Let A be a 3 \times 3 matrix such that X^T AX=O for all nonzero 3 \times 1 matrices X=[\beginarraylx y z\endarray] . If A[\beginarrayl1 1 1\endarray]=[\beginarrayc1 4 -5\endarray], A[\beginarrayl1 2 1\endarray]=[\beginarrayc0 4 -8\endarray] , and \operatornamedet(\operatornameadj(2(~A+I)))=2^\alpha 3^\beta 5^\gamma, \alpha, \beta, \gamma \in N , then \alpha^2+\beta^2+\gamma^2 is

Matrices and Determinants PYQ — JEE Main 2025 · 24 Jan · Shift 1 · Q46 with Solution

View written solutionFree

Correct answer: 44

Use the condition $X^TAX=0$ for all $X$

Given $X^TAX=0 \,\text{for all } X\in \mathbb{R}^3,$ this implies that $A$ is skew-symmetric, i.e. $A^T=-A.$

So let

0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}.$$ --- 2. **Use $A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}1\\4\\-5\end{bmatrix}$** Compute: $$A\begin{bmatrix}1\\1\\1\end{bmatrix}= \begin{bmatrix} a+b\\ -a+c\\ -b-c \end{bmatrix} = \begin{bmatrix}1\\4\\-5\end{bmatrix}.$$ Thus, $$a+b=1 \quad ...(1)$$ $$-a+c=4 \quad ...(2)$$ $$-b-c=-5 \quad ...(3)$$ --- 3. **Use $A\begin{bmatrix}1\\2\\1\end{bmatrix}=\begin{bmatrix}0\\4\\-8\end{bmatrix}$** Compute: $$A\begin{bmatrix}1\\2\\1\end{bmatrix}= \begin{bmatrix} 2a+b\\ -a+c\\ -b-2c \end{bmatrix} = \begin{bmatrix}0\\4\\-8\end{bmatrix}.$$ So, $$2a+b=0 \quad ...(4)$$ $$-a+c=4 \quad ...(5)$$ $$-b-2c=-8 \quad ...(6)$$ --- 4. **Solve for $a,b,c$** From (1) and (4): $$(2a+b)-(a+b)=0-1 \Rightarrow a=-1.$$ Then from (1): $$-1+b=1 \Rightarrow b=2.$$ From (2): $$-(-1)+c=4 \Rightarrow 1+c=4 \Rightarrow c=3.$$ Hence $$A=\begin{bmatrix} 0 & -1 & 2\\ 1 & 0 & 3\\ -2 & -3 & 0 \end{bmatrix}.$$ --- 5. **Find $A+I$ and then $2(A+I)$** $$A+I= \begin{bmatrix} 1 & -1 & 2\\ 1 & 1 & 3\\ -2 & -3 & 1 \end{bmatrix}.$$ So $$2(A+I)= \begin{bmatrix} 2 & -2 & 4\\ 2 & 2 & 6\\ -4 & -6 & 2 \end{bmatrix}.$$ --- 6. **Use determinant formula for adjugate** For an $n\times n$ matrix $M$, $$\det(\operatorname{adj}(M))=(\det M)^{n-1}.$$ Here $n=3$, so $$\det(\operatorname{adj}(M))=(\det M)^2,$$ where $$M=2(A+I).$$ Thus we first compute $\det(M)$. Also, $$\det(2(A+I))=2^3\det(A+I)=8\det(A+I).$$ So compute $\det(A+I)$. --- 7. **Compute $\det(A+I)$** $$\det\begin{bmatrix} 1 & -1 & 2\\ 1 & 1 & 3\\ -2 & -3 & 1 \end{bmatrix}$$ Expand along first row: $$=1\begin{vmatrix}1&3\\-3&1\end{vmatrix} -(-1)\begin{vmatrix}1&3\\-2&1\end{vmatrix} +2\begin{vmatrix}1&1\\-2&-3\end{vmatrix}.$$ Now, $$\begin{vmatrix}1&3\\-3&1\end{vmatrix}=1+9=10,$$ $$\begin{vmatrix}1&3\\-2&1\end{vmatrix}=1+6=7,$$ $$\begin{vmatrix}1&1\\-2&-3\end{vmatrix}=-3+2=-1.$$ Therefore, $$\det(A+I)=10+7+2(-1)=15.$$ Hence $$\det(2(A+I))=8\cdot 15=120=2^3\cdot 3\cdot 5.$$ --- 8. **Compute $\det(\operatorname{adj}(2(A+I)))$** $$\det(\operatorname{adj}(2(A+I)))=(120)^2.$$ So, $$(120)^2=(2^3\cdot 3\cdot 5)^2=2^6\cdot 3^2\cdot 5^2.$$ Thus, $$\alpha=6,\quad \beta=2,\quad \gamma=2.$$ --- 9. **Find $\alpha^2+\beta^2+\gamma^2$** $$\alpha^2+\beta^2+\gamma^2=6^2+2^2+2^2=36+4+4=44.$$ --- 10. **Comparison with stored answer** Derived answer = $44$. Stored correct answer = $44$. They agree.

