Logo

AskSia

Plus

3. Construct a 3×33 \times 3 non-triangular matrix, which has three distinct rea...
May 27, 2024
Solution by Steps
step 1
To find the eigenvalues and eigenvectors of the given matrix, we start by writing the matrix in standard form: A=(2amp;0amp;10amp;3amp;01amp;0amp;4) A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 3 & 0 \\ 1 & 0 & 4 \end{pmatrix}
step 2
We need to solve the characteristic equation det⁡(A−λI)=0\det(A - \lambda I) = 0. This involves finding the determinant of the matrix A−λIA - \lambda I: det⁡(2−λamp;0amp;10amp;3−λamp;01amp;0amp;4−λ)=0 \det \begin{pmatrix} 2 - \lambda & 0 & 1 \\ 0 & 3 - \lambda & 0 \\ 1 & 0 & 4 - \lambda \end{pmatrix} = 0
step 3
Calculate the determinant: (2−λ)∣3−λamp;00amp;4−λ∣−1∣0amp;11amp;4−λ∣=0 (2 - \lambda) \begin{vmatrix} 3 - \lambda & 0 \\ 0 & 4 - \lambda \end{vmatrix} - 1 \begin{vmatrix} 0 & 1 \\ 1 & 4 - \lambda \end{vmatrix} = 0 (2−λ)((3−λ)(4−λ))−1(0−1(4−λ))=0 (2 - \lambda)((3 - \lambda)(4 - \lambda)) - 1(0 - 1(4 - \lambda)) = 0 (2−λ)(12−7λ+λ2)+(4−λ)=0 (2 - \lambda)(12 - 7\lambda + \lambda^2) + (4 - \lambda) = 0 λ3−9λ2+23λ−15=0 \lambda^3 - 9\lambda^2 + 23\lambda - 15 = 0
step 4
Solve the cubic equation λ3−9λ2+23λ−15=0\lambda^3 - 9\lambda^2 + 23\lambda - 15 = 0 to find the eigenvalues. The roots of this equation are: λ1=3+2,λ2=3,λ3=3−2 \lambda_1 = 3 + \sqrt{2}, \quad \lambda_2 = 3, \quad \lambda_3 = 3 - \sqrt{2}
step 5
To find the eigenvectors, we solve (A−λI)v=0(A - \lambda I)\mathbf{v} = 0 for each eigenvalue. For λ1=3+2\lambda_1 = 3 + \sqrt{2}: (2−(3+2)amp;0amp;10amp;3−(3+2)amp;01amp;0amp;4−(3+2))(v1v2v3)=0 \begin{pmatrix} 2 - (3 + \sqrt{2}) & 0 & 1 \\ 0 & 3 - (3 + \sqrt{2}) & 0 \\ 1 & 0 & 4 - (3 + \sqrt{2}) \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = 0 (−1−2amp;0amp;10amp;−2amp;01amp;0amp;1+2)(v1v2v3)=0 \begin{pmatrix} -1 - \sqrt{2} & 0 & 1 \\ 0 & -\sqrt{2} & 0 \\ 1 & 0 & 1 + \sqrt{2} \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = 0 v1=−1+2,v2=0,v3=1 v_1 = -1 + \sqrt{2}, \quad v_2 = 0, \quad v_3 = 1
step 6
For λ2=3\lambda_2 = 3: (2−3amp;0amp;10amp;3−3amp;01amp;0amp;4−3)(v1v2v3)=0 \begin{pmatrix} 2 - 3 & 0 & 1 \\ 0 & 3 - 3 & 0 \\ 1 & 0 & 4 - 3 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = 0 (−1amp;0amp;10amp;0amp;01amp;0amp;1)(v1v2v3)=0 \begin{pmatrix} -1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = 0 v1=0,v2=1,v3=0 v_1 = 0, \quad v_2 = 1, \quad v_3 = 0
step 7
For λ3=3−2\lambda_3 = 3 - \sqrt{2}: (2−(3−2)amp;0amp;10amp;3−(3−2)amp;01amp;0amp;4−(3−2))(v1v2v3)=0 \begin{pmatrix} 2 - (3 - \sqrt{2}) & 0 & 1 \\ 0 & 3 - (3 - \sqrt{2}) & 0 \\ 1 & 0 & 4 - (3 - \sqrt{2}) \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = 0 (−1+2amp;0amp;10amp;2amp;01amp;0amp;1−2)(v1v2v3)=0 \begin{pmatrix} -1 + \sqrt{2} & 0 & 1 \\ 0 & \sqrt{2} & 0 \\ 1 & 0 & 1 - \sqrt{2} \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = 0 v1=−1−2,v2=0,v3=1 v_1 = -1 - \sqrt{2}, \quad v_2 = 0, \quad v_3 = 1
Answer
The eigenvalues are λ1=3+2\lambda_1 = 3 + \sqrt{2}, λ2=3\lambda_2 = 3, and λ3=3−2\lambda_3 = 3 - \sqrt{2}. The corresponding eigenvectors are v1=(−1+2,0,1)\mathbf{v}_1 = (-1 + \sqrt{2}, 0, 1), v2=(0,1,0)\mathbf{v}_2 = (0, 1, 0), and v3=(−1−2,0,1)\mathbf{v}_3 = (-1 - \sqrt{2}, 0, 1).
Key Concept
Eigenvalues and Eigenvectors
Explanation
Eigenvalues are the special set of scalars associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic roots, characteristic values, proper values, or latent roots. Eigenvectors are the non-zero vectors that change at most by a scalar factor when that linear transformation is applied.
factorize λ^2-6λ+7=0 and show all the necessary steps
Generated Graph
Solution by Steps
step 1
To factorize the quadratic equation λ2−6λ+7=0λ^2 - 6λ + 7 = 0, we start by identifying the coefficients: a=1a = 1, b=−6b = -6, and c=7c = 7
step 2
We use the quadratic formula λ=−b±b2−4ac2aλ = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} to find the roots
step 3
Substituting the coefficients into the quadratic formula: λ=−(−6)±(−6)2−4(1)(7)2(1)=6±36−282=6±82=6±222=3±2λ = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(7)}}{2(1)} = \frac{6 \pm \sqrt{36 - 28}}{2} = \frac{6 \pm \sqrt{8}}{2} = \frac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2}
step 4
The roots of the equation are λ=3+2λ = 3 + \sqrt{2} and λ=3−2λ = 3 - \sqrt{2}
step 5
Therefore, the factorized form of the quadratic equation is (λ−(3+2))(λ−(3−2))(λ - (3 + \sqrt{2}))(λ - (3 - \sqrt{2}))
Answer
(λ−(3+2))(λ−(3−2))(λ - (3 + \sqrt{2}))(λ - (3 - \sqrt{2}))
Key Concept
Quadratic Factorization
Explanation
The quadratic equation λ2−6λ+7=0λ^2 - 6λ + 7 = 0 is factorized by finding its roots using the quadratic formula and expressing the equation as a product of linear factors.
Solution by Steps
step 1
To determine if the set of all triples of real numbers (x,y,z)(x, y, z) is a vector space under the given operations, we need to verify all vector space axioms
step 2
The first operation is vector addition: (x,y,z)+(x′,y′,z′)=(x+x′+1,y+y′+2,z+z′+3)(x, y, z) + (x', y', z') = (x + x' + 1, y + y' + 2, z + z' + 3). We need to check if this operation is commutative
step 3
For commutativity, we check if (x,y,z)+(x′,y′,z′)=(x′,y′,z′)+(x,y,z)(x, y, z) + (x', y', z') = (x', y', z') + (x, y, z). Calculating both sides: (x,y,z)+(x′,y′,z′)=(x+x′+1,y+y′+2,z+z′+3) (x, y, z) + (x', y', z') = (x + x' + 1, y + y' + 2, z + z' + 3) (x′,y′,z′)+(x,y,z)=(x′+x+1,y′+y+2,z′+z+3) (x', y', z') + (x, y, z) = (x' + x + 1, y' + y + 2, z' + z + 3) Since addition is commutative in real numbers, both expressions are equal
step 4
Next, we check associativity of addition: ((x,y,z)+(x′,y′,z′))+(x′′,y′′,z′′)=(x,y,z)+((x′,y′,z′)+(x′′,y′′,z′′))((x, y, z) + (x', y', z')) + (x'', y'', z'') = (x, y, z) + ((x', y', z') + (x'', y'', z'')). Calculating both sides: ((x,y,z)+(x′,y′,z′))+(x′′,y′′,z′′)=(x+x′+1,y+y′+2,z+z′+3)+(x′′,y′′,z′′)=(x+x′+x′′+2,y+y′+y′′+4,z+z′+z′′+6) ((x, y, z) + (x', y', z')) + (x'', y'', z'') = (x + x' + 1, y + y' + 2, z + z' + 3) + (x'', y'', z'') = (x + x' + x'' + 2, y + y' + y'' + 4, z + z' + z'' + 6) (x,y,z)+((x′,y′,z′)+(x′′,y′′,z′′))=(x,y,z)+(x′+x′′+1,y′+y′′+2,z′+z′′+3)=(x+x′+x′′+2,y+y′+y′′+4,z+z′+z′′+6) (x, y, z) + ((x', y', z') + (x'', y'', z'')) = (x, y, z) + (x' + x'' + 1, y' + y'' + 2, z' + z'' + 3) = (x + x' + x'' + 2, y + y' + y'' + 4, z + z' + z'' + 6) Both expressions are equal, so associativity holds
step 5
We check the existence of an additive identity. We need a vector (0,0,0)(0, 0, 0) such that (x,y,z)+(0,0,0)=(x,y,z)(x, y, z) + (0, 0, 0) = (x, y, z). Calculating: (x,y,z)+(0,0,0)=(x+0+1,y+0+2,z+0+3)=(x+1,y+2,z+3) (x, y, z) + (0, 0, 0) = (x + 0 + 1, y + 0 + 2, z + 0 + 3) = (x + 1, y + 2, z + 3) This does not equal (x,y,z)(x, y, z), so there is no additive identity
step 6
Since there is no additive identity, the set of all triples of real numbers (x,y,z)(x, y, z) under the given operations is not a vector space
Answer
The set of all triples of real numbers (x,y,z)(x, y, z) is not a vector space under the given operations.
Key Concept
Vector Space Axioms
Explanation
To be a vector space, a set must satisfy all vector space axioms, including the existence of an additive identity. The given operations do not provide an additive identity, so the set is not a vector space.
© 2023 AskSia.AI all rights reserved