Matrices and Determinants: Previous Year Questions with Solutions

1.0Introduction

Matrices and Determinants Previous Year Questions typically cover topics such as matrix operations (addition, multiplication, transpose), types of matrices (zero, identity, diagonal, symmetric), and properties of determinants. Common problem types include finding the inverse of a matrix, solving systems of linear equations using matrices, evaluating determinants, and applying Cramer's Rule. Examples include calculating the determinant of a 3×3 matrix, checking consistency of a system of equations, or finding the adjoint and inverse of a matrix. Solutions involve using standard matrix algebra rules, cofactor expansion, and determinant properties. Practicing these questions helps in developing strong algebraic manipulation skills and understanding the structure of linear equations.

2.0Matrices and Determinants Previous Year Questions for JEE with Solutions

JEE questions in Matrices and Determinants often test concepts related to matrix operations, types of matrices, properties of determinants, and their applications in solving systems of linear equations. Some common types of problems include:

• Matrix Operations: Questions on addition, subtraction, multiplication, transpose, and properties like $(AB{)}^T=B^T{A}^T$, and checking the existence of inverses.
• Determinants: Problems involving calculation of determinants (2×2 and 3×3), using properties to simplify determinants, and evaluating them through cofactor expansion.
• Inverse and Adjoint: Finding the adjoint of a matrix and using it to determine the inverse, especially for solving linear equations.
• System of Linear Equations: Solving systems using matrices (matrix method), and checking consistency using the determinant (non-zero vs. zero).

These questions are designed to test both computational accuracy and understanding of underlying linear algebra principles.

Note: In the JEE Main Mathematics exam, you can typically expect 2 to 3 questions from the Matrices and Determinants chapter.

3.0Key Concepts to Remember – Matrices

1. Types of Matrices

• Square Matrix: Rows = Columns
• Diagonal Matrix: Non-zero elements only on the main diagonal
• Scalar Matrix: Diagonal matrix with all diagonal elements equal
• Identity Matrix (I): Diagonal elements = 1, others = 0
• Zero Matrix: All elements are zero
• Symmetric Matrix: A = A^T
• Skew-Symmetric Matrix: A^T = –A

2. Matrix Operations

• Addition/Subtraction: Only if matrices are of the same order
• Multiplication: Use row-by-column rule; not commutative
• Scalar Multiplication: Multiply every element by a scalar
• Transpose: $(A^T{)}^T=A,(AB{)}^T=B^T{A}^T$

3. Special Properties

• (AB)C = A(BC) (Associative)
• A(B + C) = AB + AC (Distributive)
• A^T A is always symmetric

4. Inverse of a Matrix

• Only defined for square, non-singular matrices

$A^{-1}=\frac{1}{\det (A)}\text{adj}(A)\text{(for }\det (A)\ne 0\text{)}$

4.0Key Concepts to Remember – Determinants

1. Basic Determinant Evaluation

• 2 × 2:

$\left|\begin{array}{c}ab\\ cd\end{array}\right|=ad-bc$

• 3×3: Use cofactor expansion or Sarrus Rule (if allowed)

2. Properties of Determinants

• Swapping two rows/columns: sign changes
• If two rows/columns are identical: determinant = 0
• If a row/column is multiplied by k, determinant is multiplied by k

$\det (A^T)=\det (A)$

3. Applications

• Area of triangle using determinant:

$\text{ Area }=\frac{1}{2}\left|\begin{array}{lll}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

• Solving system of equations:

Cramer’s Rule:

$x=\frac{\det A_1}{\det A},y=\frac{\det A_2}{\det A},z=\frac{\det A_3}{\det A}$

4. Consistency of System

• $det(A)\ne 0$: Unique solution (consistent)
• det(A) = 0: May be no solution or infinitely many (check further)

5.0JEE Mains Past Year Questions with Solutions on Matrices and Determinants

Previous Year Questions from Matrices 

1. Let $A=\left[\begin{array}{ll}1& 2\\ 0& 1\end{array}\right]$ and B = I + adj(A) + (adj A)²+…+ (adj A)¹⁰.  Then, the sum of all the elements of the matrix B is : 

(1) –110

(2) 22

(3) –88

(4) –124

Ans. (3)

Sol.  

