Maths Matrices
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The system of equation kx + y + z = 1, x + ky + z = k and x + y + zk = k2 has no solution if k is equal to:
The system of equation kx + y + z = 1, x + ky + z = k and x + y + zk = k2 has no solution if k is equal to:
The system of equations has no solution if the determinant of the coefficient matrix is zero.
Δ = |k 1|
|1 k 1|
|1 k|
Δ = k (k² - 1) - 1 (k - 1) + 1 (1 - k) = 0
Δ = k³ - k - k + 1 + 1 - k = 0
⇒ k³ - 3k + 2 = 0 ⇒ (k - 1)² (k + 2) = 0
∴ k = -2, 1
If k = 1 then all the equations are identical (infinite soluti
Upvote 
If A = [cosθ isinθ; isinθ cosθ], (θ = π/24) and A? = [a b; c d], where i = √-1, then, which one of the following is not true?
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If A = [cosθ isinθ; isinθ cosθ], (θ = π/24) and A? = [a b; c d], where i = √-1, then, which one of the following is not true?
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A² = (cos2θ isin2θ cos2θ)
Similarly, A? = (cos5θ isin5θ cos5θ) = (a b; c d)
(1) a²+b² = cos²5θ - sin²5θ = cos10θ = cos75°
(2) a²-d² = cos²5θ - cos²5θ = 0
(3) a²-b² = cos²5θ + sin²5θ = 1
(4) a²-c² = cos²5θ + sin²5θ = 1
Upvote Let A be a 3 × 3 matrix such that adj A =
2 -1 1
-1 0 2
1 -2 -1
and B = adj(adj A). If |A| = λ and |(B⁻¹)ᵀ| = μ, then the ordered pair, (|λ| ⋅ μ) is equal to:
Let A be a 3 × 3 matrix such that adj A =
2 -1 1
-1 0 2
1 -2 -1
and B = adj(adj A). If |A| = λ and |(B⁻¹)ᵀ| = μ, then the ordered pair, (|λ| ⋅ μ) is equal to:
adj A| = |A|² = 9
=> |A| = ±3 => λ = |λ| = 3
=> |B| = |adj A|² = 81
=> | (B? ¹)? | = |B? ¹| = |B|? ¹ = 1/|B| = 1/81 = µ
Upvote Let A = [[1, x],], x ∈ R and A⁴ = [aᵢⱼ]. If a₁₁ = 109, then a₂₂ is equal to
Let A = [[1, x],], x ∈ R and A⁴ = [aᵢⱼ]. If a₁₁ = 109, then a₂₂ is equal to
A = [ x 1 ]
[ 1 0 ]
A² = [ x 1 ] [ x 1 ] = [ x²+1 x ]
[ 1 0 ] [ 1 0 ] [ x 1 ]
A? = [ x²+1 x ] [ x²+1 x ]
[ x 1 ] [ x 1 ]
= [ (x²+1)²+x² x (x²+1)+x ]
[ x (x²+1)+x²+1 ]
a? = (x² + 1)² + x² = 109
⇒ x = ±3
a? = x² + 1 = 10
Upvote