Let A = X = (x, y, z)T: PX = 0 and x2 + y2 + z2 = 1 where P = [ \beginmatrix 1 & 2 & 1 - 2 & 3 & - 4 1 & 9 & - 1 \endmatrix ] , then the set A :

contains more than two elements.

Matrices and Determinants PYQ — JEE Main 2020 · 2 Sep · Shift 2 · Q36 with Solution

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Correct answer: C

We need to find all vectors $X=\begin{pmatrix}x\\y\\z\end{pmatrix}$ such that $PX=0$ and $x^2+y^2+z^2=1.$

So first, solve the homogeneous system $PX=0$ .

1&2&1\\ -2&3&-4\\ 1&9&-1 \end{pmatrix},$$ we must solve $$\begin{pmatrix} 1&2&1\\ -2&3&-4\\ 1&9&-1 \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}0\\0\\0\end{pmatrix}.$$ This gives the equations: $$x+2y+z=0 \quad ...(1)$$ $$-2x+3y-4z=0 \quad ...(2)$$ $$x+9y-z=0 \quad ...(3)$$ 3. Solve the system. From (1): $$x=-2y-z.$$ Substitute into (3): $$(-2y-z)+9y-z=0$$ $$7y-2z=0$$ $$z=\frac{7}{2}y.$$ Now from (1): $$x=-2y-\frac{7}{2}y=-\frac{11}{2}y.$$ So the solution vector is $$X=y\begin{pmatrix}-\frac{11}{2}\\1\\\frac{7}{2}\end{pmatrix}.$$ Multiplying by $2$, we can write the null-space direction as $$X=t\begin{pmatrix}-11\\2\\7\end{pmatrix}.$$ 4. Check that this indeed satisfies (2): $$-2(-11)+3(2)-4(7)=22+6-28=0,$$ so it is correct. Thus all solutions of $PX=0$ are of the form $$X=t\begin{pmatrix}-11\\2\\7\end{pmatrix}.$$ Hence the null space is a line through the origin. 5. Now impose the condition $$x^2+y^2+z^2=1.$$ For $$X=t\begin{pmatrix}-11\\2\\7\end{pmatrix},$$ we get $$x^2+y^2+z^2=t^2(11^2+2^2+7^2)=t^2(121+4+49)=174t^2.$$ So $$174t^2=1$$ $$t^2=\frac{1}{174}$$ $$t=\pm \frac{1}{\sqrt{174}}.$$ Therefore there are exactly two vectors in $A$: $$X=\pm \frac{1}{\sqrt{174}}\begin{pmatrix}-11\\2\\7\end{pmatrix}.$$ 6. Hence the set $A$ contains exactly two elements. Therefore, the correct option is: $$\boxed{\text{C}}$$ 7. Comparison with stored correct answer: Stored correct answer is C, which matches our result.

View written solutionFree

Correct answer: C

We need to find all vectors $X=\begin{pmatrix}x\\y\\z\end{pmatrix}$ such that $PX=0$ and $x^2+y^2+z^2=1.$

So first, solve the homogeneous system $PX=0$ .

1&2&1\\ -2&3&-4\\ 1&9&-1 \end{pmatrix},$$ we must solve $$\begin{pmatrix} 1&2&1\\ -2&3&-4\\ 1&9&-1 \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}0\\0\\0\end{pmatrix}.$$ This gives the equations: $$x+2y+z=0 \quad ...(1)$$ $$-2x+3y-4z=0 \quad ...(2)$$ $$x+9y-z=0 \quad ...(3)$$ 3. Solve the system. From (1): $$x=-2y-z.$$ Substitute into (3): $$(-2y-z)+9y-z=0$$ $$7y-2z=0$$ $$z=\frac{7}{2}y.$$ Now from (1): $$x=-2y-\frac{7}{2}y=-\frac{11}{2}y.$$ So the solution vector is $$X=y\begin{pmatrix}-\frac{11}{2}\\1\\\frac{7}{2}\end{pmatrix}.$$ Multiplying by $2$, we can write the null-space direction as $$X=t\begin{pmatrix}-11\\2\\7\end{pmatrix}.$$ 4. Check that this indeed satisfies (2): $$-2(-11)+3(2)-4(7)=22+6-28=0,$$ so it is correct. Thus all solutions of $PX=0$ are of the form $$X=t\begin{pmatrix}-11\\2\\7\end{pmatrix}.$$ Hence the null space is a line through the origin. 5. Now impose the condition $$x^2+y^2+z^2=1.$$ For $$X=t\begin{pmatrix}-11\\2\\7\end{pmatrix},$$ we get $$x^2+y^2+z^2=t^2(11^2+2^2+7^2)=t^2(121+4+49)=174t^2.$$ So $$174t^2=1$$ $$t^2=\frac{1}{174}$$ $$t=\pm \frac{1}{\sqrt{174}}.$$ Therefore there are exactly two vectors in $A$: $$X=\pm \frac{1}{\sqrt{174}}\begin{pmatrix}-11\\2\\7\end{pmatrix}.$$ 6. Hence the set $A$ contains exactly two elements. Therefore, the correct option is: $$\boxed{\text{C}}$$ 7. Comparison with stored correct answer: Stored correct answer is C, which matches our result.