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If a > 0 and discriminant of ax^{2}+2bx+c is –ve, then
By R_{3} → R_{3} – (xR_{1} + R_{2});
= (ax^{2} + 2bx + c)(b^{2} – ac) = (+)(–) = ve.
If the system of linear equations
x + 2ay +az = 0 ; x +3by+bz = 0 ; x +4cy+cz = 0 ; has a non  zero solution, then a, b, c.
For homogeneous system of equations to have non zero solution, Δ = 0
On simplification,
∴ a,b,c are in Harmonic Progression.
If 1, ω,ω^{2} are the cube roots of unity, then
The only correct statement about the matrix A is
If B is the inverseof matrix A, then α is
Also since, B = A^{1} ⇒ AB = I
If a_{1}, a_{2} , a_{3}, ......,a_{n}, .... are in G.P., then the value of the determinant
Let r be the common ratio, then
If A^{2} – A +I=0 , then the inverse of A is
Given A^{2}  A +I= 0
A^{1}A2  A^{1}A + A^{1}.I= A^{1}.0
(Multiplying A^{1} on both sides)
⇒ A^{1} +A^{1 }= 0 or A^{1} = 1A .
The system of equations
α x + y + z = α – 1
x + α y + z = α – 1
x + y + α z = α – 1
has infinite solutions, if α is
α x + y + z = α 1
x + α y + z = α – 1;
x + y + z α = α – 1
= α(α^{2}  1)  1(α  1) + 1(1  α)
= α (α1)(α+1)  1(α1)  1(α1)
For infinite solutions, Δ = 0
⇒ (α  1)[α^{2} + α  1  1] = 0
⇒ (α  1)[α^{2} + α  2] = 0 ⇒ α = – 2,1;
But a ≠ 1 . ∴ α = – 2
If a^{2} + b^{2 }+ c^{2} = – 2 and
then f (x) is a polynomial of degree
Applying, C_{1} → C_{1} + C_{2} + C_{3} we get
[As given that a^{2} + b^{2} + c^{2} = –2]
∴ a^{2} +b^{2}+c^{2} + 2 = 0
Applying R_{1} → R_{1}R_{2} , R_{2} → R_{2}R_{3}
f (x) = ( x  1)^{2} Hence degree = 2.
If a_{1} , a_{2},a_{3} , ............, an , ...... are in G. P., then the determinant
is equal to
∵ a_{1}, a_{2} ,a_{3} , ....... are in G.P..
∴ Using a_{n} = ar^{ n 1} ,we get the given determinant, as
Operating C_{3}  C_{2} and C_{2}C_{1} and using
= 0 (two columns being identical)
If A and B are square matrices of size n × n such that A^{2}  B^{2} = (A B)(A+B), then which of the following will be always true?
A^{2}  B^{2} = (A B)(A+B)
A^{2}  B^{2} = A^{2} + AB  BAB^{2} ⇒ AB = BA
Hence, AB = BA only when a = b
∴ There can be infinitely many B's for which AB = BA
Hence, D is divisible by both x and y
Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity matrix. Denote by tr(A), the sum of diagonal entries of a. Assume that A^{2} = I.
Statement1 : If A ≠ I and A ≠ –I, then det(A) = –1
Statement2 : If A ≠ I and A ≠ –I, then tr (A) ≠ 0.
From these four relations,
a^{2} + bc = bc + d^{2} ⇒ a^{2} = d^{2}
and b(a + d) = 0 = c( a + d) ⇒ a = – d
We can take a = 1, b = 0, c = 0, d = –1 as one possible set of values, then
Clearly A ≠ I and A ≠ – I and det A = –1 ∴ Statement 1 is true.
Also if A ≠ I then tr(A) = 0
∴ Statement 2 is false.
Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x = cy + bz, y = az + cx, and z = bx + ay. Then a^{2} + b^{2} + c^{2} + 2abc is equal to
The given equations are
–x + cy + bz = 0
cx –y + az = 0
bx + ay – z = 0
∵ x, y, z are not all zero
∴ The above system should not have unique (zero) solution
⇒ –1(1– a^{2}) – c(– c – ab) + b(ac + b) = 0
⇒–1 + a^{2} + b^{2} + c^{2} + 2abc = 0
⇒ a^{2} + b^{2} + c^{2} + 2abc = 1
Let A be a square matrix all of whose entries are integers. Then which one of the following is true?
∵ All entries of square matrix A are integers, therefore all cofactors should also be integers.
If det A = ± 1 then A^{–1} exists. Also all entries of A^{–1} are integers.