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View written solutionFree

Correct answer: 44

Use the condition $X^TAX=0$ for all $X$

Given $X^TAX=0 \,\text{for all } X\in \mathbb{R}^3,$ this implies that $A$ is skew-symmetric, i.e. $A^T=-A.$

So let

0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{bmatrix}.$$ --- 2. **Use $A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}1\\4\\-5\end{bmatrix}$** Compute: $$A\begin{bmatrix}1\\1\\1\end{bmatrix}= \begin{bmatrix} a+b\\ -a+c\\ -b-c \end{bmatrix} = \begin{bmatrix}1\\4\\-5\end{bmatrix}.$$ Thus, $$a+b=1 \quad ...(1)$$ $$-a+c=4 \quad ...(2)$$ $$-b-c=-5 \quad ...(3)$$ --- 3. **Use $A\begin{bmatrix}1\\2\\1\end{bmatrix}=\begin{bmatrix}0\\4\\-8\end{bmatrix}$** Compute: $$A\begin{bmatrix}1\\2\\1\end{bmatrix}= \begin{bmatrix} 2a+b\\ -a+c\\ -b-2c \end{bmatrix} = \begin{bmatrix}0\\4\\-8\end{bmatrix}.$$ So, $$2a+b=0 \quad ...(4)$$ $$-a+c=4 \quad ...(5)$$ $$-b-2c=-8 \quad ...(6)$$ --- 4. **Solve for $a,b,c$** From (1) and (4): $$(2a+b)-(a+b)=0-1 \Rightarrow a=-1.$$ Then from (1): $$-1+b=1 \Rightarrow b=2.$$ From (2): $$-(-1)+c=4 \Rightarrow 1+c=4 \Rightarrow c=3.$$ Hence $$A=\begin{bmatrix} 0 & -1 & 2\\ 1 & 0 & 3\\ -2 & -3 & 0 \end{bmatrix}.$$ --- 5. **Find $A+I$ and then $2(A+I)$** $$A+I= \begin{bmatrix} 1 & -1 & 2\\ 1 & 1 & 3\\ -2 & -3 & 1 \end{bmatrix}.$$ So $$2(A+I)= \begin{bmatrix} 2 & -2 & 4\\ 2 & 2 & 6\\ -4 & -6 & 2 \end{bmatrix}.$$ --- 6. **Use determinant formula for adjugate** For an $n\times n$ matrix $M$, $$\det(\operatorname{adj}(M))=(\det M)^{n-1}.$$ Here $n=3$, so $$\det(\operatorname{adj}(M))=(\det M)^2,$$ where $$M=2(A+I).$$ Thus we first compute $\det(M)$. Also, $$\det(2(A+I))=2^3\det(A+I)=8\det(A+I).$$ So compute $\det(A+I)$. --- 7. **Compute $\det(A+I)$** $$\det\begin{bmatrix} 1 & -1 & 2\\ 1 & 1 & 3\\ -2 & -3 & 1 \end{bmatrix}$$ Expand along first row: $$=1\begin{vmatrix}1&3\\-3&1\end{vmatrix} -(-1)\begin{vmatrix}1&3\\-2&1\end{vmatrix} +2\begin{vmatrix}1&1\\-2&-3\end{vmatrix}.$$ Now, $$\begin{vmatrix}1&3\\-3&1\end{vmatrix}=1+9=10,$$ $$\begin{vmatrix}1&3\\-2&1\end{vmatrix}=1+6=7,$$ $$\begin{vmatrix}1&1\\-2&-3\end{vmatrix}=-3+2=-1.$$ Therefore, $$\det(A+I)=10+7+2(-1)=15.$$ Hence $$\det(2(A+I))=8\cdot 15=120=2^3\cdot 3\cdot 5.$$ --- 8. **Compute $\det(\operatorname{adj}(2(A+I)))$** $$\det(\operatorname{adj}(2(A+I)))=(120)^2.$$ So, $$(120)^2=(2^3\cdot 3\cdot 5)^2=2^6\cdot 3^2\cdot 5^2.$$ Thus, $$\alpha=6,\quad \beta=2,\quad \gamma=2.$$ --- 9. **Find $\alpha^2+\beta^2+\gamma^2$** $$\alpha^2+\beta^2+\gamma^2=6^2+2^2+2^2=36+4+4=44.$$ --- 10. **Comparison with stored answer** Derived answer = $44$. Stored correct answer = $44$. They agree.

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Matrices and Determinants question

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