Find $x$, $y$, $z$ if $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ satisfies $A' = A^{-1}$.

Find x, y, z if A = 0 2y z x y -z x -y z satisfies A' = A⁻¹.

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Sikkim CbseLong· 3mImportance★★★★★

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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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The condition $A' = A^{-1}$ means $A$ is orthogonal, so $AA' = I$ . Equating entries gives $x = \pm\dfrac{1}{\sqrt{2}},\ y = \pm\dfrac{1}{\sqrt{6}},\ z = \pm\dfrac{1}{\sqrt{3}}$ .

The relation $A' = A^{-1}$ is equivalent to $AA' = I$ . The $(i,j)$ entry of $AA'$ is the dot product of row $i$ with row $j$ of $A$ , so the rows must be orthonormal.

The rows of $A$ are

R_1 = (0,\ 2y,\ z),\quad R_2 = (x,\ y,\ -z),\quad R_3 = (x,\ -y,\ z).

Diagonal entries (unit rows):

R_1\cdot R_1 = 4y^2 + z^2 = 1 \qquad (1)

R_2\cdot R_2 = x^2 + y^2 + z^2 = 1 \qquad (2)

R_3\cdot R_3 = x^2 + y^2 + z^2 = 1 \quad(\text{same as }(2))

Off-diagonal entries (orthogonal rows):

R_1\cdot R_2 = 2y^2 - z^2 = 0 \qquad (3)

R_1\cdot R_3 = -2y^2 + z^2 = 0 \quad(\text{same as }(3))

R_2\cdot R_3 = x^2 - y^2 - z^2 = 0 \qquad (4)

Solve the system. From $(4)$ , $x^2 = y^2 + z^2$ . Substituting into $(2)$ :

x^2 + x^2 = 1 \;\Rightarrow\; 2x^2 = 1 \;\Rightarrow\; x^2 = \tfrac12,\quad x = \pm\tfrac{1}{\sqrt2}.

From $(3)$ , $z^2 = 2y^2$ . Substituting into $(1)$ : …

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Q.Find $x$ , $y$ , $z$ if $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ satisfies $A' = A^{-1}$ .

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 condition $A'=A^{-1}$ to find unknown entries

Steps

Common Mistakes

Q.Find xxx, yyy, zzz if A=[02yzxy−zx−yz]A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}A=​0xx​2yy−y​z−zz​​ satisfies A′=A−1A' = A^{-1}A′=A−1.

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Q.Find $x$ , $y$ , $z$ if $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ satisfies $A' = A^{-1}$ .

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