$\text{ Sol. }\mathrm{Adj}(A)=\left[\begin{array}{cc}1& -2\\ 0& 1\end{array}\right]$

$(\mathrm{Adj}A{)}^2=\left[\begin{array}{cc}1& -4\\ 0& 1\end{array}\right]$

$(\mathrm{Adj}A{)}^{10}=\left[\begin{array}{cc}1& -20\\ 0& 1\end{array}\right]$

$B=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]+$$B=\left[\begin{array}{ll}1& -2\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& -4\\ 0& 1\end{array}\right]+\dots +\left[\begin{array}{cc}1& -20\\ 0& 1\end{array}\right]$

$$\begin{gathered} B=\left[\begin{array}{cc}11& -110\\ 0& 11\end{array}\right]\square \text{ sum of elements of }B \\ =-88 \end{gathered}$$

2 Let A be a 2 × 2 symmetric matrix such that  $A\left[\begin{array}{l}1\\ 1\end{array}\right]=\left[\begin{array}{l}3\\ 7\end{array}\right]$and the determinant of A be 1. If A^–1 = αA + βI, {} where I is an identity matrix of order 2 × 2, then α + β equals …..

Ans. (5)

Sol.

Let $A=\left[\begin{array}{cc}a& b\\ b& d\end{array}\right]$

$\left[\begin{array}{cc}a& b\\ b& d\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}3\\ 7\end{array}\right],ad-b^2=1$

$a+b=3,b+d=7,(3-b)(7-b)-b^2=1$

$21-10b=1\to b=2,a=1,d=5$

$A=\left[\begin{array}{cc}1& 2\\ 2& 5\end{array}\right],A^{-1}=\left[\begin{array}{cc}5& -2\\ -2& 1\end{array}\right]$

$A^{-1}=\alpha A+\beta I$

$\left[\begin{array}{cc}5& -2\\ -2& 1\end{array}\right]$

$\left[\begin{array}{cc}\alpha +\beta & 2\alpha \\ 2\alpha & 5\alpha +\beta \end{array}\right]$

$\alpha =-1,\beta =6\to \boxed{\alpha +\beta =5}$

3. Let α ∈ (0, ∞) and $A=\left[\begin{array}{ccc}1& 2& \alpha \\ 0& 1& 0\\ 1& 0& 2\end{array}\right]$If $\det (\text{adj}(2A-A^T)\cdot \text{adj}(A-2A^T))=2^8$, then $(\det A{)}^2$ is equal to :

(1) 1

(2) 49

(3) 16

(4) 36

Ans. (3)

Sol.

$$\begin{gathered} \text{adj}(A-2A^T)\cdot (2A-A^T)=28 \\ |(A-2A^T)(2A-A^T)|=24 \\ |A-2A^T|\cdot |2A-A^T|=\pm 16 \\ (A-2A^T{)}^T=A^T-2A \\ |A-2A^T|=|A^T-2A|\Rightarrow |A-2A^T{|}^2=16\Rightarrow |A-2A^T|=\pm 4 \\ \left[\begin{array}{ccc}1& 2& \alpha \\ 0& 1& 0\\ 1& 0& 2\end{array}\right]\Rightarrow A^T=\left[\begin{array}{ccc}1& 0& 1\\ 2& 1& 0\\ \alpha & 0& 2\end{array}\right] \\ A-2A^T=\left[\begin{array}{ccc}1& 2& \alpha \\ 0& 1& 0\\ 1& 0& 2\end{array}\right]-2\left[\begin{array}{ccc}1& 0& 1\\ 2& 1& 0\\ \alpha & 0& 2\end{array}\right]=\left[\begin{array}{ccc}-1& 2& \alpha -2\\ -3& -1& 0\\ 1-2\alpha & 0& -2\end{array}\right] \\ 1+3\alpha =4\Rightarrow 3\alpha =3\Rightarrow \alpha =1 \\ \alpha =1 \\ |A|=\left|\begin{array}{ccc}1& 2& 1\\ 0& 1& -1\\ -1& 3& -4\end{array}\right|=-1-3=-4 \\ |A{|}^2=16 \end{gathered}$$

