The system of equations x - 3y - 8z = -10 2x + 5y + z = 13 3x + y - 4z = 0 has infinite number of solutions if the value of is equal to: Option 1: 9 Option 2: 5 Option 3: any real number Option 4: 6

Answer: 6. Solution: For a system to have infinite solutions, the equations must be dependent and the determinant of the coefficient matrix must equal zero. The coefficient matrix is: A = bmatrix 1 & -3 & -8 \\ 2 & 5 & \\ 3 & 1 & -4 bmatrix Using first row expansion: (A) = 1 vmatrix 5 & \\ 1 & -4 vmatrix - (-3) vmatrix 2 & \\ 3 & -4 vmatrix + (-8) vmatrix 2 & 5 \\ 3 & 1 vmatrix First minor: 5(-4) - (1) = -20 - Second minor: 2(-4) - (3) = -8 - 3 Third minor: 2(1) - 5(3) = -13 Substituting: (A) = 1(-20 - ) + 3(-8 - 3) - 8(-13) = -20 - - 24 - 9 + 104 = 60 - 10 For infinite solutions, the determinant must be zero: 60 - 10 = 0 10 = 60 = 6 Therefore, = 6.

Algebra

Matrices & Determinants

Medium

applied

The system of equations

$x - 3y - 8z = -10$

$2x + 5y + \lambda z = 13$

$3x + y - 4z = 0$

has infinite number of solutions if the value of $\lambda$ is equal to:

any real number

✅ Correct Option: 4

For a system to have infinite solutions, the equations must be dependent and the determinant of the coefficient matrix must equal zero.

The coefficient matrix is:

$A = \begin{bmatrix} 1 & -3 & -8 \\ 2 & 5 & \lambda \\ 3 & 1 & -4 \end{bmatrix}$

Using first row expansion:

$\det(A) = 1 \begin{vmatrix} 5 & \lambda \\ 1 & -4 \end{vmatrix} - (-3) \begin{vmatrix} 2 & \lambda \\ 3 & -4 \end{vmatrix} + (-8) \begin{vmatrix} 2 & 5 \\ 3 & 1 \end{vmatrix}$

First minor: $5(-4) - \lambda(1) = -20 - \lambda$

Second minor: $2(-4) - \lambda(3) = -8 - 3\lambda$

Third minor: $2(1) - 5(3) = -13$

Substituting:

$\det(A) = 1(-20 - \lambda) + 3(-8 - 3\lambda) - 8(-13)$

$= -20 - \lambda - 24 - 9\lambda + 104$

$= 60 - 10\lambda$

For infinite solutions, the determinant must be zero:

$60 - 10\lambda = 0$

$10\lambda = 60$

$\lambda = 6$

Therefore, $\lambda = 6$ .

2026: 14 May Shift 2

If $A = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix}$ such that $AA^T=2I$ (where I is an identity matrix of order 3), then which of the following are True?

A. $x^2=1, 3y^2=2$

B. $6y^2=4, 2z^2=3$

C. $9z^2=6, x^2=1$

D. $x^2:y^2:z^2 = 3:1:2$

Choose the correct answer from the options given below:

2022: 17 Aug Shift 2

If $2\begin{bmatrix}3 & 4 \\ 5 & x\end{bmatrix} + \begin{bmatrix}1 & y \\ 0 & 1\end{bmatrix} = \begin{bmatrix}7 & 0 \\ 10 & 5\end{bmatrix}$ , then the value of $x-y$ is :

2023: 22 May Shift 3

If $A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$ , then the value of K for which $|2A| = K|A|$ is :

2025: 3 June Shift 1

The value of k for which the system of equations

$x + y + z = 1$

$x - ky + z = 1$

$x - y + z = 1$

has more than one solutions is

2025: 19 May Shift 1

If $C_{ij}$ represents the cofactor of element $a_{ij}$ of the matrix $A = \begin{bmatrix} 2 & -1 & 3 \\ 1 & 2 & 0 \\ 4 & 1 & 5 \end{bmatrix}$ then the value of $C_{23} + C_{31} - C_{22}$ is

2022: 25 May Shift 1

Which of the following statements is NOT CORRECT.

The system of equations x−3y−8z=−10x - 3y - 8z = -10x−3y−8z=−10 2x+5y+λz=132x + 5y + \lambda z = 132x+5y+λz=13 3x+y−4z=03x + y - 4z = 03x+y−4z=0 has infinite number of solutions if the value of λ\lambdaλ is equal to:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The system of equations

$x - 3y - 8z = -10$

$2x + 5y + \lambda z = 13$

$3x + y - 4z = 0$

has infinite number of solutions if the value of $\lambda$ is equal to: