Atrisiniet Matrix | Microsoft matemātikas risinātājs

\left( \begin{array} { l l } { 1 } & { 2 } \\ { 1 } & { 3 } \end{array} \right) \left( \begin{array} { l l } { 3 } & { a } \\ { b } & { 1 } \end{array} \right) = \left( \begin{array} { l l } { 1 } & { 0 } \\ { 0 } & { 1 } \end{array} \right)

Atrast b, a

b=-1

a=-2

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5 problēmas, kas līdzīgas:

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How do you simplify \displaystyle{5}{\left[\begin{array}{c} {1}\\-{1}\\-{3}\end{array}\right]}+{6}{\left[\begin{array}{c} -{4}\\{3}\\{5}\end{array}\right]}-{2}{\left[\begin{array}{c} -{3}\\{8}\\-{4}\end{array}\right]} ...

https://socratic.org/questions/how-do-you-simplify-5-1-1-3-6-4-3-5-2-3-8-4

I got: \displaystyle{\left[\begin{array}{c} -{13}\\-{3}\\{23}\end{array}\right]} Explanation: We can first multiply each vector matrix by the corresponding scalar to get: \displaystyle{\left[\begin{array}{c} {5}\cdot{1}\\{5}\cdot-{1}\\{5}\cdot-{3}\end{array}\right]}+{\left[\begin{array}{c} {6}\cdot-{4}\\{6}\cdot{3}\\{6}\cdot{5}\end{array}\right]}+{\left[\begin{array}{c} -{2}\cdot-{3}\\-{2}\cdot{8}\\-{2}\cdot-{4}\end{array}\right]}= ...

Skew-symmetric matrix subspace dimension and basis

https://math.stackexchange.com/questions/1285176/skew-symmetric-matrix-subspace-dimension-and-basis

A^\mathrm{T}=-A implies that A has zeroes on the main diagonal, since they change sign on the RHS, but remain unchanged on the LHS. if you incorporate that into your representation, it ...

Determine the values for a for which the set of matrices are linear independent

https://math.stackexchange.com/questions/2173693/determine-the-values-for-a-for-which-the-set-of-matrices-are-linear-independent

You have done a good job, putting them in a matrix. If you write it in a different order, you get \left( \begin{array}{ccc} 1 & -1 & 2\\ a & 1 & 3\\ 1 & a & 1\\ 0 & 0 & 0\\ \end{array} \right) \left( \begin{array}{ccc} \alpha_1\\ \alpha_2\\ \alpha_3\\ \end{array} \right)=0 ...

\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right) not diagonalizable

https://math.stackexchange.com/questions/225754/left-beginarraycc-1-1-0-1-endarray-right-not-diagonaliza

A different approach: Your matrix has a single eigenvalue \lambda=1 (with multiplicity 2) so if it was diagonalizable it would be similar to the identity, i.e. P^{-1} \begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix} P=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \Rightarrow \\ \begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix}=P \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}P^{-1}=PP^{-1}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}; ...

Describe the column vectors (a,x)^t that occur as the first column of an integer matrix A whose inverse is also an integer matrix

https://math.stackexchange.com/q/820523

HINT: What can you say about the determinant of such a matrix? Can a and c have a common factor >1?

https://math.stackexchange.com/q/582804

Long story short: suppose that \pmatrix{a&b\\-b&a} = T^{-1}AT where a,b are real numbers. If \det(T) and b have the same sign, then the trajectory will spiral clockwise. If they have ...

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