If A is a matrix, and the system AX=O has a non-trivial solution, show that there is no matrix, B, such that BA=I.

dripcima24
2022-09-29
Answered

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Cameron Wallace

Answered 2022-09-30
Author has **8** answers

If $V$ is a nontrivial solution and $BA=I$, then

$$V=IV=(BA)V=B(AV)=BO=O,$$

which contradicts $V$ being nontrivial.

$$V=IV=(BA)V=B(AV)=BO=O,$$

which contradicts $V$ being nontrivial.

KesseTher12

Answered 2022-10-01
Author has **6** answers

π΄ multiplied on the right by π gets 0 and multiplied on the left by π΅ gets πΌ. Multiply on both sides and use associativity, that is, you can first multiply on the right by π, then on the left by π΅, you get the same as: first multiply on the left by π΅ then multiply on the right by π.

asked 2022-07-04

Consider the system of equations in real numbers $\text{\Beta}x,y,z\text{\Beta}$ satisfying

$\frac{x}{4\text{\Beta}\sqrt{{x}^{2}+1}}\text{\Beta}=\text{\Beta}\frac{y}{5\text{\Beta}\sqrt{{y}^{2}+1}}\text{\Beta}=\text{\Beta}\frac{z}{6\text{\Beta}\sqrt{{z}^{2}+1}}$

and $\text{\Beta}x+y+z\text{\Beta}=\text{\Beta}xyz\text{\Beta}$. Find $\text{\Beta}x,y,z\text{\Beta}.$.

I substituted $\text{\Beta}x,y,z\text{\Beta}$ as $\text{\Beta}\mathrm{tan}\beta \x81\u2018({\mathrm{\Xi \u0388}}_{1})\text{\Beta},\text{\Beta}\mathrm{tan}\beta \x81\u2018({\mathrm{\Xi \u0388}}_{2})\text{\Beta},\text{\Beta}\mathrm{tan}\beta \x81\u2018({\mathrm{\Xi \u0388}}_{3})\text{\Beta}$, where

${\mathrm{\Xi \u0388}}_{1}\text{\Beta}+\text{\Beta}{\mathrm{\Xi \u0388}}_{2}\text{\Beta}+\text{\Beta}{\mathrm{\Xi \u0388}}_{3}\text{\Beta}=\text{\Beta}180{\rm B}\u038a\text{\Beta}\text{\Beta}.$

$\frac{x}{4\text{\Beta}\sqrt{{x}^{2}+1}}\text{\Beta}=\text{\Beta}\frac{y}{5\text{\Beta}\sqrt{{y}^{2}+1}}\text{\Beta}=\text{\Beta}\frac{z}{6\text{\Beta}\sqrt{{z}^{2}+1}}$

and $\text{\Beta}x+y+z\text{\Beta}=\text{\Beta}xyz\text{\Beta}$. Find $\text{\Beta}x,y,z\text{\Beta}.$.

I substituted $\text{\Beta}x,y,z\text{\Beta}$ as $\text{\Beta}\mathrm{tan}\beta \x81\u2018({\mathrm{\Xi \u0388}}_{1})\text{\Beta},\text{\Beta}\mathrm{tan}\beta \x81\u2018({\mathrm{\Xi \u0388}}_{2})\text{\Beta},\text{\Beta}\mathrm{tan}\beta \x81\u2018({\mathrm{\Xi \u0388}}_{3})\text{\Beta}$, where

${\mathrm{\Xi \u0388}}_{1}\text{\Beta}+\text{\Beta}{\mathrm{\Xi \u0388}}_{2}\text{\Beta}+\text{\Beta}{\mathrm{\Xi \u0388}}_{3}\text{\Beta}=\text{\Beta}180{\rm B}\u038a\text{\Beta}\text{\Beta}.$

asked 2022-07-01

Suppose I were to be given 3 equations involving variables a, b, and c. The goal of the problem is to find the number of solutions (they do not necessarily have to be real) for the system of equations. Would it be more logical to use Cramer's Rule to solve the linear system or to use basic algebra? A roadmap of the following example problem or the like would be helpful.

A basic system taken from the book would be:

$\{\begin{array}{l}3a+4y\beta \x88\x92z=4\\ 3a+2b\beta \x88\x9211c=\beta \x88\x9213\\ a+2b+3c=7\end{array}$

A basic system taken from the book would be:

$\{\begin{array}{l}3a+4y\beta \x88\x92z=4\\ 3a+2b\beta \x88\x9211c=\beta \x88\x9213\\ a+2b+3c=7\end{array}$

asked 2022-09-05

Consider the coupled nonlinear system of equations given by

$${x}^{3}+{e}^{y}=s\text{\Beta}\text{\Beta}\text{\Beta}\text{\Beta}\text{\Beta}\text{\Beta}\text{\Beta}\mathrm{cos}\beta \x81\u2018x+xy=t$$

which we wish to be able to solve uniquely for $$(x,y)$$ in terms of $$(s,t)$$. Show this cannot be done at$$(x,y)=(0,0)$$.

$${x}^{3}+{e}^{y}=s\text{\Beta}\text{\Beta}\text{\Beta}\text{\Beta}\text{\Beta}\text{\Beta}\text{\Beta}\mathrm{cos}\beta \x81\u2018x+xy=t$$

which we wish to be able to solve uniquely for $$(x,y)$$ in terms of $$(s,t)$$. Show this cannot be done at$$(x,y)=(0,0)$$.

asked 2022-10-21

For real numbers π,π, if π+ππ+π=3, then find the range of π=πβππ+π. Is there any inequalities here to use?

asked 2022-09-27

You have an 24 cm long string. Examine if you can cut in two parts and create

a) Two squares

b) Two circles

whose total area is 20 cm$${}^{2}$$. (The entire length must be used)

It says the string is cut into 2 parts, and not 2 equal parts.

So for 2 squares: The sum of perimeters will be 24 cm. That's,

$4{l}_{1}+4{l}_{2}=24$ and ${l}_{1}^{2}+{l}_{2}^{2}=20$

Similarly, For 2 circles:

$2\mathrm{{\rm O}\x80}{r}_{1}+2\mathrm{{\rm O}\x80}{r}_{2}=24$ and $\mathrm{{\rm O}\x80}{r}_{1}^{2}+\mathrm{{\rm O}\x80}{r}_{2}^{2}=20$

I get 2 equations and 2 unknowns, how do I solve these equations?

a) Two squares

b) Two circles

whose total area is 20 cm$${}^{2}$$. (The entire length must be used)

It says the string is cut into 2 parts, and not 2 equal parts.

So for 2 squares: The sum of perimeters will be 24 cm. That's,

$4{l}_{1}+4{l}_{2}=24$ and ${l}_{1}^{2}+{l}_{2}^{2}=20$

Similarly, For 2 circles:

$2\mathrm{{\rm O}\x80}{r}_{1}+2\mathrm{{\rm O}\x80}{r}_{2}=24$ and $\mathrm{{\rm O}\x80}{r}_{1}^{2}+\mathrm{{\rm O}\x80}{r}_{2}^{2}=20$

I get 2 equations and 2 unknowns, how do I solve these equations?

asked 2022-09-09

Algebraic process to find numbers so that $xy=45$ and $x+y=18$

The sum of two numbers is 18 and their product is 45. Find the numbers.

The sum of two numbers is 18 and their product is 45. Find the numbers.

asked 2021-12-21

Solve using substitution:

35x-5y=20 and y=7x+4

35x-5y=20 and y=7x+4