If $A = \begin{bmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}$, and $A^{-1} = A'$, find value of $\alpha$.

If A = cosα sinα\-sinα cosα , and A⁻¹ = A', find value of α.

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A = \begin{bmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}

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Concept understanding — Orthogonal Matrix Verification

Orthogonal Matrix Verification – From Intuition to Precision

An orthogonal matrix is a square matrix that, multiplied by its own transpose, gives back the identity. Why care? Think of a rotation in 2D: rotating the coordinate axes preserves the lengths of vectors and the angles between them. A matrix that preserves both lengths and angles is orthogonal. "Verification" just means checking whether a given matrix has this property.

The Intuition: What Does "Orthogonal" Mean Here?

"Orthogonal" means "at right angles." For matrices it refers to the columns:

Each column vector has length 1 (a unit vector).
Any two different columns are perpendicular (their dot product is zero).

So the columns form an orthonormal set. The same holds for the rows.

Note

That is why the matrix is called orthogonal: its columns are orthogonal to each other and each is normalized to length 1.

The Precise Definition

A square $n \times n$ matrix $A$ is orthogonal if and only if:

A^T A = I

where $A^T$ is the transpose and $I$ the $n \times n$ identity.

Verification: How to Check

Compute $A^T A$ and check whether it equals the identity.

Example: Check $A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ .

A^T A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} = \begin{pmatrix} \cos^2\theta + \sin^2\theta & 0 \\ 0 & \sin^2\theta + \cos^2\theta \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I

So this rotation matrix is orthogonal.

Why This Works: The Column Interpretation

Let the columns of $A$ be $c_1, \dots, c_n$ . The $(i,j)$ entry of $A^T A$ is the dot product $c_i \cdot c_j$ .

When $i = j$ : the entry is $\|c_i\|^2$ . For it to equal 1, each column must have length 1.
When $i \neq j$ : the entry is $c_i \cdot c_j$ . For it to equal 0, different columns must be orthogonal.

So $A^T A = I$ is exactly the condition that the columns are orthonormal.

Important

For an orthogonal $A$ , also $A A^T = I$ (rows are orthonormal too), and $A^{-1} = A^T$ — the inverse is just the transpose, a huge computational advantage.

Common Mistakes to Avoid …

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$A$ is an orthogonal (rotation) matrix, so $A^{-1}=A'$ holds automatically for every $\alpha$ . The condition imposes no restriction: it is satisfied for all real $\alpha$ .

Step 1 — Write $A$ and its transpose.

A=\begin{bmatrix} \cos\alpha & \sin\alpha \\-\sin\alpha & \cos\alpha \end{bmatrix},\qquad A'=\begin{bmatrix} \cos\alpha & -\sin\alpha \\\sin\alpha & \cos\alpha \end{bmatrix}.

Step 2 — Compute $A^{-1}$ . The determinant is

\det A=\cos^2\alpha+\sin^2\alpha=1,

so, using $\begin{bmatrix} a & b \\c & d \end{bmatrix}^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix} d & -b \\-c & a \end{bmatrix}$ ,

$A^{-1}=\begin{bmatrix} \cos\alpha & -\sin\alpha \\\sin\alpha & \cos\alpha \end{bmatrix}.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2023Set ANNUAL1 markMCQ
$A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$
(a) $I$
(b) $\begin{bmatrix} \cos^2\alpha & \sin^2\alpha \\ -\sin^2\alpha & \cos^2\alpha \end{bmatrix}$
(c) $\begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix}$
(d) $1$
›Reveal solutionSolution

$A$ is an orthogonal (rotation) matrix, so $A'A=I$ .

$A=\begin{bmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}$ , so its transpose is $A'=\begin{bmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{bmatrix}$ .

$A'A=\begin{bmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}$

…

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CBSE 2019Set 65/2/11 mark
$A'A = I$
›Reveal solutionSolution
For an orthogonal matrix $A$ , the defining property $A'A = I$ forces the determinant to satisfy $|A|^2 = 1$ , so $|A| = \pm 1$ .

The key here is not just to compute, but to understand why the determinant must be $\pm 1$ . The condition $A'A = I$ defines an orthogonal matrix — a matrix whose columns (and rows) are orthonormal vectors. Geometrically, such a matrix represents a rotation or a reflection, both of which preserve lengths and angles. A rotation has determinant $+1$ , a reflection has determinant $-1$ .

Let’s see how this emerges algebraically.

Start with the given condition.

We have $A'A = I$ , where $A$ is a square matrix and $I$ is the identity matrix of the same order.

Take determinants on both sides.

The determinant of a product is the product of determinants:

$|A'A| = |I|.$

Use the property $|A'A| = |A'| \cdot |A|$ . Also, the determinant of a transpose equals the determinant of the original matrix: $|A'| = |A|$ . So:

$|A| \cdot |A| = |I|.$

The determinant of the identity matrix is $1$ . Therefore:

$|A|^2 = 1.$

Solve for $|A|$ . …
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Q.If $A = \begin{bmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}$ , and $A^{-1} = A'$ , find value of $\alpha$ .

Concept understanding — Orthogonal Matrix Verification

Orthogonal Matrix Verification – From Intuition to Precision

The Intuition: What Does "Orthogonal" Mean Here?

The Precise Definition

Verification: How to Check

Why This Works: The Column Interpretation

Common Mistakes to Avoid …

Method: Using the orthogonality of a rotation matrix ( $A^{-1}=A'$ )

Steps

Common Mistakes

Q.If A=[cos⁡αsin⁡α−sin⁡αcos⁡α]A = \begin{bmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}A=[cosα−sinα​sinαcosα​], and A−1=A′A^{-1} = A'A−1=A′, find value of α\alphaα.

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Q.If $A = \begin{bmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}$ , and $A^{-1} = A'$ , find value of $\alpha$ .

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