Let \( A \) be a \( 3 \times 3 \) real matrix such that**
\(A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \quad A \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = 4 \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, \quad A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}.\)
Then, the system \( (A - 3I) \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \) has
\(A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \quad A \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = 4 \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, \quad A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}.\)
Then, the system \( (A - 3I) \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \) has
- unique solution
- exactly two solutions
- no solution
- infinitely many solutions
The Correct Option is A
Solution and Explanation
Define the matrix \( A \) with elements. Let
\[ A = \begin{pmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{pmatrix}. \]
Use the given conditions to form equations. Using the dot product notation for matrix multiplication, we have the following conditions:
From
\[ A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} : \]
\[ (x_1 \ y_1 \ z_1) \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 2, \quad (x_2 \ y_2 \ z_2) \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 0, \quad (x_3 \ y_3 \ z_3) \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 1. \]
Expanding each dot product:
\[ x_1 + z_1 = 2, \quad x_2 + z_2 = 0, \quad x_3 + z_3 = 1. \quad (1) \]
From
\[ A \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ 1 \end{pmatrix} : \]
\[ (x_1 \ y_1 \ z_1) \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 4, \quad (x_2 \ y_2 \ z_2) \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 0, \quad (x_3 \ y_3 \ z_3) \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 1. \]
Expanding each dot product:
\[ x_1 + y_1 + z_1 = 4, \quad x_2 + y_2 + z_2 = 0, \quad x_3 + y_3 + z_3 = 1. \quad (2) \]
From
\[ A \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} : \]
\[ (x_1 \ y_1 \ z_1) \cdot \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = 2, \quad (x_2 \ y_2 \ z_2) \cdot \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = 1, \quad (x_3 \ y_3 \ z_3) \cdot \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = 0. \]
Expanding each dot product:
\[ y_1 + z_1 = 2, \quad y_2 + z_2 = 1, \quad y_3 + z_3 = 0. \quad (3) \]
Solve for elements of \( A \). Using equations (1), (2), and (3), we can solve for the individual elements of \( A \):
From equation (2): \( x_1 + y_1 + z_1 = 4 \) and \( y_1 + z_1 = 2 \) from equation (3). Substitute \( z_1 = 2 - y_1 \) into equation (1) to find \( x_1, y_1, z_1 \).
Similarly, solve for \( x_2, y_2, z_2 \) and \( x_3, y_3, z_3 \) to complete the matrix \( A \).
Set up the system:
\[ (A - 3I) \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}. \]
Now, calculate \( A - 3I \) and substitute to find the unique solution for the system.
Therefore, the answer is: unique solution.
Top Questions on Matrices
- Let $ A $ be a matrix of order $ 3 \times 3 $ and $ |A| = 5 $. If $ |2 \, \text{adj}(3A \, \text{adj}(2A))| = 2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma}, \quad \alpha, \beta, \gamma \in \mathbb{N} $ then $ \alpha + \beta + \gamma $ is equal to
Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to
Let $ A $ be a $ 3 \times 3 $ matrix such that $ | \text{adj} (\text{adj} A) | = 81.
$ If $ S = \left\{ n \in \mathbb{Z}: \left| \text{adj} (\text{adj} A) \right|^{\frac{(n - 1)^2}{2}} = |A|^{(3n^2 - 5n - 4)} \right\}, $ then the value of $ \sum_{n \in S} |A| (n^2 + n) $ is:Let \( A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix} , \ \alpha > 0 \), such that \( \det(A) = 0 \) and \( \alpha + \beta = 1. \) If \( I \) denotes the \( 2 \times 2 \) identity matrix, then the matrix \( (I + A)^8 \) is:
- Let \( A = \begin{bmatrix} 1 & -2 & -1 \\ 0 & 4 & -1 \\ -3 & 2 & 1 \end{bmatrix}, B = \begin{bmatrix} -5 \\ -2 \end{bmatrix}, C = [9 \ \ 7], \) which of the following is defined?
Questions Asked in JEE Main exam
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to: - Assume a living cell with 0.9%(\(w/w\)) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data up to first decimal place only) The cell will:
- JEE Main - 2025
- osmosis
Given below are two statements:
Statement (I):
are isomeric compounds.
Statement (II):
are functional group isomers.
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Isomerism
- Let \( [x] \) denote the greatest integer function, and let \( m \) and \( n \) respectively be the numbers of the points, where the function \( f(x) = [x] + |x - 2| \), \( -2<x<3 \), is not continuous and not differentiable. Then \( m + n \) is equal to:
- JEE Main - 2025
- Differential Equations
The effect of temperature on the spontaneity of reactions are represented as: Which of the following is correct?
- JEE Main - 2025
- Thermodynamics and Work Done