Let A be a 2 × 2 matrix
Statement 1 : adj (adj A) = A
Statement 2 : adj A = A
We know that  adj (adj A)  = adj A^{2–1}
=  A  2–1= A
∴ Both the statements are true and statement 2 is a correct explantion for statement1 .
Let a, b, c be such that b(a + c) ≠ 0 if
then the value of n is :
(Taking transpose of second determinant)
⇒ n should be an odd integer.
The number of 3 × 3 nonsingular matrices, with four entries as 1 and all other entries as 0, is
are 6 nonsingular matrices because 6 blanks will be filled by 5 zeros and 1 one.
Similarly, are 6 nonsingular matrices.
So, required cases are more than 7, nonsingular 3 × 3 matrices.
Let A be a 2 × 2 matrix with nonzero entries and let A^{2} = I , where I is 2 × 2 identity matrix. Define Tr(A) = sum of diagonal elements of A and A = determinant of matrix A.
Statement  1 : Tr(A) = 0.
Statement 2 : A = 1.
Consider the system of linear equations ;
x_{1} + 2x_{2} + x_{3} = 3
2x_{1} + 3x_{2} + x_{3} = 3
3x_{1} + 5x_{2} + 2x_{3} = 1
The system has
⇒ Given system, does not have any solution.
⇒ No solution
The number of values of k for which the linear equations 4x + ky + 2z = 0 , kx + 4y + z = 0 and 2x + 2y + z = 0 possess a nonzero solution is
Let A and B be two symmetric matrices of order 3.
Statement1: A(BA) and (AB)A are symmetric matrices.
Statement2: AB is symmetric matrix if matrix multiplication of A with B is commutative.
∴ A' = A, B' = B
Now (A(BA))' = (BA)'A'
= (A'B')A' = (AB)A = A(BA)
Similarly ((AB)A)' = (AB)A
So, A(BA) and (AB)A are symmetric matrices.
Again (AB)' = B'A' = BA
Now if BA = AB, then AB is symmetric matrix.
If u_{1} and u_{2} are column matrices such that then u_{1} + u_{2} is equal to :
..(1)
We know,
Now, from equation (1), we have
Let P and Q be 3 x 3 matrices P ≠ Q. If P^{3} = Q^{3} and P^{2}Q = Q^{2}P then determinant of (P^{2} + Q^{2}) is equal to :
Given P^{3} = Q^{3} ...(1)
and P^{2}Q = Q^{2}P ...(2)
Subtracting (1) and (2), we get
P^{3} – P^{2}Q = Q^{3} – Q^{2}P
⇒ P^{2} (P–Q) + Q^{2} (P – Q) = 0
⇒ (P^{2} + Q2) (P–Q) = 0 ⇒ P^{2} + Q^{2} = 0 as P ≠ Q
If is the adjoint of a 3 × 3 matrix A and A = 4, then α is equal to :
If α, β ≠ 0, and f (n) = α^{n} +β^{n} and
then K is equal to:
Consider
If A is a 3 × 3 nonsingular matrix such that AA' = A'A and B = A^{–1}A', then BB' equals:
BB' = B(A^{1} A')' = B(A ')'(A^{1})' = BA (A^{–1})'
= (A^{ 1} A')(A(A^{1})')
= A^{–1}A .A'.(A^{–1})' {as AA' = A'A}
= I(A^{–1}A)'
= I.I = I^{2} = I
The set of all values of l for which the system of linear equations :
2x_{1} – 2x_{2} + x_{3} = λx_{1}
2x_{1} – 3x_{2} + 2x_{3} = λx_{2}
–x_{1} + 2x_{2} = λx_{3}
has a nontrivial solution
a matrix satisfying the equation AA^{T} = 9I, where I is 3 × 3 identity matrix, then the ordered pair (a, b) is equal to:
⇒ a + 4 + 2b = 0 ⇒ a + 2b = – 4 ...(i)
2a + 2 – 2b = 0 ⇒ 2a – 2b = – 2 ⇒ a – b = –1 ...(ii)
On solving (i) and (ii) we get
– 1 + b + 2b = – 4 ...(i)
b = – 1 and a = – 2
(a, b) = (–2, –1)
The system of linear equations
x + λy – z = 0
λx – y – z = 0
x + y – λz = 0
has a nontrivial solution for:
For trivial solution,
then 5a + b is equalto :
A(adj A) = A A^{T}
⇒ A^{–1}A (adj A) = A^{–1}A A^{T}
adj A = A^{T}
⇒ 5a + b=5
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