Prática de Shadowing: Lec 1 | MIT 18.06 Linear Algebra, Spring 2005 - Aprenda a falar inglês com vídeo

Hi.

This is the first lecture in MIT's course 1806, Linear Algebra, and I'm Gilbert Strang.

The text for the course is this book, Introduction to Linear Algebra, and the course web page,

which has got a lot of exercises from the past,

MATLAB codes, the syllabus for the course, is web.mit.edu slash 18.06.

And this is the first lecture, lecture one.

So, and later we'll say where, which will give the web address for viewing these videotapes.

Okay, so what's in the first lecture?

This is my plan.

The fundamental problem of linear algebra, which is to solve a system of linear equations.

So let's start with the case when we have some number of equations, say n equations and n unknowns.

So an equal number of equations and unknowns.

That's the normal NISC case.

And what I want to do is, with examples, of course, to describe first what I call the row picture.

That's the picture of one equation at a time.

It's the picture you've seen before in two by two equations where lines meet.

So in a minute you'll see lines meeting.

The second picture, I'll put a star beside that, because that's such an important one.

and maybe new to you, is the picture a column at a time.

And those are the rows and columns of a matrix.

So the third, the algebra way to look at the problem is the matrix form using a matrix that I'll call A.

OK.

So can I do an example?

The whole semester will be examples and then see what's going on with the example.

So take an example.

Two equations, two unknowns.

So let me take 2x minus y equals zero,

let's say, and minus x plus, say, 2y equals three.

OK.

Let me, I can even say right away, what's the matrix that, that is, what's the coefficient matrix?

The matrix that involves these numbers?

A matrix is just a rectangular array of numbers.

Here the, here it's two rows and two columns.

So two and minus one in the first row, minus one and two in the second row, that's the matrix.

And the right-hand, the unknown, well, we've got two unknowns.

So we've got a vector, x with two components, x and y, and we've got two right-hand sides that go into a vector .

I couldn't resist writing the matrix form right even before the pictures.

So I always will think of this as the matrix A, the matrix of coefficients.

Then there's a vector of unknowns.

Here we've only got two unknowns.

Later we'll have any number of unknowns.

And that vector of unknowns, well, I'll often, I'll make that x extra bold.

And the right-hand side is also a vector that I'll always call b.

So linear equations are Ax equal b,

and the idea now is to solve this particular example and then step back to see the bigger picture.

OK, what's the picture for this example?

The row picture.

OK, so here comes the row picture.

So that means I take one row at a time and I'm drawing here the xy plane and I'm going to put,

plot all the points that satisfy that first equation.

So I'm looking at all the points that satisfy 2x minus y equals zero.

I'm often, it's often good to start with, which point on the horizontal line?

On this horizontal line, y is zero.

The x-axis has y is zero, and that, in this case, actually, then x is zero.

So the point, the origin, the point with coordinates zero, zero, is on the line.

It solves that equation.

Okay, tell me an, well, I guess I have to tell you another point that solves this same equation.

Let me suppose x is one, so I'll take x to be one, then y should be two, right?

So there's the point that also solves this equation.

Then I could put in more points.

But let me put in all the points at once, because they all lie on a straight line.

This is a linear equation, and that word linear got the letters for line in it.

That's the equation, this is the line of solutions to 2x minus y equals zero.

My first row, first equation.

So typically maybe x equal a half, y equal one will work.

And sure enough, it does.

Okay.

That's the first one.

Now the second one is not going to go through the origin.

It's always important, do we go through the origin or not?

In this case, yes, because there's a zero over there.

In this case, we don't go through the origin because if x and y are zero, we don't get three.

So let me again say, suppose y is zero, what x do we actually get?

If y is zero, then I get x is minus three.

So if y is zero, I go along minus three.

So there's one point on this second line.

Now let me say, well, suppose x is minus one.

Just to take another x.

If x is minus one, then this is a one.

And I think y should be a one.

Because if x is minus one, then I think y should be a one and we'll get that point.

Is that right?

If x is minus one, that's a one.

If y is a one, that's a two.

And the one and the two make three, and that point's on the equation.

Okay.

Now I should just draw the line, right?