4. Let αβ ≠ 0 and . If $A=\left[\begin{array}{ccc}\beta & \alpha & 3\alpha \\ \alpha & \alpha & \beta \\ -\beta & \alpha & 2\alpha \end{array}\right],B=\left[\begin{array}{ccc}3\alpha & -9& 3\alpha \\ -\alpha & 7& -2\alpha \\ -2\alpha & 5& -2\beta \end{array}\right],\alpha \beta \ne 0$ is the matrix of cofactors of the elements of A, then det(AB) is equal to :

(1) 343

(2) 125

(3) 64

(4) 216

Ans. (4)

Sol. Equating co-factor fo A₂₁ 

$$\begin{gathered} A_{21}=(2{\alpha }^2-3\alpha )=\alpha \\ \Rightarrow \alpha =0,2\text{(accept } \\ 2{\alpha }^2-\alpha \beta =3\alpha \\ \Rightarrow \alpha =2,\beta =1 \\ |AB|=|A\cdot \text{cof}(A)|=|A{|}^3 \\ A=\left[\begin{array}{ccc}1& 2& 3\\ 2& 2& 1\\ -1& 2& 4\end{array}\right] \\ \Rightarrow |A|=6-2(9)+3(6)=6 \end{gathered}$$

5. If A is a square matrix of order 3 such that det(A) = 3 and det(adj(–4 adj(–3 adj(3 adj((2A)^–1))))) = 2^m3ⁿ, then m + 2n is equal to:

(1) 3

(2) 2

(3) 4

(4) 6

Ans. (3)

Sol.

$$\begin{gathered} |A|=3 \\ |\text{adj}(-4\text{adj}(-3\text{adj}(3\text{adj}((2A{)}^T))))| \\ =|-4\text{adj}(-3\text{adj}(3\text{adj}(2A{)}^T)){|}^2 \\ =4^6|\text{adj}(-3\text{adj}(3\text{adj}(2A{)}^T)){|}^2 \\ =2^{12}\cdot 3^{12}\cdot |\text{adj}(2A{)}^T{|}^8 \\ {2}^{12}\cdot 3^{12}\cdot 3^{24}{\left|\text{adj}(2A{)}^T\right|}^8 \\ =2^{12}\cdot 3^{36}\cdot |(2A{)}^{-1}{|}^{16} \\ =2^{12}\cdot 3^{36}\cdot \frac{1}{|2A{|}^{16}} \\ =2^{12}\cdot 3^{36}\cdot \frac{1}{2^{48}|\det A{|}^{16}} \\ =2^{12}\cdot 3^{36}\cdot \frac{1}{2^{48}\cdot 3^{16}} \\ =\frac{3^{20}}{2^{36}}=2^{-36}\cdot 3^{20} \end{gathered}$$

6. Let $A=\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]$. If the sum of the diagonal elements of A¹³ is 3ⁿ, then n is equal to ______. 

Ans. (7)

Sol.

$$\begin{gathered} A=\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right] \\ {A}^2=\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}3& -3\\ 3& 0\end{array}\right] \\ {A}^3=\left[\begin{array}{cc}3& -3\\ 3& 0\end{array}\right]\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}3& -6\\ 6& -3\end{array}\right] \\ {A}^4=\left[\begin{array}{cc}3& -6\\ 6& -3\end{array}\right]\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}0& -9\\ 9& -9\end{array}\right] \\ {A}^5=\left[\begin{array}{cc}0& -9\\ 9& -9\end{array}\right]\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}-9& -9\\ 9& -18\end{array}\right] \\ {A}^6=\left[\begin{array}{cc}-9& -9\\ 9& -18\end{array}\right]\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}-27& 0\\ 0& -27\end{array}\right] \\ {A}^7=\left[\begin{array}{ccc}-27& -0& -27\\ 0& -27& -27\\ -54& 27& -27\end{array}\right]\left[\begin{array}{c}{3}^6\times 2\\ 27^2\\ -27^2\end{array}\right] \\ {3}^7=3^n\Rightarrow n=7 \end{gathered}$$

7. Let $B=\left[\begin{array}{cc}1& 3\\ 1& 5\end{array}\right]$ and A be a 2 × 2 matrix such that AB^–1 = A^–1. If BCB^–1 = A and C⁴ + αC² + βI = O, then 2β – α is equal to : 

(1) 16

(2) 2

(3) 8

(4) 10

Ans. (4)

