Let A be a 3 \times 3 real matrix such that A(\beginarrayl 1 0 1 \endarray)=2(\beginarrayl 1 0 1 \endarray), A(\beginarrayl -1 0 1 \endarray)=4(\beginarrayl -1 0 1 \endarray), A(\beginarrayl 0 1 0 \endarray)=2(\beginarrayl 0 1 0 \endarray) . Then, the system (A-3 I)(\beginarraylx y z\endarray)=(\beginarrayl1 2 3\endarray) has :

Matrices and Determinants PYQ — JEE Main 2024 · 31 Jan · Shift 2 · Q33 with Solution

View written solutionFree

Correct answer: C

Identify eigenvectors and eigenvalues

We are given:

A\begin{pmatrix}-1\\0\\1\end{pmatrix}=4\begin{pmatrix}-1\\0\\1\end{pmatrix},\quad A\begin{pmatrix}0\\1\\0\end{pmatrix}=2\begin{pmatrix}0\\1\\0\end{pmatrix}.$$ So the vectors $$v_1=\begin{pmatrix}1\\0\\1\end{pmatrix},\quad v_2=\begin{pmatrix}-1\\0\\1\end{pmatrix},\quad v_3=\begin{pmatrix}0\\1\\0\end{pmatrix}$$ are eigenvectors of $A$ with eigenvalues $2,4,2$ respectively. 2. **Check whether these vectors are linearly independent** Let $$c_1v_1+c_2v_2+c_3v_3=0.$$ Then $$c_1\begin{pmatrix}1\\0\\1\end{pmatrix}+c_2\begin{pmatrix}-1\\0\\1\end{pmatrix}+c_3\begin{pmatrix}0\\1\\0\end{pmatrix} =\begin{pmatrix}c_1-c_2\\c_3\\c_1+c_2\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.$$ Thus, $$c_1-c_2=0,\quad c_3=0,\quad c_1+c_2=0.$$ From the first and third equations, $c_1=c_2$ and $c_1=-c_2$, so $c_1=c_2=0$, and also $c_3=0$. Hence $v_1,v_2,v_3$ are linearly independent, so they form a basis of $\mathbb R^3$. 3. **Find eigenvalues of $A-3I$** If $Av=\lambda v$, then $$(A-3I)v=(\lambda-3)v.$$ So for the above basis vectors, the eigenvalues of $A-3I$ are: - for $v_1$: $2-3=-1$ - for $v_2$: $4-3=1$ - for $v_3$: $2-3=-1$ Thus the eigenvalues of $A-3I$ are $-1,1,-1$. 4. **Determine invertibility of $A-3I$** Since none of the eigenvalues of $A-3I$ is $0$, the matrix $A-3I$ is invertible. Equivalently, $$\det(A-3I)=(-1)(1)(-1)=1\neq 0.$$ 5. **Conclude about the system** The system $$(A-3I)\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix}$$ has a coefficient matrix $A-3I$ which is invertible. Therefore, it has **exactly one solution**. 6. **Check options** - A: exactly two solutions — false - B: infinitely many solutions — false - C: unique solution — true - D: no solution — false Hence, the correct option is: $$\boxed{\text{C}}$$

More from Matrices and Determinants

Let A be a $3 \times 3$ matrix and $det (A) = 2$ . If $n = det (2024 - times adj (adj (\dots .. (adj A))))$ , then the remainder when $n$ is divided…2024 · Numerical
Let $S$ denote the set of all real values of $λ$ such that the system of equations $λ x + y + z = 1 x + λ y + z = 1 x + y + λ z = 1$ is inconsistent, then $\sum_{λ \in S} (∣ λ ∣^{2} + ∣ λ ∣)$ is equal to2023 · MCQ
For the system of linear equations $α x + y + z = 1, x + α y + z = 1, x + y + α z = β$ , which one of the following statements is NOT correct?2023 · MCQ
If $A = \frac{1}{2} [1 - 3 31]$ , then :2023 · MCQ
If the system of equations $x + y + a z = b 2 x + 5 y + 2 z = 6 x + 2 y + 3 z = 3$ has infinitely many solutions, then $2 a + 3 b$ is equal to :2023 · MCQ
Let $A = [a_{ij}]_{2 \times 2}$ , where $a_{ij} e q 0$ for all $i, j$ and $A^{2} = I$ . Let a be the sum of all diagonal elements of $A$ and $b = ∣ A ∣$ …2023 · MCQ
Let $P$ be a square matrix such that $P^{2} = I - P$ . For $α, β, γ, δ \in N$ , if $P^{α} + P^{β} = γ I - 29 P$ and $P^{α} - P^{β} = δ I - 13 P$ , then $α + β + γ - δ$ is equal to :2023 · MCQ
For the system of equations $x + y + z = 6 x + 2 y + α z = 10 x + 3 y + 5 z = β$ , which one of the following is NOT true?2023 · MCQ

View written solutionFree

Correct answer: C

Identify eigenvectors and eigenvalues

We are given:

A\begin{pmatrix}-1\\0\\1\end{pmatrix}=4\begin{pmatrix}-1\\0\\1\end{pmatrix},\quad A\begin{pmatrix}0\\1\\0\end{pmatrix}=2\begin{pmatrix}0\\1\\0\end{pmatrix}.$$ So the vectors $$v_1=\begin{pmatrix}1\\0\\1\end{pmatrix},\quad v_2=\begin{pmatrix}-1\\0\\1\end{pmatrix},\quad v_3=\begin{pmatrix}0\\1\\0\end{pmatrix}$$ are eigenvectors of $A$ with eigenvalues $2,4,2$ respectively. 2. **Check whether these vectors are linearly independent** Let $$c_1v_1+c_2v_2+c_3v_3=0.$$ Then $$c_1\begin{pmatrix}1\\0\\1\end{pmatrix}+c_2\begin{pmatrix}-1\\0\\1\end{pmatrix}+c_3\begin{pmatrix}0\\1\\0\end{pmatrix} =\begin{pmatrix}c_1-c_2\\c_3\\c_1+c_2\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.$$ Thus, $$c_1-c_2=0,\quad c_3=0,\quad c_1+c_2=0.$$ From the first and third equations, $c_1=c_2$ and $c_1=-c_2$, so $c_1=c_2=0$, and also $c_3=0$. Hence $v_1,v_2,v_3$ are linearly independent, so they form a basis of $\mathbb R^3$. 3. **Find eigenvalues of $A-3I$** If $Av=\lambda v$, then $$(A-3I)v=(\lambda-3)v.$$ So for the above basis vectors, the eigenvalues of $A-3I$ are: - for $v_1$: $2-3=-1$ - for $v_2$: $4-3=1$ - for $v_3$: $2-3=-1$ Thus the eigenvalues of $A-3I$ are $-1,1,-1$. 4. **Determine invertibility of $A-3I$** Since none of the eigenvalues of $A-3I$ is $0$, the matrix $A-3I$ is invertible. Equivalently, $$\det(A-3I)=(-1)(1)(-1)=1\neq 0.$$ 5. **Conclude about the system** The system $$(A-3I)\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix}$$ has a coefficient matrix $A-3I$ which is invertible. Therefore, it has **exactly one solution**. 6. **Check options** - A: exactly two solutions — false - B: infinitely many solutions — false - C: unique solution — true - D: no solution — false Hence, the correct option is: $$\boxed{\text{C}}$$

Matrices and Determinants question

Matrices and Determinants question

More from Matrices and Determinants

Matrices and Determinants question

Matrices and Determinants question

More from Matrices and Determinants