If A is a square matrix satisfying A'A = I, write the value of |A|.

Question

A'A = I

CBSECBSE Class XII Board 2019Subjective· 1mImportance★★★★★

✓ Free question

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

Not checking both conditions: columns can be orthogonal but not unit length. $\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$ has orthogonal columns that aren't unit vectors — not orthogonal.
Confusing with "symmetric": symmetric means $A^T = A$ , completely different from $A^T A = I$ .
Forgetting it must be square: only square matrices are called orthogonal.

Quick Verification Steps

Compute $A^T A$ .
Check every diagonal entry is exactly 1.
Check every off-diagonal entry is exactly 0.
If both hold, $A$ is orthogonal — its columns (and rows) form an orthonormal set.

Verifying whether a matrix is orthogonal by checking AᵀA = I goes beyond the core CBSE Class 12 Matrices syllabus, making it an important topic for JEE Advanced and other competitive exams that build on the NCERT Class 12 Mathematics curriculum on matrix transpose and identity matrices. "How to check if a matrix is orthogonal" is a commonly searched topic among students preparing for these advanced-level questions.

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 $|A|$ . Taking square roots gives:

|A| = \pm 1.

Watch out

A common mistake is to forget that $|A'| = |A|$ and write $|A'A| = |A'||A| = |A|^2$ incorrectly. Also, do not assume $|A| = 1$ — both $+1$ and $-1$ are possible.

Tip

This result holds for any real square matrix satisfying $A'A = I$ . If $A$ is complex and satisfies $A^*A = I$ (a unitary matrix), the same reasoning gives $|\det A| = 1$ , but the determinant itself can be any complex number on the unit circle.

✓Final answer

|A|

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

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

Entry (1,1): $\cos^2\alpha+\sin^2\alpha=1$ . Entry (1,2): $\cos\alpha\sin\alpha-\sin\alpha\cos\alpha=0$ .

Entry (2,1): $\sin\alpha\cos\alpha-\cos\alpha\sin\alpha=0$ . Entry (2,2): $\sin^2\alpha+\cos^2\alpha=1$ .

So $A'A=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I$ .

✓Final answer
$A'A=I$
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 $|A|$ . Taking square roots gives:

$|A| = \pm 1.$
Watch out

A common mistake is to forget that $|A'| = |A|$ and write $|A'A| = |A'||A| = |A|^2$ incorrectly. Also, do not assume $|A| = 1$ — both $+1$ and $-1$ are possible.

Tip

This result holds for any real square matrix satisfying $A'A = I$ . If $A$ is complex and satisfies $A^*A = I$ (a unitary matrix), the same reasoning gives $|\det A| = 1$ , but the determinant itself can be any complex number on the unit circle.

✓Final answer
$|A|$

Q.If A is a square matrix satisfying A′A=IA'A = IA′A=I, write the value of ∣A∣|A|∣A∣.

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Q.If A is a square matrix satisfying $A'A = I$ , write the value of $|A|$ .

Unlock everything free for 14 days