$\left[\begin{array}{ccc}2x+y& x-2y& 4x\\ 3x& 3y& x+y\end{array}\right]+\left[\begin{array}{ccc}5x& 8y& 3x+y\\ 5x+5y& x& 2x\end{array}\right]$

Gorlandint
2022-08-02
Answered

Perform the indicated matrix operation.

$\left[\begin{array}{ccc}2x+y& x-2y& 4x\\ 3x& 3y& x+y\end{array}\right]+\left[\begin{array}{ccc}5x& 8y& 3x+y\\ 5x+5y& x& 2x\end{array}\right]$

$\left[\begin{array}{ccc}2x+y& x-2y& 4x\\ 3x& 3y& x+y\end{array}\right]+\left[\begin{array}{ccc}5x& 8y& 3x+y\\ 5x+5y& x& 2x\end{array}\right]$

You can still ask an expert for help

Barbara Klein

Answered 2022-08-03
Author has **19** answers

$\left[\begin{array}{ccc}2x+y& x-2y& 4x\\ 3x& 3y& x+y\end{array}\right]+\left[\begin{array}{ccc}5x& 8y& 3x+y\\ 5x+5y& x& 2x\end{array}\right]$

$\left[\begin{array}{ccc}2x+y+5x& x-y+8y& 4x+3x+y\\ 3x+5x+5y& 3y+x& x+y+2x\end{array}\right]=\left(\begin{array}{ccc}7x+y& x+6y& 7x+y\\ 8x+5y& x+3y& 3x+y\end{array}\right)$

$\left[\begin{array}{ccc}2x+y+5x& x-y+8y& 4x+3x+y\\ 3x+5x+5y& 3y+x& x+y+2x\end{array}\right]=\left(\begin{array}{ccc}7x+y& x+6y& 7x+y\\ 8x+5y& x+3y& 3x+y\end{array}\right)$

asked 2021-09-18

Find an explicit description of Nul A by listing vectors that span the null space.

asked 2021-09-13

Assume that A is row equivalent to B. Find bases for Nul A and Col A.

asked 2021-06-13

For the matrix A below, find a nonzero vector in Nul A and a nonzero vector in Col A.

$A=\left[\begin{array}{cccc}2& 3& 5& -9\\ -8& -9& -11& 21\\ 4& -3& -17& 27\end{array}\right]$

Find a nonzero vector in Nul A.

$A=\left[\begin{array}{c}-3\\ 2\\ 0\\ 1\end{array}\right]$

Find a nonzero vector in Nul A.

asked 2022-05-12

Two 3 dimensional points. $A[{x}_{1},{y}_{1},{z}_{1}]$ and $B[{x}_{2},{y}_{2},{z}_{2}]$. Need to find a transformation matrix which when multiplied to A will give me B and when multiplied by B give me A. The transformation needs to be a reflection against the plane that's perpendicular to the middle of the AB segment and passing through the midpoint of the AB.

asked 2022-06-23

Give an example of non linear transformation matrix? What is the difference between linear and non linear transformation matrix?

asked 2022-07-15

Let V be a 3 dimensional vector space over a field F and fix $({\mathbf{v}}_{\mathbf{1}},{\mathbf{v}}_{\mathbf{2}},{\mathbf{v}}_{\mathbf{3}})$ as a basis. Consider a linear transformation $T:V\to V$. Then we have

$T({\mathbf{v}}_{\mathbf{1}})={a}_{11}{\mathbf{v}}_{\mathbf{1}}+{a}_{21}{\mathbf{v}}_{\mathbf{2}}+{a}_{31}{\mathbf{v}}_{\mathbf{3}}$

$T({\mathbf{v}}_{\mathbf{2}})={a}_{12}{\mathbf{v}}_{\mathbf{1}}+{a}_{22}{\mathbf{v}}_{\mathbf{2}}+{a}_{32}{\mathbf{v}}_{\mathbf{3}}$

$T({\mathbf{v}}_{\mathbf{3}})={a}_{13}{\mathbf{v}}_{\mathbf{1}}+{a}_{23}{\mathbf{v}}_{\mathbf{2}}+{a}_{33}{\mathbf{v}}_{\mathbf{3}}$

So that we can identify T by the matrix

$\left(\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right)$

But then when I read several linear algebra book, it said if $T({\mathbf{v}}_{\mathbf{i}})=\sum _{j}{a}_{ij}{\mathbf{v}}_{\mathbf{j}}$ , then we can identify T by the matrix $({a}_{ij})$. My problem is: isn't the matrix is $({a}_{ji})$ instead of $({a}_{ij})$?

$T({\mathbf{v}}_{\mathbf{1}})={a}_{11}{\mathbf{v}}_{\mathbf{1}}+{a}_{21}{\mathbf{v}}_{\mathbf{2}}+{a}_{31}{\mathbf{v}}_{\mathbf{3}}$

$T({\mathbf{v}}_{\mathbf{2}})={a}_{12}{\mathbf{v}}_{\mathbf{1}}+{a}_{22}{\mathbf{v}}_{\mathbf{2}}+{a}_{32}{\mathbf{v}}_{\mathbf{3}}$

$T({\mathbf{v}}_{\mathbf{3}})={a}_{13}{\mathbf{v}}_{\mathbf{1}}+{a}_{23}{\mathbf{v}}_{\mathbf{2}}+{a}_{33}{\mathbf{v}}_{\mathbf{3}}$

So that we can identify T by the matrix

$\left(\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right)$

But then when I read several linear algebra book, it said if $T({\mathbf{v}}_{\mathbf{i}})=\sum _{j}{a}_{ij}{\mathbf{v}}_{\mathbf{j}}$ , then we can identify T by the matrix $({a}_{ij})$. My problem is: isn't the matrix is $({a}_{ji})$ instead of $({a}_{ij})$?

asked 2021-11-05

Find an explicit description off Nul A by listing vectors that span the null space.

$$A=\left[\begin{array}{cccc}1& 3& 5& 0\\ 0& 1& 4& -2\end{array}\right]$$