Q.Find the values of $x, y, z$ if the matrix $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ satisfy the equation $A'A = I$.

Concept: Orthogonal Matrix Verification — for a real matrix $A$, $A'A = I$ means $A$ is orthogonal, so its columns (and rows) form an orthonormal set.

Step 1 – Compute $A'A$.

Since $A'A = I$, each diagonal entry must be $1$ and each off-diagonal entry $0$.

Step 2 – Write the conditions from $A'A$.

First column dot with itself:

$0^2 + x^2 + x^2 = 2x^2 = 1 \implies x = \pm \frac{1}{\sqrt{2}}$.

Second column dot with itself:

$(2y)^2 + y^2 + (-y)^2 = 4y^2 + y^2 + y^2 = 6y^2 = 1 \implies y = \pm \frac{1}{\sqrt{6}}$.

Third column dot with itself:

$z^2 + (-z)^2 + z^2 = 3z^2 = 1 \implies z = \pm \frac{1}{\sqrt{3}}$.

Step 3 – Check orthogonality between columns.

First and second columns dot product:

$0(2y) + x(y) + x(-y) = xy - xy = 0$ — automatically satisfied.

First and third: $0(z) + x(-z) + x(z) = -xz + xz = 0$ — automatically satisfied.

Second and third: $(2y)(z) + y(-z) + (-y)(z) = 2yz - yz - yz = 0$ — automatically satisfied.

No further constraints; signs are independent.

The condition $A'A = I$ is the definition of an orthogonal matrix — its columns (and rows) form an orthonormal set. That means each column vector has length 1, and any two distinct columns are perpendicular (dot product zero). This is a powerful geometric check: we don't need to multiply the full matrices blindly; we can work column by column.

Let’s denote the columns of $A$ as:

The condition $A'A = I$ is equivalent to:

We’ll solve systematically.

1. Column 1 with itself (norm squared = 1):

2. Column 2 with itself:

3. Column 3 with itself:

So far we have magnitudes. Now we need the sign relationships from dot products between different columns.

4. Column 1 dot Column 2 = 0:

This is automatically 0 for any $x, y$ — no constraint. Good.

5. Column 1 dot Column 3 = 0:

Also automatically satisfied. So column 1 is orthogonal to both 2 and 3 regardless of signs.

6. Column 2 dot Column 3 = 0 — this is the only nontrivial orthogonality condition:

Again, it simplifies to $0 = 0$ — automatically satisfied! So all off-diagonal dot products vanish for any choice of signs.

Thus the only constraints are the three norm equations. So the signs of $x$, $y$, and $z$ are independent of each other.

Method: Solving for unknowns from $A'A = I$ (orthogonality by columns)

Use this whenever a matrix with unknown entries is stated to satisfy $A'A = I$ (or $AA' = I$) and you must find those entries.

Steps

Step 1: Translate $A'A = I$ into column conditions.

The $(i,j)$ entry of $A'A$ is the dot product of column $i$ and column $j$ of $A$. So $A'A = I$ is exactly:

Each column has length $1$ (norm condition) and distinct columns are perpendicular (orthogonality condition).

Step 2: Write the norm equations to fix magnitudes.

Set each $C_i \cdot C_i = 1$. Every such equation is a sum of squares of the unknowns in that column, so it pins down the magnitude of each unknown as $\pm$ a value.

Step 3: Use the off-diagonal equations to fix the signs / relations.

Set each $C_i \cdot C_j = 0$ ($i \neq j$). These either link the signs of the unknowns or (as often happens with a symmetric layout) vanish identically, meaning the signs are independent. Only keep the constraints these equations genuinely impose — do not assume they force a relation.

Common Mistakes

Mistake 1: Checking only the norm (diagonal) conditions and stopping.

Why it's wrong: $A'A = I$ also demands every off-diagonal dot product be $0$; skipping them can leave sign relations undetermined or unnoticed. Correct approach: always test $C_i \cdot C_j = 0$ for $i \neq j$ as well — here they happen to vanish, so the signs are independent, but you must verify that rather than assume it.

Mistake 2: Reporting only the positive roots.

Why it's wrong: $2x^2 = 1$ gives $x = \pm\tfrac{1}{\sqrt2}$, not just $\tfrac{1}{\sqrt2}$. Correct approach: keep both signs for each of $x, y, z$ since each square root is double-valued.

Mistake 3: Using rows when the condition is $A'A$ (or vice versa).

Why it's wrong: $A'A = I$ is the columns-orthonormal condition; mixing it up with $AA' = I$ (rows) can give a different set of equations for a non-symmetric layout. Correct approach: for $A'A$, dot the columns of $A$.