Q: Let S= \ A= ( array lll 0 & 1 & c \\ 1 & a & d \\ 1 & b & e array ): a, b, c, d, e \ 0,1\ . and .|A| \ -1,1\ \ , where |A| denotes the determinant of A. Then the number of elements in S is

$|A|=-(e-d)+c(b-a)= 1$ Case (i) : $c=0 (e, d)=(1,0),(0,1) 2$ ways b and a can be each 2 ways  Total $=8$ ways Case (ii) : $c 0 c=1$ $ d-e+b-a= 1$ $ . array llll 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 array \ 4 2=8$ ways Total $=16$ ways

Q: Let α , β and γ be real numbers. Consider the following system of linear equations x + 2 y + z = 7 x + α z = 11 2 x - 3 y + β z = γ Match each entry in List-I to t

Given, System of equations, x + 2 y + z = 7 x + α z = 11 2 x - 3 y + β z = γ Now finding Δ = 1 2 1 1 0 α 2 - 3 β = 0 ⇒ 3 α - 2 β - 2 α - 3 = 0 ⇒ 7 α - 2 β = 3 ⇒ β = 1 2 7 α - 3 Now finding, Δ 3 = 1 2 7 1 0 11 2 - 3 γ Now equating, Δ 3 = 0 we get, ⇒ 33 - 2 γ - 22 + 7 - 3 = 0 ⇒ γ = 28 And Δ 1 = 7 2 1 11 0 α γ - 3 β ⇒ Δ

Q: Let p , q , r be non-zero real numbers that are, respectively, the 10 th , 100 th and 1000 th terms of a harmonic progression. Consider the system of linear equa

Given, x + y + z = 1 ⋯ 1 10 x + 100 y + 1000 z = 0 ⋯ 2 q r x + p r y + p q z = 0 ⋯ 3 Now equation 3 can be re-written as x p + y q + z r = 0 ∵ p , q , r ≠ 0 Now given p , q & r are 10 t h , 100 t h & 1000 t h term of a

Q: Let , and be real numbers such that the system of linear equations array c x+2 y+3 z= \\ 4 x+5 y+6 z= \\ 7 x+8 y+9 z= -1 array is consistent. Let |M| represent the determinant of the matrix M= [ array

x + 2 y + 3 z = α 4 x + 5 y + 6 z = β 7 x + 8 y + 9 z = γ - 1 The system of linear equations is consistent it means it has unique solution or infinite solutions Here, Δ = 1 2 3 4 5 6 7 8 9 = 0 Hence, there are infinitely many solutions. If equations have infinitely many solutions, then the equations are linearly connected, i.e., L 1 + λ L 2 = L 3 ⇒ x + 2 y + 3 z - α + λ ( 4 x + 5 y + 6 z - β ) = 7 x + 8 y + 9 z - γ + 1 1 + 4 λ 7 = 2 +

Q: Let , and be real numbers such that the system of linear equations array c x+2 y+3 z= \\ 4 x+5 y+6 z= \\ 7 x+8 y+9 z= -1 array is consistent. Let |M| represent the determinant of the matrix M= [ array

x + 2 y + 3 z = α 4 x + 5 y + 6 z = β 7 x + 8 y + 9 z = γ - 1 The system of linear equations is consistent it means it has unique solution or infinite solutions Here, Δ = 1 2 3 4 5 6 7 8 9 = 0 Hence, there are infinitely many solutions. If equations have infinitely many solutions, then the equations are linearly connected, i.e., L 1 + λ L 2 = L 3 ⇒ x + 2 y + 3 z - α + λ ( 4 x + 5 y + 6 z - β ) = 7 x + 8 y + 9 z - γ + 1 1 + 4 λ 7 = 2 +

Question 1

Let S= \ A= ( array lll 0 & 1 & c \ 1 & a & d \ 1 & b & e array ): a, b, c, d, e \ 0,1\ . and .|A| \ -1,1\ \ , where |A| denotes the determinant of A. Then the number of elements in S is

Accepted Answer

$|A|=-(e-d)+c(b-a)= 1$ Case (i) : $c=0 (e, d)=(1,0),(0,1) 2$ ways b and a can be each 2 ways  Total $=8$ ways Case (ii) : $c 0 c=1$ $ d-e+b-a= 1$ $ . array llll 1 & 1 & 1 & 0 \ 1 & 1 & 0 & 1 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 array \ 4 2=8$ ways Total $=16$ ways

Question 2

Let &#945; , &#160; &#946; and &#947; be real numbers. Consider the following system of linear equations x + 2 y + z = 7 x + &#945; z = 11 2 x - 3 y + &#946; z = &#947; Match each entry in List-I to t

