Maths Matrices
Get insights from 117 questions on Maths Matrices, answered by students, alumni, and experts. You may also ask and answer any question you like about Maths Matrices
- 117
Questions
- 0
Discussions
- 0
Active Users
- 7
Followers
Have a question about Maths Matrices? Ask our experts
Keep it short & simple. Type complete word. Avoid abusive language.
Let a, b, c ∈ R be all non-zero and satisfy a³ + b³ + c³ = 2. If the matrix A = [a, b, c; b, c, a; c, a, b] satisfies A?A = I, then a value of abc can be
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Let a, b, c ∈ R be all non-zero and satisfy a³ + b³ + c³ = 2. If the matrix A = [a, b, c; b, c, a; c, a, b] satisfies A?A = I, then a value of abc can be
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
A? A = I
⇒ a²+b²+c²=1 and ab+bc+ca=0
Now, (a+b+c)²=1 ⇒ a+b+c=±1
So, a³+b³+c³-3abc = (a+b+c) (a²+b²+c²-ab-bc-ca) = (±1) (1-0)=±1
⇒ 3abc = 2±1 = 3,1
⇒ abc = 1, 1/3
Upvote 
Let S be the set of all λ∈R for which the system of linear equations
2x-y+2z=2
x-2y+λz=-4
x+λy+z=4
has no solution. Then the set S
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Let S be the set of all λ∈R for which the system of linear equations
2x-y+2z=2
x-2y+λz=-4
x+λy+z=4
has no solution. Then the set S
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
2x-y+2z=2
x-2y+λz=-4
x+λy+z=4
For no solution:
D=|2, -1,2; 1, -2, λ 1,1,1|=0
⇒ 2 (-2-λ²)+1 (1-λ)+2 (λ+2)=0
⇒ -2λ²+λ+1=0
⇒ λ=1, -1/2
D? =|2, -1,2; -4, -2, λ 4,1,1|
=2 (-2-λ)+1 (-4-4λ)+2 (-4+8)
=2 (1+λ) which is not equal to zero for λ=1, -1/2
Upvote Let A be a 2x2 real matrix with entries from {0,1} and |A| ≠ 0. Consider the following two statements;
(P) If A ≠ I₂, then |A| = -1
(Q) If |A| = 1, then tr(A) = 2
where I₂ denotes 2x2 identity matrix and tr(A) denotes the sum of the diagonal entries of A. Then:
Let A be a 2x2 real matrix with entries from {0,1} and |A| ≠ 0. Consider the following two statements;
(P) If A ≠ I₂, then |A| = -1
(Q) If |A| = 1, then tr(A) = 2
where I₂ denotes 2x2 identity matrix and tr(A) denotes the sum of the diagonal entries of A. Then:
|A|≠0
For (P): A≠I?
So, A = [1 0; 0 1] or [1; 0 1] or [1 0; 1]
or [1; 1 0]
So (P) is false.
A = [1 0; 1 0] or [1; 0 1] or [1 0; 1]
⇒ tr (A)=2
⇒ Q is true
Upvote The values of a and b, for which the system of equations
2x + 3y + 6z = 8
x + 2y + az = 5
3x + 5y + 9z = b
Has no solution, are:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
The values of a and b, for which the system of equations
2x + 3y + 6z = 8
x + 2y + az = 5
3x + 5y + 9z = b
Has no solution, are:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
System of equations can be written as
[2 3 6; 1 2 a; 3 5 9] [x; y; z] = [8; 5; b]
R? →R? -1/2R? , R? →R? -3/2R?
. the system will have no solution if 3-a=0 and b-13≠0.
i.e. for a=3 & b≠13.
Upvote For real numbers α and β, consider the following system of linear equations:
x + y - z = 2, x + 2y + αz = 1, 2x - y + z = β.
If the system has infinite solutions, then α + β is equal to ________.
For real numbers α and β, consider the following system of linear equations:
x + y - z = 2, x + 2y + αz = 1, 2x - y + z = β.
If the system has infinite solutions, then α + β is equal to ________.
Δ = |1,1, -1; 1,2, α 2, -1,1| = 1 (2+α)-1 (1-2α)-1 (-1-4) = 2+α+2α-1+5 = 3α+6=0 ⇒ α=-2.
Δ? = |2,1, -1; 1,2, α β, -1,1| = 2 (2+α)-1 (1-αβ)-1 (-1-2β) = 4+2α-1+αβ+1+2β = 4+2α+αβ+2β=0.
4-4-2β+2β=0. This holds.
Δ? = |1,2, -1; 1,1, α 2, β,1| = 1 (1-αβ)-2 (1-2α)-1 (β-2) = 1-αβ-2+4α-β+2 = 1+4α-αβ-β=0.
1-8+2β-β=
Upvote Let y = y(x) be solution of the differential equation logₑ(dy/dx) = 3x + 4y, with y(0) = 0.
If y(-2/3 logₑ2) = α logₑ2, then the value of α is equal to:
Let y = y(x) be solution of the differential equation logₑ(dy/dx) = 3x + 4y, with y(0) = 0.
If y(-2/3 logₑ2) = α logₑ2, then the value of α is equal to:
A = [, [-1, 4]. |A| = 2 - 1 = 1.
Characteristic equation: λ² - tr (A)λ + |A| = 0 ⇒ λ² - 3λ + 1 = 0.
By Cayley-Hamilton, A² - 3A + I = 0. A? ¹ (A² - 3A + I) = A - 3I + A? ¹ = 0.
A? ¹ = 3I - A.
Comparing with A? ¹ = αI + βA, we get α=3, β=-1.
4 (α - β) = 4 (3 - (-1) = 16.
Upvote If P = 
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
If P =
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
