Q.If $A = \begin{bmatrix} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{bmatrix}$ and $I$ is the identity matrix of order 2, show that $I + A = (I - A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$.

Concept and Intuition

Rotating coordinate axes by angle $\theta$ transforms a general second-degree term $Ax^2+Bxy+Cy^2$ so that the new $xy$-coefficient is $B\cos2\theta-(A-C)\sin2\theta$. Setting this to zero (to "remove the xy term") gives the classic relation $\tan2\theta=\dfrac{B}{A-C}$, connecting the rotation angle to the original coefficients.

Step-by-Step Solution

1. Equation: $ax^2+bxy+y^2=0$, so $A=a$ (coeff. of $x^2$), $B=b$ (coeff. of $xy$), $C=1$ (coeff. of $y^2$).
2. Condition to eliminate the $xy$ term after rotating by $\theta$: $\tan2\theta = \dfrac{B}{A-C} = \dfrac{b}{a-1}$.
3. Given $\theta=\dfrac{\pi}{8}$, so $2\theta=\dfrac{\pi}{4}$ and $\tan\dfrac{\pi}{4}=1$. …

Concept and Intuition

Rotating the coordinate axes by $\theta$ replaces $(x,y)$ with $x=x'\cos\theta-y'\sin\theta,\ y=x'\sin\theta+y'\cos\theta$. A carefully chosen rotation angle can remove the $xy$ cross-term from a general second-degree curve, revealing its standard form in the rotated frame — exactly the technique used here.

Step-by-Step Solution

1. $\tan\theta=2 \Rightarrow \sin\theta=\dfrac{2}{\sqrt5},\ \cos\theta=\dfrac{1}{\sqrt5}$ (taking the acute rotation).
2. $x=\dfrac{x'-2y'}{\sqrt5},\quad y=\dfrac{2x'+y'}{\sqrt5}$.
3. $x^2=\dfrac{(x'-2y')^2}{5}=\dfrac{x'^2-4x'y'+4y'^2}{5}$.
4. $xy=\dfrac{(x'-2y')(2x'+y')}{5}=\dfrac{2x'^2-3x'y'-2y'^2}{5}$.
5. $3x^2 = \dfrac{3x'^2-12x'y'+12y'^2}{5}$, and $4xy=\dfrac{8x'^2-12x'y'-8y'^2}{5}$. …

Concept and Intuition

Translating axes to the centre of a central conic removes the linear terms. Then rotating axes removes the cross ($xy$) term; the required angle satisfies $\tan 2\theta = \dfrac{2h}{a-b}$ for $ax^2+2hxy+by^2$. When $a=b$, this ratio blows up, forcing $2\theta = \pi/2$, i.e. $\theta=\pi/4$ — the diagonal directions are the natural axes of symmetry when the two quadratic coefficients match.

Step-by-Step Solution

1. Substitute $x=X+2,\ y=Y+3$ into $3x^2+2xy+3y^2-18x-22y+50=0$.
2. Expanding: $3x^2=3X^2+12X+12$, $2xy=2XY+6X+4Y+12$, $3y^2=3Y^2+18Y+27$, $-18x=-18X-36$, $-22y=-22Y-66$.
3. Collect $X$ terms: $12+6-18=0$. Collect $Y$ terms: $4+18-22=0$ (confirming $(2,3)$ is the centre). Constant: $12+12+27-36-66+50=-1$.
4. So the equation becomes $3X^2+2XY+3Y^2-1=0$, i.e. $3X^2+2XY+3Y^2=1$, with $a=3,\ 2h=2\Rightarrow h=1,\ b=3$.
5. Since $a=b$, the rotation angle satisfies $\tan 2\theta = \dfrac{2h}{a-b} \to \infty$, so $2\theta = \dfrac{\pi}{2}$, giving $\theta = \dfrac{\pi}{4}$. …

Concept and Intuition

Rotating axes by $\theta$ to eliminate the $xy$-term from $ax^2+2hxy+by^2(+\ldots)=0$ requires $\tan2\theta = \dfrac{2h}{a-b}$. Here that formula is first used on the auxiliary homogeneous equation to fix $\theta$, then the same rotation (same $\theta$) is applied to transform the second equation into new coordinates.