Sol. BCB^–1 = A 

⇒ (BCB^–1) (BCB^–1) = A.A

⇒ BCI CB^–1 = A²

⇒ BC²B^–1 = A²

⇒ B^–1(BC²B^–1)B = B^–1(A.A)B

From equation (1)

C² = A^–1.A.B 

C² = B

Also AB^–1 = A^–1

⇒ AB^–1.A = A^–1 A = I

⇒ A^–1(AB^–1A) =A^–1I

B^–1A = A^–1 

Now characteristics equation of C² is

|C₂–λI| = 0

|B – λI| = 0

$\Rightarrow \left|\begin{array}{cc}1-\lambda & 3\\ 1& 5-\lambda \end{array}\right|=0$

⇒ (1 – λ) (5 – l) –3 = 0 ⇒ (λ² – 6λ +5) – 3 = 0

⇒ λ² – 6λ + 2 = 0

⇒ β² – 6B + 2I = 0

⇒ C⁴ – 6C² + 2I = 0

α = – 6

β = 2

∴ 2β – α = 4 + 6 = 10

8. If $A=\left[\begin{array}{cc}\sqrt{2}& 1\\ -1& \sqrt{2}\end{array}\right],B=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right],C=AB{A}^T,X=A^T{C}^{2}A$. then, det X is equal to :

(1) 243

(2) 729

(3) 27

(4) 891

Ans. (2)

Sol.

$$\begin{gathered} A=\left[\begin{array}{cc}\sqrt{2}& 1\\ -1& \sqrt{2}\end{array}\right]\Rightarrow \det (A)=3 \\ B=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\Rightarrow \det (B)=1 \\ \text{Now }C=AB{A}^T\Rightarrow \det (C)=(\det (A){)}^2\times \det (B) \\ |C|=9 \\ \text{Now }|X|=|A^T{C}^{2}A| \\ =|A^T|\cdot |C^2|\cdot |A| \end{gathered}$$

= |A|² |C|²

= 9 x 81

= 729

9. Consider the matrix $f(x)=\left[\begin{array}{ccc}\cos x& -\sin x& 0\\ \sin x& \cos x& 0\\ 0& 0& 1\end{array}\right]$.Given below are two statements :

Statement I: f(–x) is the inverse of the matrix f(x).

Statement II: f(x) f(y) = f(x + y).

In the light of the above statements, choose the correct answer from the options given below 

(1) Statement I is false but Statement II is true

(2) Both Statement I and Statement II are false

(3) Statement I is true but Statement II is false

(4) Both Statement I and Statement II are true

Ans. (4)

Sol.

$$\begin{gathered} f(-x)=\left[\begin{array}{ccc}\cos x& \sin x& 0\\ -\sin x& \cos x& 0\\ 0& 0& 1\end{array}\right] \\ f(x)\cdot f(-x)=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I \\ f(y)=\left[\begin{array}{ccc}\cos y& -\sin y& 0\\ \sin y& \cos y& 0\\ 0& 0& 1\end{array}\right] \\ f(x)\cdot f(y)=\left[\begin{array}{ccc}\cos (x+y)& -\sin (x+y)& 0\\ \sin (x+y)& \cos (x+y)& 0\\ 0& 0& 1\end{array}\right] \end{gathered}$$

10. $A=\left[\begin{array}{ccc}2& 0& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right]$Let , B = [B₁, B₂, B₃], where B₁, B₂, B₃ are column matrices, and $A{B}_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],A{B}_2=\left[\begin{array}{c}3\\ 0\\ 0\end{array}\right],A{B}_3=\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]$If α = |B| and β is the sum of all the diagonal elements of B, then α³ + β³  is equal to _____.

Ans. (28)

Sol.

$$\begin{gathered} A=\left[\begin{array}{ccc}2& 0& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right],B=[B_1,B_2,B_3] \\ {B}_1=\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\end{array}\right],B_2=\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\end{array}\right],B_3=\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\end{array}\right] \\ A{B}_1=\left[\begin{array}{ccc}2& 0& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \\ {x}_1=1,y_1=-1,z_1=-1 \\ A{B}_2=\left[\begin{array}{ccc}2& 0& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\end{array}\right]=\left[\begin{array}{c}2\\ 3\\ 0\end{array}\right] \\ {x}_2=2,y_2=1,z_2=-2 \\ A{B}_3=\left[\begin{array}{ccc}2& 0& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\end{array}\right]=\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right] \\ {x}_3=2,y_3=0,z_3=-1 \\ B=\left[\begin{array}{ccc}1& 2& 2\\ -1& 1& 0\\ -1& -2& -1\end{array}\right] \\ \alpha =|B|=3,\beta =1 \\ {\alpha }^3+{\beta }^3=27+1=28 \end{gathered}$$

