Q.Show that if $A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$, then $A^{n} = \begin{pmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{pmatrix}$ for every positive integer $n$.

We start with the general second-degree equation in $x$ and $y$:

When we rotate the axes by an angle $\theta$ (here $\theta = \pi/4$), the coefficients change, but certain quantities remain invariant. The key idea: the trace $a+b$ and the determinant $ab - h^2$ of the symmetric coefficient matrix are invariant under rotation. Also, the constant term $c$ transforms in a simple way because it is the value of the quadratic form at the point, and rotation doesn't change the "size" of the ellipse — it only reorients it.

The rotated equation is given as:

This is an ellipse centered at the origin with semi-axes along the new coordinate axes. Our job: recover $a$, $h$, $b$, and $c$ from the invariants and the specific rotation.

1. Identify the invariants. For the quadratic form $ax^2 + 2hxy + by^2$, the matrix is

Under any rotation, the trace $\operatorname{tr}(M) = a+b$ and the determinant $\det(M) = ab - h^2$ are unchanged.

For the rotated form $25x^2 + 9y^2$, the matrix is diagonal:

Hence:

2. Use the rotation angle to relate $a$, $b$, $h$. When rotating by $\theta = \pi/4$, the new coefficients are given by the transformation law for a quadratic form. In particular, the off-diagonal term $2hxy$ vanishes after rotation because the ellipse is aligned with the axes. The condition for the $xy$-term to disappear is:

With $\theta = \pi/4$, we have $2\theta = \pi/2$, so $\tan(\pi/2)$ is undefined — this means $a - b = 0$.

Thus $a = b$.

3. Solve for $a$, $b$, $h$.

   From $a = b$ and $a+b = 34$, we get $a = b = 17$.

   Then $ab - h^2 = 17 \cdot 17 - h^2 = 289 - h^2 = 225$, so $h^2 = 64$, hence $h = \pm 8$.

   The sign of $h$ depends on the orientation; it doesn't affect the final expression because we need $2h$ and it will be squared later.

4. Find $c$.

   The constant term $c$ is not invariant under rotation — but note: the rotated equation is $25x^2 + 9y^2 = 225$. The original equation is $ax^2 + 2hxy + by^2 = c$.

   Under rotation, the value of the quadratic form at any point is preserved (since it's just a change of coordinates). In particular, the maximum and minimum values of the quadratic form on the unit circle are the eigenvalues of $M$, and the constant $c$ scales the ellipse.

   A simpler invariant: the ratio of the constant term to the product of eigenvalues is preserved up to rotation. Actually, the ellipse's size is determined by $c$ relative to the eigenvalues.

   For the rotated form, divide both sides by 225:

The eigenvalues of the original matrix are the same as those of the rotated matrix: $25$ and $9$. The original equation $ax^2 + 2hxy + by^2 = c$ can be written in the principal axes as $25X^2 + 9Y^2 = c$, because the eigenvalues are $25$ and $9$ (they are invariants under rotation).

But the given rotated equation has $c' = 225$ on the right. Wait — careful: the rotated equation is $25x^2 + 9y^2 = 225$, so the constant term after rotation is $225$. Since the constant term is just the value of the quadratic form at the point, and rotation doesn't change the set of points satisfying the equation, the constant $c$ must equal $225$ as well.

Thus $c = 225$, so $\sqrt{c} = 15$ (positive root, since $c>0$).

5. Compute the desired expression. We have $a = 17$, $b = 17$, $h = \pm 8$, $c = 225$. Then:

This gives either $3$ or $35$. Squaring:

Which one is correct? The problem likely expects a unique answer. Notice that $h$ could be positive or negative depending on the rotation direction; but the expression $(a+2h+b-\sqrt{c})^2$ would then have two possible values. However, the rotation is specified as "through an angle $\pi/4$ in the positive direction". The positive direction (counterclockwise) determines the sign of $h$ relative to the transformation.

So the value is $1225$.

When you rotate the coordinate axes, the equation of a conic changes form, but certain combinations of its coefficients stay the same. This is the key idea: rotation is an orthogonal transformation, so the trace (sum of coefficients of $x^2$ and $y^2$) and the determinant (the discriminant $ab - h^2$) of the quadratic part are invariant. The constant term $c$ is also unchanged because rotation doesn't shift the origin.

Here the rotation angle is $\pi/4$ (positive, i.e. anticlockwise). The given original equation is

and after rotation it becomes

Notice the transformed equation has no $xy$ term — that's because the rotation has aligned the axes with the conic's principal axes.

We need $\left(a + 2h + b - \sqrt{c}\right)^2$. Let's find each piece.

1. Invariance of $a+b$ (the trace)

   Under any rotation of axes, the sum of the coefficients of $x^2$ and $y^2$ remains the same. For the original: $a + b$. For the rotated form: $25 + 9 = 34$.

   So $a + b = 34$.

2. Invariance of $ab - h^2$ (the discriminant)

   This quantity is also rotation-invariant. For the original: $ab - h^2$. For the rotated form, since there is no $xy$ term, $h' = 0$, so $a'b' - (h')^2 = 25 \times 9 - 0 = 225$.

   Hence $ab - h^2 = 225$.

3. The constant term $c$

   Rotation does not affect the constant term. The rotated equation has constant $225$ on the right (after moving everything to one side: $25x^2 + 9y^2 - 225 = 0$). So $c = 225$.

