If the system of equations x + 2y + 3z = 10 -x + y + z = 20 2x + 3y + z = 0 does not possess a unique solution, then is equal to Option 1: 1 Option 2: 15/4 Option 3: 4/15 Option 4: -1

Answer: 15/4. Solution: For a system of equations to not possess a unique solution, the determinant of the coefficient matrix must equal zero. The coefficient matrix from the three equations is: A = pmatrix 1 & 2 & 3 \\ -1 & 1 & \\ 2 & 3 & pmatrix Using first row expansion: (A) = 1 vmatrix 1 & \\ 3 & vmatrix - 2 vmatrix -1 & \\ 2 & vmatrix + 3 vmatrix -1 & 1 \\ 2 & 3 vmatrix For the first term: 1 vmatrix 1 & \\ 3 & vmatrix = 1[(1)() - ()(3)] = 1[ - 3] = -2 For the second term: -2 vmatrix -1 & \\ 2 & vmatrix = -2[(-1)() - ()(2)] = -2[- - 2] = -2[-3] = 6 For the third term: 3 vmatrix -1 & 1 \\ 2 & 3 vmatrix = 3[(-1)(3) - (1)(2)] = 3[-3 - 2] = 3[-5] = -15 Combining all terms: (A) = -2 + 6 - 15 = 4 - 15 Setting the determinant equal to zero: 4 - 15 = 0 4 = 15 = 15/4 Therefore, = 15/4

Algebra

Matrices & Determinants

Medium

common

If the system of equations

$x + 2y + 3z = 10$

$-x + y + \lambda z = 20$

$2x + 3y + \lambda z = 0$

does not possess a unique solution, then $\lambda$ is equal to

$\frac{15}{4}$

$\frac{4}{15}$

-1

✅ Correct Option: 2

For a system of equations to not possess a unique solution, the determinant of the coefficient matrix must equal zero.

The coefficient matrix from the three equations is:

$A = \begin{pmatrix} 1 & 2 & 3 \\ -1 & 1 & \lambda \\ 2 & 3 & \lambda \end{pmatrix}$

Using first row expansion:

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

For the first term:

$1 \begin{vmatrix} 1 & \lambda \\ 3 & \lambda \end{vmatrix} = 1[(1)(\lambda) - (\lambda)(3)]$

$= 1[\lambda - 3\lambda]$

$= -2\lambda$

For the second term:

$-2 \begin{vmatrix} -1 & \lambda \\ 2 & \lambda \end{vmatrix} = -2[(-1)(\lambda) - (\lambda)(2)]$

$= -2[-\lambda - 2\lambda]$

$= -2[-3\lambda]$

$= 6\lambda$

For the third term:

$3 \begin{vmatrix} -1 & 1 \\ 2 & 3 \end{vmatrix} = 3[(-1)(3) - (1)(2)]$

$= 3[-3 - 2]$

$= 3[-5]$

$= -15$

Combining all terms:

$\det(A) = -2\lambda + 6\lambda - 15$

$= 4\lambda - 15$

Setting the determinant equal to zero:

$4\lambda - 15 = 0$

$4\lambda = 15$

$\lambda = \dfrac{15}{4}$

Therefore, $\lambda = \dfrac{15}{4}$

2025: 3 June Shift 1

Let A be a matrix such that $A = \begin{bmatrix} 1 & 2 \\ -2 & 3 \end{bmatrix}$ . Then which of the following are TRUE?

(A) A is non-singular matrix

(B) $A^T = A$

(D) A is not skew-symmetric matrix

Choose the correct answer from the options given below:

2025: 29 May Shift 2

Which of the following statement are correct?

(A) $A = [a_{ij}]_{n \times n}$ is a diagonal matrix if $a_{ij} = 0$ when $i = j$

(B) A square matrix $A = [a_{ij}]$ is called a symmetric matrix if $a_{ij} = a_{ji}$ for all $i, j$

(D) For every square matrix $A$ , there exist an identity matrix of the same order such that $IA = AI = I$

Choose the correct answer from the options given below:

2025: 30 May Shift 1

If $A = \begin{bmatrix} 0 & 1 & 3 \\ 1 & 2 & x \\ 2 & 3 & 1 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} \frac{1}{2} & -4 & \frac{5}{2} \\ -\frac{1}{2} & 3 & -\frac{3}{2} \\ \frac{1}{2} & y & \frac{1}{2} \end{bmatrix}$ , then the value of $8x + 5y$ is:

2026: 19 May Shift 1

Given a matrix $A = \begin{bmatrix} 0 & 3 & -8 \\ -3 & 0 & 13 \\ 8 & -13 & 0 \end{bmatrix}$ . Which of the following statements is/are TRUE?

(A) The matrix A is a skew-symmetric matrix.

(B) The matrix A is a symmetric matrix.

(D) The matrix A is a non-singular matrix.

Choose the correct answer from the options given below:

2026: 31 May Shift 2

Let $A = [a_{ij}]_{3\times3}$ be a matrix such that $a_{ij} = \dfrac{i-j}{i+j}$ , then which of the following statements is/are TRUE?

A. A is a symmetric matrix

B. A is a skew symmetric matrix

C. $|A| = 0$

D. $|A| \neq 0$

Choose the correct answer from the options given below:

2026: 18 May Shift 2

If $\begin{vmatrix} m! & (m+1)! & (m+2)! \\ (m+1)! & (m+2)! & (m+3)! \\ (m+2)! & (m+3)! & (m+4)! \end{vmatrix} = 2k$ , where $m$ is a positive integer then which of the following are factors of $k$ ?

A. $m!$

B. $(m+1)!$

C. $(m+2)!$

D. $(m+3)!$

Choose the correct answer from the options given below:

If the system of equations x+2y+3z=10x + 2y + 3z = 10x+2y+3z=10 −x+y+λz=20-x + y + \lambda z = 20−x+y+λz=20 2x+3y+λz=02x + 3y + \lambda z = 02x+3y+λz=0 does not possess a unique solution, then λ\lambdaλ is equal to

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If the system of equations

$x + 2y + 3z = 10$

$-x + y + \lambda z = 20$

$2x + 3y + \lambda z = 0$

does not possess a unique solution, then $\lambda$ is equal to