11. Let A be 3 x 3 real matrix such that $A\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)=2\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right),A\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)=4\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right),A\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)=2\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)$Then, the system $(A-3I)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)$

(1) unique solution

(2) exactly two solutions

(3) no solution

(4) infinitely many solutions

Ans. (1)

Sol.-

$$\begin{gathered} A=\left[\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_3& z_3\end{array}\right] \\ A\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}2\\ 0\\ 2\end{array}\right] \\ Given\left[\begin{array}{c}{x}_1+z_1\\ x_2+z_2\\ x_3+z_3\end{array}\right]=\left[\begin{array}{c}2\\ 0\\ 2\end{array}\right] \\ \therefore x_1+z_1=\mathrm{2.....}(2) \\ {x}_2+z_2=\mathrm{0......}(3) \\ {x}_3+z_3=\mathrm{0......}(4) \\ A=\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right],\text{Given }\left[\begin{array}{c}-x_1+z_1\\ -x_2+z_2\\ -x_3+z_3\end{array}\right]=\left[\begin{array}{c}4\\ 0\\ 4\end{array}\right] \\ \Rightarrow -x_1+z_1=\mathrm{4.......}(5) \\ -x_2+z_2=\mathrm{0.......}(6) \\ -x_3+z_3=\mathrm{4.....}(7) \\ A=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],\text{Given }\left[\begin{array}{c}{y}_1\\ y_2\\ y_3\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ 0\end{array}\right] \\ \therefore y_1=0,y_2=2,y_3=0 \\ \therefore \text{from (2), (3), (4), (5), (6), and (7):} \\ {x}_1=3,x_2=0,x_3=-1 \\ {y}_1=0,y_2=2,y_3=0 \\ {z}_1=-1,z_2=0,z_3=3 \\ \therefore A=\left[\begin{array}{ccc}3& 0& -1\\ 3& 2& 0\\ -1& 0& 3\end{array}\right] \\ \therefore (A-3I)\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right] \\ \left[\begin{array}{ccc}0& 0& -1\\ -1& -1& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}-z\\ -y\\ -x\end{array}\right]=\left[\begin{array}{c}1\\ -2\\ -3\end{array}\right]\Rightarrow [z=-1],[y=-2],[x=-3] \\ {2}^{2024}=9m+4,m\leftarrow \text{even} \\ {2}^{9m+4}=16\cdot (2^3{)}^{3m}\equiv 16(\mathrm{mod}9) \end{gathered}$$

Previous Year Questions from Determinants

1. If the system of equations

$$\begin{gathered} x+(\sqrt{2}\sin \alpha )y+(\sqrt{2}\cos \alpha )z=0 \\ x+(\cos \alpha )y+(\sin \alpha )z=0 \\ x+(\sin \alpha )y-(\cos \alpha )z=0 \end{gathered}$$

has a non-trivial solution, then $\alpha \in \left(0,\frac{\pi }{2}\right)$ is equal to :

$$\begin{gathered} (1)\frac{3\pi }{4} \\ (2)\frac{7\pi }{24} \\ (3)\frac{5\pi }{24} \\ (4)\frac{11\pi }{24} \end{gathered}$$

Ans. (3)

Sol.

$$\begin{gathered} \left|\begin{array}{ccc}1& \sqrt{2}\sin \alpha & \sqrt{2}\cos \alpha \\ 1& \sin \alpha & -\cos \alpha \\ 1& \cos \alpha & \sin \alpha \end{array}\right|=0 \\ \Rightarrow 1-\sqrt{2}\sin \alpha (\sin \alpha +\cos \alpha )+\sqrt{2}\cos \alpha (\cos \alpha -\sin \alpha )=0 \\ \Rightarrow 1+\sqrt{2}\cos 2\alpha -\sqrt{2}\sin 2\alpha =0 \\ \cos 2\alpha -\sin 2\alpha =-\frac{1}{\sqrt{2}} \\ \cos \left(2\alpha +\frac{\pi }{4}\right)=-\frac{1}{2} \\ 2\alpha +\frac{\pi }{4}=2n\pi \pm \frac{2\pi }{3} \\ \alpha +\frac{\pi }{8}=n\pi \pm \frac{\pi }{3} \\ n=0, \\ x=\frac{\pi }{3}-\frac{\pi }{8}=\frac{5\pi }{24} \end{gathered}$$

