Q.If $A = \begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}$, then show that $A^2 = \begin{bmatrix}\cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$.

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:

$6^3 + 2\cdot6 = 216 + 12 = 228$ …

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.

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 $9$. …