x - y =2z=13

2x +2y -z =-6

-x + 3y +z = -7

a) Provide a coefficient matrix corresponding to the system oflinear equations

b) what is the inverse of this matrix?

c) What is the transpose of the matrix?

d) find the determinant for this matrix?

Livia Cardenas
2022-07-31
Answered

This has four parts to it. the following system of linear equations

x - y =2z=13

2x +2y -z =-6

-x + 3y +z = -7

a) Provide a coefficient matrix corresponding to the system oflinear equations

b) what is the inverse of this matrix?

c) What is the transpose of the matrix?

d) find the determinant for this matrix?

x - y =2z=13

2x +2y -z =-6

-x + 3y +z = -7

a) Provide a coefficient matrix corresponding to the system oflinear equations

b) what is the inverse of this matrix?

c) What is the transpose of the matrix?

d) find the determinant for this matrix?

You can still ask an expert for help

coolng90qo

Answered 2022-08-01
Author has **14** answers

a) $\left[\begin{array}{cccc}1& -1& -2& 13\\ 2& 2& -1& -6\\ -1& 3& 1& -7\end{array}\right]$

b) $\left[\begin{array}{cccccc}1& -1& -2& 1& 0& 0\\ 2& 2& -1& 0& 1& 0\\ -1& 3& 1& 0& 0& 1\end{array}\right]=\left[\begin{array}{cccccc}1& -1& -2& 1& 0& 0\\ 0& 4& 3& -2& 1& 0\\ -1& 3& 1& 0& 0& 1\end{array}\right]-R1+R2\to R2$

$\left[\begin{array}{cccccc}1& -1& -2& 1& 0& 0\\ 0& 4& 3& -2& 1& 0\\ 0& 2& -1& 1& 0& 1\end{array}\right]R3+R1\to R3$

$\left[\begin{array}{cccccc}1& -1& -2& 1& 0& 0\\ 0& 4& 3& -2& 1& 0\\ 0& 0& 5& -4& 1& -2\end{array}\right]-R3+R2\to R3$

$\left[\begin{array}{cccccc}1& 0& -5/4& 1/2& 1/4& 0\\ 0& 1& 3/4& -1/2& 1/4& 0\\ 0& 0& 1& -4/5& 1/5& -2/5\end{array}\right]-R1+R2\to R1$

$\left[\begin{array}{cccccc}1& 0& -5/4& 1/2& 1/4& 0\\ 0& 1& 0& 1/10& 1/10& 3/10\\ 0& 0& 1& -4/5& 1/5& -2/5\end{array}\right]-R1-(3/4)R3\to R2$

Finally

$R1+(5/4)R3\to R1$

that is your inverse

$\left[\begin{array}{cccccc}1& 0& 0& -1/2& 1/2& -1/2\\ 0& 1& 0& 1/10& 1/10& 3/10\\ 0& 0& 1& -4/5& 1/5& -2/5\end{array}\right]$

$A=\left[\begin{array}{ccc}1& -1& -2\\ 2& 2& -1\\ -1& 3& 1\end{array}\right],{A}^{-1}=\left[\begin{array}{ccc}-1/2& 1/2& -1/2\\ 1/10& 1/10& 3/10\\ -4/5& 1/5& -2/5\end{array}\right]$

c) transpose: $A={A}^{T}$

d) Det(A)=-10

Det(A)=1(2+3)+1(2-1)-2(6+2)=-10

b) $\left[\begin{array}{cccccc}1& -1& -2& 1& 0& 0\\ 2& 2& -1& 0& 1& 0\\ -1& 3& 1& 0& 0& 1\end{array}\right]=\left[\begin{array}{cccccc}1& -1& -2& 1& 0& 0\\ 0& 4& 3& -2& 1& 0\\ -1& 3& 1& 0& 0& 1\end{array}\right]-R1+R2\to R2$

$\left[\begin{array}{cccccc}1& -1& -2& 1& 0& 0\\ 0& 4& 3& -2& 1& 0\\ 0& 2& -1& 1& 0& 1\end{array}\right]R3+R1\to R3$

$\left[\begin{array}{cccccc}1& -1& -2& 1& 0& 0\\ 0& 4& 3& -2& 1& 0\\ 0& 0& 5& -4& 1& -2\end{array}\right]-R3+R2\to R3$

$\left[\begin{array}{cccccc}1& 0& -5/4& 1/2& 1/4& 0\\ 0& 1& 3/4& -1/2& 1/4& 0\\ 0& 0& 1& -4/5& 1/5& -2/5\end{array}\right]-R1+R2\to R1$

$\left[\begin{array}{cccccc}1& 0& -5/4& 1/2& 1/4& 0\\ 0& 1& 0& 1/10& 1/10& 3/10\\ 0& 0& 1& -4/5& 1/5& -2/5\end{array}\right]-R1-(3/4)R3\to R2$

Finally

$R1+(5/4)R3\to R1$

that is your inverse

$\left[\begin{array}{cccccc}1& 0& 0& -1/2& 1/2& -1/2\\ 0& 1& 0& 1/10& 1/10& 3/10\\ 0& 0& 1& -4/5& 1/5& -2/5\end{array}\right]$

$A=\left[\begin{array}{ccc}1& -1& -2\\ 2& 2& -1\\ -1& 3& 1\end{array}\right],{A}^{-1}=\left[\begin{array}{ccc}-1/2& 1/2& -1/2\\ 1/10& 1/10& 3/10\\ -4/5& 1/5& -2/5\end{array}\right]$

c) transpose: $A={A}^{T}$

d) Det(A)=-10

Det(A)=1(2+3)+1(2-1)-2(6+2)=-10

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Determine whether the given set S is a subspace of the vector space V.