2. If the system of equations 

$$\begin{gathered} 11x+y+\lambda z=-5 \\ 2x+3y+5z=3 \\ 8x-19y-39z=\mu \end{gathered}$$

has infinitely many solutions, then λ⁴ – µ is equal to : 

(1) 49

(2) 45 

(3) 47

(4) 51 

Ans. (3)

Sol.

$$\begin{gathered} 11x+y+\lambda z=-5 \\ 2x+3y+5z=3 \\ 8x-19y-39z=\mu \\ D=\left|\begin{array}{ccc}11& 1& \lambda \\ 2& 3& 5\\ 8& -19& -39\end{array}\right|=0 \\ \Rightarrow 11(-117+95)-1(-78-40)+\lambda (-38-24) \\ \Rightarrow 11(-22)+118-\lambda (62)=0 \\ \Rightarrow -242+118-62\lambda =0\Rightarrow 62\lambda =118-242 \\ \lambda =\frac{-124}{62}=-2 \\ {D}_1=\left|\begin{array}{ccc}-5& 1& -2\\ 3& 3& 5\\ \mu & -19& -39\end{array}\right|=0 \\ \Rightarrow -5(-117+95)-1(-117-5\mu )-2(-57-3\mu )=0 \\ \Rightarrow 5(-22)+117+5\mu +114+6\mu =0 \\ \Rightarrow 11\mu =-110-231=-341\Rightarrow \mu =-31 \\ {\lambda }^4-\mu =(-2{)}^4-(-31)=16+31=47 \end{gathered}$$

3. For α, β ∈ R and a natural number n, let $A_n=\left[\begin{array}{ccc}r& 1& \frac{n^2+\alpha }{2}\\ 2r& 2& n^2-\beta \\ 3r-2& 3& \frac{n(3n-1)}{2}\end{array}\right]$. Then 2A₁₀ – A₈ is 

(1) 4α + 2β

(2) 2α + 4β

(3) 2n

(4) 0

Ans. (1)

Sol.

$$\begin{gathered} A_r=\left|\begin{array}{ccc}r& 1& \frac{n^2}{2}+\alpha \\ 2r& 2& n^2-\beta \\ 3r-2& 3& \frac{n(3n-1)}{2}\end{array}\right| \\ 2A_{10}-A_8=\left|\begin{array}{ccc}20& 1& \frac{n^2}{2}+\alpha \\ 40& 2& n^2-\beta \\ 56& 3& \frac{n(3n-1)}{2}\end{array}\right|-\left|\begin{array}{ccc}8& 1& \frac{n^2}{2}+\alpha \\ 16& 2& n^2-\beta \\ 22& 3& \frac{n(3n-1)}{2}\end{array}\right| \\ \Rightarrow \left|\begin{array}{ccc}12& 1& \frac{n^2}{2}+\alpha \\ 24& 2& n^2-\beta \\ 34& 3& \frac{n(3n-1)}{2}\end{array}\right| \\ \Rightarrow \left|\begin{array}{ccc}0& 1& \frac{n^2}{2}+\alpha \\ 0& 2& n^2-\beta \\ -2& 3& \frac{n(3n-1)}{2}\end{array}\right| \\ \Rightarrow -2\left((n^2-\beta )-\left(n^2+2\alpha \right)\right) \\ \Rightarrow -2(-\beta -2\alpha )\Rightarrow 4\alpha +2\beta \end{gathered}$$

4. Let αβγ = 45 ; α,β, γ ∈ R. If x(α, 1, 2) +  y(1, β, 2) + z(2, 3, γ) = (0, 0, 0) for some x, y, z ∈ R, xyz ≠ 0 , then 6α + 4β + γ is equal to_______  

Ans. (55)

Sol.

