Consider the system of equations x+y+z=2, 2x+3y+2z=5, x+2y+ z= then which of the following statements are TRUE? A. The system of equations has unique solution if =2 and R (set of real number) B. The system of equations is inconsistent if =1 and =1 C. The system of equations is consistent if =1 and =1 D. The system of equations is consistent if =1 and =3 Choose the correct answer from the options given below: Option 1: A and C only Option 2: A and B only Option 3: A, B and D only Option 4: B and D only

Answer: A, B and D only. Solution: The coefficient determinant is -1, so the solution is unique whenever ≠1; in particular for =2 (A true). If =1, subtracting twice the first equation from the second gives y=1 and x+z=1, so the third equation forces =3. Hence it is inconsistent for =1 (B true, C false) and consistent for =3 (D true).

Algebra

Matrices & Determinants

Hard

core

Consider the system of equations $x+y+z=2, 2x+3y+2z=5, x+2y+\lambda z=\mu$ then which of the following statements are TRUE?

A. The system of equations has unique solution if $\lambda=2$ and $\mu\in\mathbb R$ (set of real number)

B. The system of equations is inconsistent if $\lambda=1$ and $\mu=1$

C. The system of equations is consistent if $\lambda=1$ and $\mu=1$

D. The system of equations is consistent if $\lambda=1$ and $\mu=3$

Choose the correct answer from the options given below:

A and C only

A and B only

A, B and D only

B and D only

✅ Correct Option: 3

The coefficient determinant is $\lambda-1$ , so the solution is unique whenever $\lambda\neq1$ ; in particular for $\lambda=2$ (A true).

If $\lambda=1$ , subtracting twice the first equation from the second gives $y=1$ and $x+z=1$ , so the third equation forces $\mu=3$ .

Hence it is inconsistent for $\mu=1$ (B true, C false) and consistent for $\mu=3$ (D true).

2025: 14 May Shift 1

Match List-I with List-II

List-I	List-II
(Matrix)	(Inverse of the Matrix)
(A) $\begin{pmatrix} 1 & 7 \\ 4 & -2 \end{pmatrix}$	(I) $\begin{pmatrix} 2/15 & 1/10 \\ -1/15 & 1/5 \end{pmatrix}$
(B) $\begin{pmatrix} 6 & -3 \\ 2 & 4 \end{pmatrix}$	(II) $\begin{pmatrix} 1/5 & -2/15 \\ -1/10 & 7/30 \end{pmatrix}$
(C) $\begin{pmatrix} 5 & 2 \\ -5 & 4 \end{pmatrix}$	(III) $\begin{pmatrix} 1/15 & 7/30 \\ 2/15 & -1/30 \end{pmatrix}$
(D) $\begin{pmatrix} 7 & 4 \\ 3 & 6 \end{pmatrix}$	(IV) $\begin{pmatrix} 2/15 & -1/15 \\ 1/6 & 1/6 \end{pmatrix}$

Choose the correct answer from the options given below:

2022: 4 Aug Shift 1

If $A = \begin{bmatrix} 6 & 4 \\ 5 & 3 \end{bmatrix}$ and $B = adj(A)$ , then $|B|$ is equal to:

2025: 30 May Shift 1

If $A = \begin{bmatrix} 0 & 0 & \sqrt{7} \\ 0 & \sqrt{7} & 0 \\ \sqrt{7} & 0 & 0 \end{bmatrix}$ , then $|\text{adj } A|$ is equal to

2025: 3 June Shift 2

If $A = \begin{bmatrix} 3 & 2a \\ 1 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 \\ b & 5 \end{bmatrix}$ both are singular matrices, then $a + b$ is equal to

2025: 21 May Shift 1

Let $M_{ij}$ and $A_{ij}$ denote respectively minors and co-factors of the element in the $i^{th}$ row and $j^{th}$ column of the matrix $A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 2 & 3 \\ 4 & -1 & 0 \end{bmatrix}$ . Then:

(A) $M_{32} = 6$

(B) $M_{23} = 9$

(D) $A_{23} = -9$

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:

Solution

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