Accepted Answer

Given, System of equations, x + 2 y + z = 7 x + &#945; z = 11 2 x - 3 y + &#946; z = &#947; Now finding  &#916; = 1 2 1 1 0 &#945; 2 - 3 &#946; = 0 &#8658; 3 &#945; - 2 &#946; - 2 &#945; - 3 = 0 &#8658; 7 &#945; - 2 &#946; = 3 &#8658; &#946; = 1 2 7 &#945; - 3 Now finding, &#916; 3 = 1 2 7 1 0 11 2 - 3 &#947; Now equating,  &#916; 3 = 0  we get, &#8658; 33 - 2 &#947; - 22 + 7 - 3 = 0 &#8658; &#947; = 28 And  &#916; 1 = 7 2 1 11 0 &#945; &#947; - 3 &#946;   &#8658; &#916;

Question 3

Let p , q , r  be non-zero real numbers that are, respectively, the 10 th &#160; , 100 th &#160;  and 1000 th &#160;  terms of a harmonic progression. Consider the system of linear equa

Accepted Answer

Given, x + y + z = 1 &#160; &#160; &#160; &#160; &#160; &#160; &#8943; 1 10 x + 100 y + 1000 z = 0 &#160; &#160; &#160; &#160; &#160; &#160; &#8943; 2 q r x + p r y + p q z = 0 &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#8943; 3 Now equation 3  can be re-written as x p + y q + z r = 0 &#160; &#160; &#160; &#160; &#8757; &#160; p , q , r &#8800; 0 Now given  p , &#160; q &#160; &#38; &#160; r &#160;are&#160;10 t h ,&#160;100 t h &#160;&#38;&#160;1000 t h &#160;term&#160;of&#160;a

Question 4

Let , and be real numbers such that the system of linear equations array c x+2 y+3 z= \ 4 x+5 y+6 z= \ 7 x+8 y+9 z= -1 array is consistent. Let |M| represent the determinant of the matrix M= [ array

Accepted Answer

x + 2 y + 3 z = &#945; 4 x + 5 y + 6 z = &#946; 7 x + 8 y + 9 z = &#947; - 1 The system of linear equations is consistent it means it has unique solution or infinite solutions Here,  &#916; = 1 2 3 4 5 6 7 8 9 = 0 Hence, there are infinitely many solutions. If equations have infinitely many solutions, then the equations are linearly connected, i.e., L 1 + &#955; L 2 = L 3 &#8658; x + 2 y + 3 z - &#945; + &#955; ( 4 x + 5 y + 6 z - &#946; ) = 7 x + 8 y + 9 z - &#947; + 1 1 + 4 &#955; 7 = 2 +

Question 5

Let , and be real numbers such that the system of linear equations array c x+2 y+3 z= \ 4 x+5 y+6 z= \ 7 x+8 y+9 z= -1 array is consistent. Let |M| represent the determinant of the matrix M= [ array

Accepted Answer

x + 2 y + 3 z = &#945; 4 x + 5 y + 6 z = &#946; 7 x + 8 y + 9 z = &#947; - 1 The system of linear equations is consistent it means it has unique solution or infinite solutions Here,  &#916; = 1 2 3 4 5 6 7 8 9 = 0 Hence, there are infinitely many solutions. If equations have infinitely many solutions, then the equations are linearly connected, i.e., L 1 + &#955; L 2 = L 3 &#8658; x + 2 y + 3 z - &#945; + &#955; ( 4 x + 5 y + 6 z - &#946; ) = 7 x + 8 y + 9 z - &#947; + 1 1 + 4 &#955; 7 = 2 +

Question 6

Let x &#8712; R and let P = 1 1 1 0 2 2 0 0 3 , &#160; Q = 2 x x 0 4 0 x x 6 and R = P Q P - 1 . Then which of the following options is/are correct?

Accepted Answer

P = 1 1 1 0 2 2 0 0 3 Q = 2 x x 0 4 0 x x 6 Option ( C ) Now R = P Q P - 1 det &#8289; R = det &#8289; R det &#8289; Q det &#8289; P - 1 det &#8289; R = det &#8289; Q det &#8289; P - 1 = 1 det &#8289; P det &#8289; R = det &#8289; 2 x x 0 4 0 x x 6 det &#8289; R = 48 - 4 x 2 now det &#8289; 2 x x 0 4 0 x x 5 = 40 - 4 x 2 det &#8289; R = det &#8289; 2 x x 0 4 0 x x 5 + 8 Option A R &#945; &#946; &#947; = 0 0 0 must have not trivial solution So det &#8289; R = 0 48 - 4 x 2 = 0 &#8658; x = &#177; 2

Determinants — Previous Year Questions & Solutions