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Q9:
If $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ is such that $A^2 = I$, then
If $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ is such that $A^2 = I$, then
Solution :
Step 1: Establishing the Matrix Equation
Given the matrix $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$ and the condition $A^2 = I$, where $I$ is the identity matrix of order $2 \times 2$, we define:
$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
[By the definition of Identity Matrix for a $2 \times 2$ space]
Step 2: Computing $A^2$
Perform matrix multiplication $A \times A$:
$A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$
Calculating the elements of the resulting matrix:
- $R_1C_1: (\alpha \cdot \alpha) + (\beta \cdot \gamma) = \alpha^2 + \beta\gamma$
- $R_1C_2: (\alpha \cdot \beta) + (\beta \cdot -\alpha) = \alpha\beta - \beta\alpha = 0$
- $R_2C_1: (\gamma \cdot \alpha) + (-\alpha \cdot \gamma) = \gamma\alpha - \alpha\gamma = 0$
- $R_2C_2: (\gamma \cdot \beta) + (-\alpha \cdot -\alpha) = \gamma\beta + \alpha^2$
[By Matrix Multiplication row-by-column rule]
Step 3: Setting the Equation and Equating Corresponding Elements
We have:
$\begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \alpha^2 + \beta\gamma \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
By comparing the elements of the two matrices:
$\alpha^2 + \beta\gamma = 1$
[By Definition of Equality of Matrices]
Step 4: Logical Simplification
The equation can be rearranged to represent the relationship between the variables:
$1 - \alpha^2 - \beta\gamma = 0$
Or expressed in terms of the identity unit:
$\alpha^2 + \beta\gamma + 1 = 0$ is incorrect based on the sign; rather, the standard form is $1 - \alpha^2 - \beta\gamma = 0$.
Visual Representation of Matrix Relation
Final Solution: $1 - \alpha^2 - \beta\gamma = 0$ (or $\alpha^2 + \beta\gamma = 1$)
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.
- 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 $A′A = I$.
- 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 $X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$
- 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
CBSE Solutions for Class 12 Mathematics Matrices
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