Question:
For the system of linear equations
\(2x-y+3z=5\)
\(3x+2y-z=7\)
\(4x+5y+\alpha z=\beta\),
which of the following is NOT correct?
For the system of linear equations
\(2x-y+3z=5\)
\(3x+2y-z=7\)
\(4x+5y+\alpha z=\beta\),
which of the following is NOT correct?
\(2x-y+3z=5\)
\(3x+2y-z=7\)
\(4x+5y+\alpha z=\beta\),
which of the following is NOT correct?
Show Hint
Compute the determinant of the coefficient matrix: it is \(7\alpha+35\), so the system has a unique solution iff \(\alpha\neq-5\). For \(\alpha=-5\) check consistency using \(\beta\).
Updated On: Nov 2, 2026
- The system has infinitely many solutions for \(\alpha=-6\) and \(\beta=9\)
- The system has a unique solution for \(\alpha\neq-5\) and \(\beta=8\)
- The system is inconsistent for \(\alpha=-5\) and \(\beta=8\)
- The system has infinitely many solutions for \(\alpha=-5\) and \(\beta=9\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Coefficient determinant: \(\Delta=\begin{vmatrix}2&-1&3\\3&2&-1\\4&5&\alpha\end{vmatrix}=2(2\alpha+5)+1(3\alpha+4)+3(15-8)=7\alpha+35\).
So \(\Delta=0\) iff \(\alpha=-5\).
For \(\alpha=-5\): the third row of coefficients \((4,5,-5)=-1(2,-1,3)+2(3,2,-1)\). Hence the system is consistent only if \(\beta=-1(5)+2(7)=9\).
So for \(\alpha=-5\): \(\beta=9\) gives infinitely many solutions, \(\beta\neq9\) (e.g. 8) gives no solution.
Check options: (1) \(\alpha=-6\) gives \(\Delta=-7\neq0\), a unique solution, so "infinitely many solutions" is false. (2) \(\alpha\neq-5\) gives a unique solution, true. (3) \(\alpha=-5,\beta=8\) inconsistent, true. (4) \(\alpha=-5,\beta=9\) infinitely many, true.
The statement that is NOT correct is option (1).
Answer: (1) The system has infinitely many solutions for \(\alpha=-6\) and \(\beta=9\)
So \(\Delta=0\) iff \(\alpha=-5\).
For \(\alpha=-5\): the third row of coefficients \((4,5,-5)=-1(2,-1,3)+2(3,2,-1)\). Hence the system is consistent only if \(\beta=-1(5)+2(7)=9\).
So for \(\alpha=-5\): \(\beta=9\) gives infinitely many solutions, \(\beta\neq9\) (e.g. 8) gives no solution.
Check options: (1) \(\alpha=-6\) gives \(\Delta=-7\neq0\), a unique solution, so "infinitely many solutions" is false. (2) \(\alpha\neq-5\) gives a unique solution, true. (3) \(\alpha=-5,\beta=8\) inconsistent, true. (4) \(\alpha=-5,\beta=9\) infinitely many, true.
The statement that is NOT correct is option (1).
Answer: (1) The system has infinitely many solutions for \(\alpha=-6\) and \(\beta=9\)
Was this answer helpful?
0
0
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
View More Questions
Top JEE Main Determinants Questions
- If \[ f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 + 2 \cos^4 x & 2 \sin^4 x & \sin^2 2x \\ 2 \cos^4 x & 3 + 2 \sin^4 x & \sin^2 2x \end{vmatrix} \] then \( \frac{1}{5} f'(0) \) is equal to
- If \( \sin\left(\frac{y}{x}\right) = \log_e |x| + \frac{\alpha}{2} \) is the solution of the differential equation \[x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x\]and \( y(1) = \frac{\pi}{3} \), then \( \alpha^2 \) is equal to
- If \[ f(x) = \begin{vmatrix} x^3 & 2x^2 + 1 & 1 + 3x \\ 3x^2 + 2 & 2x & x^3 + 6 \\ x^3 - x & 4 & x^2 - 2 \end{vmatrix} \] for all \( x \in \mathbb{R} \), then \( 2f(0) + f'(0) \) is equal to
- Let \( A \) be a \( 3 \times 3 \) matrix and \( \det(A) = 2 \). If \(n = \det(\text{adj}(\text{adj}(\ldots(\text{adj}(A))\ldots)))\) with adjoint applied 2024 times, then the remainder when \( n \) is divided by 9 is equal to \(\_\_\_\_\_.\)
If $ A = \begin{pmatrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2p & 8 + 3p + 2q \\ 6 & 12 + 3p & 20 + 6p + 3q \end{pmatrix} $, then the value of $ \det(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n $, then $ m + n $ is equal to:
View More Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions