A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}3\\3\\3\end{bmatrix}=3\begin{bmatrix}1\\1\\1\end{bmatrix}.$$ Also, $$\det(A)=\det(3I)=3^3=27.$$ So the upper bound is attained. --- 5. **Conclusion** Hence the maximum value of $\det(A)$ is $$\boxed{27}.$$