A. V=${P}_{5}$ , and S is the subset of ${P}_{5}$ consisting of those polynomials satisfying p(1)>p(0).

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C.$V={R}^{n}$ , and S is the set of solutions to the homogeneous linear system Ax=0 where A is a fixed m×n matrix.

D. V=${C}^{2}(I)$ , and S is the subset of V consisting of those functions satisfying the differential equation y″−4y′+3y=0.

E. V is the vector space of all real-valued functions defined on the interval [a,b], and S is the subset of V consisting of those functions satisfying f(a)=5.

F. V=${P}_{n}$ , and S is the subset of ${P}_{n}$ consisting of those polynomials satisfying p(0)=0.

G.$V={M}_{n}(R)$ , and S is the subset of all symmetric matrices

A. V=

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Solve the system

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${x}_{1}-2{x}_{2}+2{x}_{3}=1$

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On the one hand, we always use the multiplication by transformation matrix when we want to receive a matrix in a new basis:

- ${A}^{\prime}=CA$ , where ${A}^{\prime}$ is a matrix in a new basis; $C$ is a transformation matrix; $A$ is a matrix in the standard basis.

However, when we speak about getting the matrix in its eigenbasis, we use the other formula.

- $A={C}^{-1}{A}^{\prime}C$ where $A$ is a matrix in a standard basis; ${A}^{\prime}$ is a matrix in the eigenbasis; $C$ is the transformation matrix.

Why do we have different formulas while doing the same thing (matrix transformations) or do I get something wrong?

- ${A}^{\prime}=CA$ , where ${A}^{\prime}$ is a matrix in a new basis; $C$ is a transformation matrix; $A$ is a matrix in the standard basis.

However, when we speak about getting the matrix in its eigenbasis, we use the other formula.

- $A={C}^{-1}{A}^{\prime}C$ where $A$ is a matrix in a standard basis; ${A}^{\prime}$ is a matrix in the eigenbasis; $C$ is the transformation matrix.

Why do we have different formulas while doing the same thing (matrix transformations) or do I get something wrong?

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Suppose $T$ is an element of $L({P}_{3}(\mathbb{R}),{P}_{2}(\mathbb{R}))$ is the differentiation map defined by $Tp={p}^{\prime}$. Find a basis of ${P}_{3}(\mathbb{R})$ and a basis of ${P}_{2}(\mathbb{R})$ such that the matrix of T with respect to these basis is

$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\end{array}\right)$

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Linear transformation and its matrix

two bases:

$A=\{{v}_{1},{v}_{2},{v}_{3}\}$ and $B=\{2{v}_{1},{v}_{2}+{v}_{3},-{v}_{1}+2{v}_{2}-{v}_{3}\}$

There is also a linear transformation: $T:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$

Matrix in base $A$:

${M}_{T}^{A}=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 1& 1& 0\end{array}\right]$

Now I am to find matrix of the linear transformation $T$ in base $B$.

I have found two transition matrixes (from base $A$ to $B$ and from $B$ to $A$):

${P}_{A}^{B}=\left[\begin{array}{ccc}2& 0& -1\\ 0& 1& 2\\ 0& 1& -1\end{array}\right]$

$({P}_{A}^{B}{)}^{-1}={P}_{B}^{A}=\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{6}& \frac{-1}{6}\\ 0& \frac{1}{3}& \frac{2}{3}\\ 0& \frac{1}{3}& \frac{-1}{3}\end{array}\right]$

How can I find ${M}_{T}^{B}$?

two bases:

$A=\{{v}_{1},{v}_{2},{v}_{3}\}$ and $B=\{2{v}_{1},{v}_{2}+{v}_{3},-{v}_{1}+2{v}_{2}-{v}_{3}\}$

There is also a linear transformation: $T:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$

Matrix in base $A$:

${M}_{T}^{A}=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 1& 1& 0\end{array}\right]$

Now I am to find matrix of the linear transformation $T$ in base $B$.

I have found two transition matrixes (from base $A$ to $B$ and from $B$ to $A$):

${P}_{A}^{B}=\left[\begin{array}{ccc}2& 0& -1\\ 0& 1& 2\\ 0& 1& -1\end{array}\right]$

$({P}_{A}^{B}{)}^{-1}={P}_{B}^{A}=\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{6}& \frac{-1}{6}\\ 0& \frac{1}{3}& \frac{2}{3}\\ 0& \frac{1}{3}& \frac{-1}{3}\end{array}\right]$

How can I find ${M}_{T}^{B}$?