$$\begin{gathered} \alpha \beta \gamma =45,\alpha ,\beta ,\gamma \in \mathbb{R} \\ x(\alpha ,1,2)+y(1,\beta ,2)+z(2,3,\gamma )=(0,0,0) \\ x,y,z\in \mathbb{R},xyz\ne 0 \\ \Rightarrow \alpha x+y+2z=0 \\ x+\beta y+3z=0 \\ 2x+2y+\gamma z=0 \\ xyz\ne 0\Rightarrow \text{non-trivial} \\ \left|\begin{array}{ccc}\alpha & 1& 2\\ 1& \beta & 3\\ 2& 2& \gamma \end{array}\right|=0 \\ \Rightarrow \alpha (\beta \gamma -6)-1(\gamma -6)+2(2-2\beta )=0 \\ \Rightarrow \alpha \beta \gamma -6\alpha -\gamma +6+4-4\beta =0 \\ \Rightarrow 6\alpha +4\beta +\gamma =55 \end{gathered}$$

5. Let the system of equations x + 2y +3z = 5, 2x + 3y + z = 9, 4x + 3y + λz = μ have infinite number of solutions. Then λ + 2μ is equal to :

(1) 28

(2) 17

(3) 22

(4) 15

Ans. (2)

Sol.

$$\begin{gathered} x+2y+3z=5 \\ 2x+3y+z=9 \\ 4x+3y+\lambda z=\mu \\ \text{For infinite solutions, the following must hold:}\Delta ={\Delta }_1={\Delta }_2={\Delta }_3=0 \\ \Delta =\left|\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\\ 4& 3& \lambda \end{array}\right|=0\Rightarrow \lambda =-13 \\ {\Delta }_1=\left|\begin{array}{ccc}5& 2& 3\\ 9& 3& 1\\ \mu & 3& -13\end{array}\right|=0\Rightarrow \mu =15 \\ {\Delta }_2=\left|\begin{array}{ccc}1& 5& 3\\ 2& 9& 1\\ 4& 15& -13\end{array}\right|=0 \\ {\Delta }_3=\left|\begin{array}{ccc}1& 2& 5\\ 2& 3& 9\\ 4& 3& 15\end{array}\right|=0 \\ for\lambda =-13,\mu =15,\text{the system of equations has infinite solutions.} \\ \Rightarrow \lambda +2\mu =-13+30=17 \end{gathered}$$

6. The values of α, for which $\left|\begin{array}{ccc}1& \frac{3}{2}& \alpha +\frac{3}{2}\\ 1& \frac{1}{3}& \alpha +\frac{1}{3}\\ 2\alpha +3& 3\alpha +1& 0\end{array}\right|=0$, lie in the interval

$$\begin{gathered} (1)(–2,1) \\ (2)(–3,0) \\ (3)\left(-\frac{3}{2},\frac{3}{2}\right) \\ (4)(0,3) \end{gathered}$$

Ans. (2)

Sol.

$$\begin{gathered} \left|\begin{array}{ccc}1& \frac{3}{2}& \alpha +\frac{3}{2}\\ 1& 1& \alpha +1\\ 2\alpha +3& 3\alpha +1& 0\end{array}\right|=0 \\ \Rightarrow (2\alpha +3)\left|\begin{array}{cc}\frac{3}{2}& \alpha +\frac{3}{2}\\ 1& \alpha +1\end{array}\right| \\ -(3\alpha +1)\left|\begin{array}{cc}1& \alpha +\frac{3}{2}\\ 1& \alpha +1\end{array}\right|=0 \\ \Rightarrow (2\alpha +3)\cdot \frac{7\alpha }{6}+(3\alpha +1)\cdot \frac{-7}{6}=0 \\ \Rightarrow 2{\alpha }^2+3\alpha +3\alpha +1=0 \\ \Rightarrow 2{\alpha }^2+6\alpha +1=0 \\ \Rightarrow \alpha =\frac{-3+\sqrt{7}}{2},\frac{-3-\sqrt{7}}{2} \end{gathered}$$

7. Consider the system of linear equations x + y + z = 5, x + 2y +λ²z = 9, x + 3y +λz = μ, where λ, μ ∈ R. Then, which of the following statement is NOT correct?