4. Finding $a + 2h + b$

   We know $a+b = 34$, but we need $a + 2h + b = (a+b) + 2h = 34 + 2h$.

   To find $h$, use the discriminant: $ab - h^2 = 225$. But we don't know $ab$ individually. However, we can find $(a-b)^2$ from the fact that the rotation angle $\theta = \pi/4$ relates the coefficients.

   There is a standard formula: when rotating by $\theta$, the new coefficient of $xy$ becomes zero if $\tan 2\theta = \frac{2h}{a-b}$. Here $\theta = \pi/4$, so $2\theta = \pi/2$, and $\tan(\pi/2)$ is undefined — this means $a-b = 0$.

   Let's check: $\tan 2\theta = \frac{2h}{a-b}$. For $\theta = \pi/4$, $2\theta = \pi/2$, and $\tan(\pi/2) \to \infty$, so the denominator $a-b$ must be $0$. Hence $a = b$.

   Tip

   When the rotation angle is $45^\circ$, the $xy$ term vanishes only if $a = b$. This is a quick shortcut: for $\theta = \pi/4$, the condition for no $xy$ term in the rotated equation is $a = b$.

   So $a = b$. Then from $a+b = 34$, we get $2a = 34$, so $a = b = 17$.

   Now use $ab - h^2 = 225$: $17 \times 17 - h^2 = 225 \Rightarrow 289 - h^2 = 225 \Rightarrow h^2 = 64 \Rightarrow h = \pm 8$.

   Therefore $a + 2h + b = 17 + 2(\pm 8) + 17 = 34 \pm 16$.

   That gives two possibilities: $50$ or $18$. Which one is correct? The problem likely expects a unique answer. Notice the expression we need is squared: $(a+2h+b - \sqrt{c})^2$. If $c=225$, then $\sqrt{c}=15$.

   For $h=8$: $50 - 15 = 35$, square is $1225$.

   For $h=-8$: $18 - 15 = 3$, square is $9$.

   Both $1225$ and $9$ appear as options. Which one is intended? The rotation is through a positive angle $\pi/4$. The sign of $h$ determines the orientation of the original conic. Without additional data, both are mathematically possible. However, in standard exam problems, the positive sign is often taken (or the larger value). Let's check the options: (A) $3$, (B) $1225$, (C) $225$, (D) $9$. Both $1225$ and $9$ are present. But note: $3$ is not $3^2$ but $3$ itself — so (A) is $3$, not $9$. So $9$ is option (D). Which one is more plausible?

   Watch out

   A common mistake is to forget that $h$ can be positive or negative. The problem statement doesn't specify the sign, so both $h=8$ and $h=-8$ satisfy the given conditions. However, the expression $(a+2h+b - \sqrt{c})^2$ yields different squares. The intended answer is likely the one that matches a single option unambiguously — but here two options match. Let's re-read: the rotation is "through an angle $\pi/4$ in the positive direction". That fixes the rotation matrix, but the sign of $h$ in the original equation is not determined by that alone — it depends on the original orientation. In many such problems, they implicitly take $h>0$ for convenience, giving $1225$.

   Given that $1225$ is option (B) and $9$ is option (D), and the problem asks for a single value, the more common result in such rotation problems (with $45^\circ$ and $a=b$) is $a+2h+b = 50$, leading to $1225$.

   Important

   The invariance of $a+b$ and $ab-h^2$ under rotation is the central tool. Always use these to avoid solving for individual coefficients when only combinations are needed.

5. Final computation

   Taking $h = 8$ (the positive root), we have

so

We have a $2\times 2$ matrix $A$ with a parameter $k$. The key idea is to compute $A^3$ explicitly in terms of $k$, then match the given entry $d=228$ to solve for $k$. Once $k$ is known, the sum $b+c$ follows directly.

Why compute $A^3$ directly? Because $A$ is small, matrix multiplication is straightforward. There is no need for diagonalization or Cayley-Hamilton here — simple repeated multiplication gives us everything.

Let’s go step by step.

1. Write $A$ and compute $A^2$ first.

Multiply:

2. Now compute $A^3 = A^2 \cdot A$.

Multiply row by column:

• $(1,1)$ entry: $1\cdot0 + k\cdot1 = k$
• $(1,2)$ entry: $1\cdot1 + k\cdot k = 1 + k^2$
• $(2,1)$ entry: $k\cdot0 + (1+k^2)\cdot1 = 1 + k^2$
• $(2,2)$ entry: $k\cdot1 + (1+k^2)\cdot k = k + k(1+k^2) = k + k + k^3 = 2k + k^3$

So:

Here $a = k$, $b = 1+k^2$, $c = 1+k^2$, $d = 2k + k^3$.

3. Use the given condition $d = 228$. We have $2k + k^3 = 228$. This is a cubic in $k$:

We need a real root. Try small integer factors of 228: $k=6$ gives $216 + 12 = 228$, so $k=6$ works. Check:

So $k=6$ is a root. The cubic has only one real root (the other two are complex, since the derivative $3k^2+2 > 0$ always), so $k=6$ is the only real solution.

4. Find $b$ and $c$. From above, $b = 1+k^2 = 1+36 = 37$, and $c = 1+k^2 = 37$ as well. So $b+c = 37+37 = 74$.