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Q3:
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 $A′A = I$.
Solution :
Step 1: Establishing the Transpose Matrix $A'$
Given the matrix $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$, the transpose $A'$ is obtained by interchanging rows and columns [Per the definition of Transpose of a Matrix].
$A' = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix}$
Step 2: Formulating the Equation $A'A = I$
We set up the product $A'A$ equal to the Identity Matrix $I_3$:
$\begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix} \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Step 3: Executing Matrix Multiplication
Performing the multiplication row-by-column:
Row 1, Column 1: $(0)(0) + (x)(x) + (x)(x) = 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$
Row 2, Column 2: $(2y)(2y) + (y)(y) + (-y)(-y) = 4y^2 + y^2 + y^2 = 6y^2 = 1 \implies y^2 = \frac{1}{6} \implies y = \pm \frac{1}{\sqrt{6}}$
Row 3, Column 3: $(z)(z) + (-z)(-z) + (z)(z) = z^2 + z^2 + z^2 = 3z^2 = 1 \implies z^2 = \frac{1}{3} \implies z = \pm \frac{1}{\sqrt{3}}$
Step 4: Verification of Off-Diagonal Elements
To ensure equality, the product of row vectors must satisfy the orthogonality condition (off-diagonal elements must be zero):
| Element Position | Calculation | Result |
|---|---|---|
| (1,2) | $0(2y) + x(y) + x(-y) = 0$ | $0 = 0$ (Verified) |
| (1,3) | $0(z) + x(-z) + x(z) = 0$ | $0 = 0$ (Verified) |
| (2,3) | $2y(z) + y(-z) + (-y)(z) = 2yz - yz - yz = 0$ | $0 = 0$ (Verified) |
SVG Representation of Matrix Orthogonality
Final Analysis
The values are derived from the squares of the variables necessitated by the diagonal identity requirements. Since the off-diagonal constraints are satisfied for any signs of $x, y, z$, the values are:
Final Solution: $x = \pm \frac{1}{\sqrt{2}}, y = \pm \frac{1}{\sqrt{6}}, z = \pm \frac{1}{\sqrt{3}}$
More Questions from Class 12 Mathematics Matrices Miscellaneous Exercise on Chapter 3
- Q1: If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.
- Q2: Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.
- Q4: For what values of $x$ : $\begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = O$?
- Q5: If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 – 5A + 7I = 0$.
- Q6: Find $x$, if $\begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = O$
- Q7(a): A manufacturer produces three products $x, y, z$ which he sells in two markets. Annual sales are indicated below: (a) If unit sale prices of $x, y$ and $z$ are ` 2.50, ` 1.50 and ` 1.00, respectively, find the total revenue in each market with the help of matrix algebra.
- Q7(b): A manufacturer produces three products $x, y, z$ which he sells in two markets. Annual sales are indicated below: (b) If the unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively. Find the gross profit.
- Q8: Find the matrix $X$ so that $X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$
- Q9: If $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ is such that $A^2 = I$, then
- Q10: If the matrix A is both symmetric and skew symmetric, then
- Q11: If A is square matrix such that $A^2 = A$, then $(I + A)^3 – 7 A$ is equal to
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