(1) System has infinite number of solution if λ= 1  and μ =13

(2) System is  inconsistent if λ = 1 and μ  ≠ 13

(3) System is  consistent if λ ≠ 1 and μ = 13

(4) System has unique solution if λ ≠ 1 and μ ≠ 13

Ans. (4)

Sol.

$$\begin{gathered} \left|\begin{array}{ccc}1& 1& 1\\ 1& 2& {\lambda }^2\\ 1& 3& \lambda \end{array}\right|=0 \\ \Rightarrow 2{\lambda }^2-\lambda -1=0 \\ \lambda =1,-\frac{1}{2} \\ \left|\begin{array}{ccc}1& 1& 5\\ 2& {\lambda }^2& 9\\ 3& \lambda & \mu \end{array}\right|=0\Rightarrow \mu =13 \\ \text{Infinite solution}\lambda =1\mu =13 \\ \text{For unique soln}\lambda \ne 1 \\ \text{For no soln}\lambda =1\mu \ne 13 \\ If\lambda \ne 1and\mu \ne 13 \\ \text{Considering the case when}\lambda =-\frac{1}{2}and\mu \ne 13\text{ this will generate no solution case} \end{gathered}$$

8. If then $f(x)=\left|\begin{array}{ccc}2{\cos }^{4}x& 2{\sin }^{4}x& 3+{\sin }^{2}2x\\ 3+2{\cos }^{4}x& 2{\sin }^{4}x& {\sin }^{2}2x\\ 2{\cos }^{4}x& 3+2{\sin }^{4}x& {\sin }^{2}2x\end{array}\right|\Rightarrow \frac{1}{5}{f}^{\prime }(0)$ is equal to ______

(1) 0

(2) 1

(3) 2

(4) 6

Ans. (1)

Sol.

$$\begin{gathered} \left|\begin{array}{ccc}2{\cos }^{4}x& 2{\sin }^{4}x& 3+{\sin }^{2}2x\\ 3+2{\cos }^{4}x& 2{\sin }^{4}x& {\sin }^{2}2x\\ 2{\cos }^{4}x& 3+2{\sin }^{4}x& {\sin }^{2}2x\end{array}\right| \\ Rowoperations:R_2\to R_2-R_1,R_3\to R_3-R_1 \\ \Rightarrow \left|\begin{array}{ccc}2{\cos }^{4}x& 2{\sin }^{4}x& 3+{\sin }^{2}2x\\ 3& 0& -3\\ 0& 3& -3\end{array}\right| \end{gathered}$$

f(x) = 45

f'(x)=0

9. If the system of linear equations

$$\begin{gathered} x-2y+z=-4 \\ 2x+\alpha y+3z=5 \\ 3x-y+\beta z=3 \end{gathered}$$

has infinitely many solutions, then 12α + 13β is equal to 

(1) 60

(2) 64

(3) 54

(4) 58

Ans. (4)

Sol.  

$$\begin{gathered} D=\left|\begin{array}{ccc}1& -2& 1\\ 2& \alpha & 3\\ 3& -1& \beta \end{array}\right| \\ =1(\alpha \beta +3)+2(2\beta -9)+1(-2-3\alpha ) \\ =\alpha \beta +3+4\beta -18-2-3\alpha \\ =\alpha \beta -3\alpha +4\beta -\mathrm{17.......}(1) \\ {D}_1=\left|\begin{array}{ccc}-4& -2& 1\\ 5& \alpha & 3\\ 3& -1& \beta \end{array}\right|=0 \\ {D}_2=\left|\begin{array}{ccc}1& -4& 1\\ 2& 5& 3\\ 3& 3& \beta \end{array}\right|=0 \\ \Rightarrow 1(5\beta -9)+4(2\beta -9)+1(6-15)=0 \\ 5\beta -9+8\beta -36-9=0 \\ 13\beta =54\Rightarrow \beta =\frac{54}{13}\text{put in (1)} \\ \frac{54}{13}\alpha -3\alpha +4\left(\frac{54}{13}\right)=17 \\ \frac{54\alpha }{13}-\frac{39\alpha }{13}+\frac{216}{13}=17 \\ \frac{15\alpha +216}{13}=17\Rightarrow 15\alpha =5\Rightarrow \alpha =\frac{1}{3} \\ \text{Now, }12\alpha +13\beta =12\cdot \frac{1}{3}+13\cdot \frac{54}{13} \\ =4+54=58 \end{gathered}$$