Step-by-Step Solution

1. For $3x^2+2\sqrt3\,xy+y^2=0$: $a=3,\ 2h=2\sqrt3\Rightarrow h=\sqrt3,\ b=1$.

2. Rotation formulas (old coords in terms of new, rotating axes by $\theta$):

3. Compute $x^2+y^2$: the cross terms cancel (rotation preserves $x^2+y^2=x'^2+y'^2$). …

Concept and Intuition

Rotating coordinate axes by angle $\theta$ replaces $(x,y)$ with $(x'\cos\theta-y'\sin\theta,\ x'\sin\theta+y'\cos\theta)$ in the equation. For a conic, there's always some angle that removes the $x'y'$ cross term (the principal-axes angle); here that angle happens to be $30°$, so the transformed equation comes out especially clean. Two rotation invariants — $A+C$ and $B^2-4AC$ — must match before and after, which is a fast way to check the algebra.

Step-by-Step Solution

1. Use $x=\frac{\sqrt3}{2}x'-\frac12y'$, $y=\frac12x'+\frac{\sqrt3}{2}y'$.
2. Compute $x^2 = \frac34x'^2-\frac{\sqrt3}{2}x'y'+\frac14y'^2$.
3. Compute $xy = \frac{\sqrt3}{4}x'^2+\frac12x'y'-\frac{\sqrt3}{4}y'^2$.
4. Compute $y^2 = \frac14x'^2+\frac{\sqrt3}{2}x'y'+\frac34y'^2$.
5. Assemble $x^2+\frac{\sqrt3}{2}xy+\frac12y^2$: collecting $x'^2$-coefficients gives $\frac34+\frac38+\frac18=\frac54$; the $x'y'$-coefficients cancel to $0$; the $y'^2$-coefficients give $\frac14-\frac38+\frac38=\frac14$. …

Concept and Intuition

Under rotation of axes by $\alpha$ (about the origin), $x^2+y^2$ is invariant, and the linear part $Dx+Ey$ transforms into new coefficients $D'=D\cos\alpha+E\sin\alpha$ and $E'=-D\sin\alpha+E\cos\alpha$, which is just a rotation of the vector $(D,E)$ — its magnitude $\sqrt{D^2+E^2}$ never changes. Only a translation (shift of origin), not a rotation, can remove linear terms.

Step-by-Step Solution

1. Here $D=2, E=2$, so $(D,E)=(2,2)\ne(0,0)$.
2. For no linear terms we'd need $D'=0$ AND $E'=0$ simultaneously, i.e. $2\cos\alpha+2\sin\alpha=0$ and $-2\sin\alpha+2\cos\alpha=0$.
3. The first gives $\tan\alpha=-1$; the second gives $\tan\alpha=1$ — these can never hold at the same $\alpha$. …

Concept and Intuition

When axes are rotated by angle $\theta$, a general second-degree term $Ax^2+Bxy+Cy^2$ transforms into $A'x'^2+B'x'y'+C'y'^2$ where the new coefficients are given by the standard rotation formulas. Two useful facts simplify this kind of problem:

• The trace $A+C$ is invariant under rotation (it equals the sum of the eigenvalues of the associated symmetric matrix, which rotation doesn't change).
• The cross-term coefficient transforms as $B' = (C-A)\sin2\theta + B\cos2\theta$.

So instead of doing a full substitution with $x=x'\cos\theta-y'\sin\theta$, $y=x'\sin\theta+y'\cos\theta$, we can use these invariants directly.

Step-by-Step Solution

1. Original equation: $x^2+2\sqrt3\,xy-y^2=2a^2$, so $A=1$, $B=2\sqrt3$, $C=-1$.
2. Trace invariance: $A'+C' = A+C = 1+(-1) = 0$.
3. New cross-term coefficient at $\theta=30°$ (so $2\theta=60°$, $\sin60°=\tfrac{\sqrt3}2$, $\cos60°=\tfrac12$):

So the cross term vanishes after this rotation — as expected, since $30°$ is precisely the angle that diagonalizes this particular conic.