Connecting those two points, that will give me the whole line, and if I've done this reasonably well,

I think it's going to happen to go through, well, not happen, it was arranged to go through that point.

So I think that the second line is this one, and this is the all-important point that lies on both lines.

So we just check that that point, which is the point x equal one and y was two, right?

That's the point there.

And that, I believe, solves both equations.

Let's just check this.

If x is one, I have a minus one plus four equals three.

OK.

Apologies for drawing this picture that you've seen before.

But this seeing the row picture, first of all, for n equal two, two equations and two unknowns, it's the right place to start.

OK.

100

So we've got the solution, the point that lies on both lines.

101

Now can I come to the column picture?

102

Pay attention, this is the key point.

103

So the column picture.

104

I'm now going to look at the columns of the matrix.

105

I'm going to look at this part and this part.

106

I'm going to say that the x part is really, so, so is really x times, you see I'm putting the two,

107

I'm kind of getting the two equations at once.

108

That part, and then I have a y, and in the first equation it's multiplying a minus one, and in the second equation a two,

109

and on the right-hand side, zero and three.

110

You see the columns of the matrix?

111

The columns of A are here and the right-hand side B is there.

112

And now what is the equation asking for?

113

It's asking us to find,

114

somehow to combine that vector and this one in the right amounts to get that one.

115

It's asking us to find the right linear combination.

116

This is called a linear combination.

117

And it's the most fundamental operation in the whole course.

118

It's a linear combination of the columns.

119

That's what we're seeing on the left side.

120

Again, I don't want to write down a big definition.

121

You can see what it is.

122

There's column one, there's column two.

123

I multiply by some numbers and I add.

124

That's a combination, a linear combination, and I want to make those numbers the right numbers to produce zero three.

125

OK.

126

Now I want to draw a picture that represents what this, this is algebra, what's the geometry, what's the picture that goes with it?

127

OK.

128

So again, these vectors have two components, so I better draw a picture like that.

129

So can I put down these columns?

130

I'll draw these columns as they are, and then I'll do a combination of them.

131

So the first column is over two and down one, right?

132

So there's the first column.

133

The first column, column one.

134

It's the vector two minus one.

135

The second column is, let's see, I go over minus one is the first component and up two, it's here.

136

There's column two.

137

So this, again, you see what its components are.

138

Its components are minus one, two.

139

Good.

140

That's this guy.

141

Now what is, what, now I have to take a combination.

142

What combination shall I take?

143

Why not the right combination?

144

What the hell.

145

OK.

146

So the combination I'm going to take is the right one to produce zero three, and then we'll see it happen in the picture.

147

So the right combination is to take x as one of those and two of these.

148

It's because we already know that that's the right x and y, so why not take the correct combination here and see it happen?

149

Okay, so how do I picture this linear combination?

150

So I start with this vector, that's already here, so that's one of column one, that's one times column one, right there.

151

And now I want to add on, so I'm going to hook the next vector on to the front of the arrow will start the next vector,

152

and it'll go this way.

153

So let's see, can I do it right?

154

If I added on one of these vectors, it would go left one and up two, so it would go left one and up two, so it would probably get us to there.

155

Maybe I'll do dotted line for that.

156

OK?

157

That's one of column two tucked on to the end, but I wanted to tuck on two of column two.

158

So the second one will go up left one and up two also.

159

It'll probably end there.

160

And there's another one.

161

So what I've put in here is two of column two.

162

Added on.

163

And where did I end up?

164

What are the coordinates of this result?

165

What do I get when I take one of this plus two of that?

166

I do get that, of course.

167

There it is, there it is x is zero, y is three, that's b.

168

That's the answer we wanted.

169

And how do I do it?

170

You see I do it just like the first component, I have a two and a minus two that produces a zero.

171

And in the second component I have a minus one and a four, they combine to give the three.

172

But look at this picture.

173

So here's our key picture.

174

I combined this column and this column to get, maybe I better, to get this guy.

175

That was the B.

176

That's the zero three.

177

Okay.

178

So that idea of linear combinations is crucial, and also, do we want to think about this question?

179

Sure, why not.

180

What are all the combinations?

181

If I took, can I go back to x's and y's?