4. New $A'$ coefficient (using $\cos30°=\tfrac{\sqrt3}2,\sin30°=\tfrac12$, so $\cos^2\theta=\tfrac34,\sin^2\theta=\tfrac14,\sin\theta\cos\theta=\tfrac{\sqrt3}{4}$): …

Concept and Intuition

When coordinate axes are rotated through angle $\theta$ (keeping the origin fixed), a point's coordinates transform as $X = x\cos\theta + y\sin\theta$ and $Y = -x\sin\theta + y\cos\theta$, where $(x,y)$ are old coordinates and $(X,Y)$ are new. This is the standard formula for expressing the same physical point in a rotated frame.

Step-by-Step Solution

1. Here $\theta=45^\circ$, so $\cos\theta=\sin\theta=\dfrac{1}{\sqrt2}$.
2. Old coordinates: $x=2\sqrt2$, $y=-3\sqrt2$.
3. $X = x\cos\theta+y\sin\theta = \dfrac{2\sqrt2}{\sqrt2}+\dfrac{-3\sqrt2}{\sqrt2} = 2-3=-1$. …

Concept and Intuition

Rotating coordinate axes by angle $\theta$ replaces $(x,y)$ with $x=x'\cos\theta-y'\sin\theta$, $y=x'\sin\theta+y'\cos\theta$. This is used to remove the cross ($xy$) term from a conic, turning a tilted ellipse/hyperbola equation into its standard axis-aligned form.

Step-by-Step Solution

1. For $\theta=45°$: $x=\dfrac{x'-y'}{\sqrt2}$, $y=\dfrac{x'+y'}{\sqrt2}$.
2. $x^2 = \dfrac{(x'-y')^2}{2} = \dfrac{x'^2-2x'y'+y'^2}{2}$; $y^2=\dfrac{(x'+y')^2}{2}=\dfrac{x'^2+2x'y'+y'^2}{2}$.
3. $x^2+y^2 = \dfrac{2x'^2+2y'^2}{2} = x'^2+y'^2$.
4. $xy = \dfrac{(x'-y')(x'+y')}{2} = \dfrac{x'^2-y'^2}{2}$.
5. Substitute into $3(x^2+y^2)+2xy=2$: $3(x'^2+y'^2) + 2\cdot\dfrac{x'^2-y'^2}{2} = 3x'^2+3y'^2+x'^2-y'^2 = 4x'^2+2y'^2$. …

Concept and Intuition

When axes are rotated by angle $\theta$, a point's coordinates transform via $x = x'\cos\theta - y'\sin\theta$, $y = x'\sin\theta + y'\cos\theta$, where $(x,y)$ are original-system coordinates and $(x',y')$ are new-system coordinates. This is just expressing the same point in a rotated basis.

Step-by-Step Solution

1. $\theta=135^\circ$: $\cos135^\circ=-\dfrac{1}{\sqrt2}$, $\sin135^\circ=\dfrac{1}{\sqrt2}$.
2. Given new coordinates $(x',y')=(4,-3)$.
3. $x = x'\cos\theta - y'\sin\theta = 4\left(-\dfrac{1}{\sqrt2}\right) - (-3)\left(\dfrac{1}{\sqrt2}\right) = -\dfrac{4}{\sqrt2}+\dfrac{3}{\sqrt2} = -\dfrac{1}{\sqrt2}$. …

Concept and Intuition

When axes are rotated by an angle $\theta$, a point's old coordinates $(x,y)$ relate to new coordinates $(x',y')$ by $x=x'\cos\theta-y'\sin\theta$, $y=x'\sin\theta+y'\cos\theta$. Substituting into the original curve's equation and simplifying yields the curve's equation in the new (rotated) axes — the shape is unchanged, only its algebraic description changes.

Step-by-Step Solution

1. Here $\theta=\pi/3$, so $\cos\theta=\dfrac12,\ \sin\theta=\dfrac{\sqrt3}{2}$.
2. Substitute $x=\dfrac12x'-\dfrac{\sqrt3}{2}y'$ and $y=\dfrac{\sqrt3}{2}x'+\dfrac12y'$ into $x^2+2\sqrt3xy-y^2=8$.
3. Compute each piece: $x^2=\dfrac14x'^2-\dfrac{\sqrt3}{2}x'y'+\dfrac34y'^2$ $y^2=\dfrac34x'^2+\dfrac{\sqrt3}{2}x'y'+\dfrac14y'^2$ $xy=\dfrac{\sqrt3}{4}x'^2-\dfrac12x'y'-\dfrac{\sqrt3}{4}y'^2$ …