182

This is a question for really, it's going to come up over and over,

183

but why don't we see it once now.

184

If I took all the x's and all the y's, all the combinations, what would be all the results?

185

And actually the result would be that I could get any right hand side at all.

186

The combinations of this and this would fill the whole plane.

187

You can tuck that away, we'll explore it further.

188

But this idea of what linear combination gives b, and what do all the linear combinations give,

189

what are all the possible achievable right-hand sides b, that's going to be basic.

190

Okay, can I move to three equations and three unknowns?

191

Because it's easy to picture the two by two case.

192

Let me do a three by three example.

193

Okay, I'll sort of start it the same way, say maybe 2x minus y and maybe I'll take no z's as a zero

194

and maybe a minus x and a two y and maybe a minus z is a, oh, let me make that a minus one.

195

And just for variety, let me take minus three z,

196

minus three y's, I should keep the y's in that line, and four z's is, say, four.

197

OK.

198

That's three equations.

199

I'm in three dimensions, x, y, z, and I don't have a solution yet.

200

So I want to understand the equations and then solve them.

201

Okay.

202

So how do I, how do you understand them?

203

The row picture is one way, the column picture is another very important way.

204

Just let's remember the matrix form here, because that's easy.

205

The matrix form, what's our matrix A?

206

Our matrix A is this right-hand side, the two and the minus one and the zero from the first row,

207

the minus one and the two and the minus one from the second row,

208

the zero, the minus three and the four from the third row.

209

So it's a three by three matrix.

210

Three equations, three unknowns.

211

And what's our right-hand side?

212

Of course, it's the vector, zero, minus one, four.

213

OK.

214

So that's the way, well, that's the shorthand to write out the three equations.

215

But it's the picture that I'm looking for today.

216

OK.

217

So the row picture.

218

All right.

219

So I'm in three dimensions, x, y, and z,

220

and I want to take those equations one at a time

221

and ask and make a picture of all the points that satisfy, let's take equation number two.

222

If I make a picture of all the points that satisfy all the x, y, z points that solve this equation,

223

well, first of all, The origin is not one of them.

224

x, y, z being zero, zero, zero would not solve that equation.

225

So what are some points that do solve the equation?

226

Let's see, maybe if x is one, y and z could be zero.

227

That would work, right?

228

So there's one point.

229

I'm looking at this second equation here just to start with.

230

Let's see, also I guess if z, z could be one, x and y could be zero, so that would just go straight up that axis.

231

And probably I want a third point here.

232

Let me take x to be zero.

233

Let's say x to be zero, z to be zero, then y would be minus a half, right?

234

So there's a third point somewhere.

235

Oh, my.

236

OK.

237

Let's see.

238

I want to put in all the points that satisfy that equation.

239

Do you know what that bunch of points will be?

240

It's a plane.

241

If we have a linear equation, then fortunately the graph of the thing, the plot of all the points that solve it, are a plane.

242

So I, and three, these three points determine a plane, but your lecture is not Rembrandt,

243

and this, the art is going to be the weak point here.

244

So I'm just going to draw a plane, right?

245

There's a plane somewhere.

246

That's my plane.

247

That that plane is all the points that solves this guy.

248

Then what about this one?

249

2x minus y plus zero z.

250

So z actually can be anything.

251

Again, it's going to be another plane.

252

Each row in a three by three problem gives us a plane in three dimensions.

253

So this one is going to be some other plane, maybe it, maybe I'll try to draw it like this.

254

And those two planes meet in a line.

255

So if I have two equations, just the first two equations in three dimensions, those give me a line,

256

the line where those two planes meet.

257

And now the third guy is a third plane,

258

And it goes some, somewhere.

259

OK, those three things meet in a point.

260

Now I don't know where that point is, frankly.

261

But linear algebra will find it.

262

The main point is that there is, that the three planes, because they're, you know, they're not parallel,

263

they're not special, they're, they do meet in one point, and that's the solution.

264

But maybe you can see that this row picture is getting a little hard to see.

265

The row picture was a cinch when we looked at two lines meeting.

266

When we look at three planes meeting, it's not so clear, and in four dimensions, probably a little less clear.

267

So can I quit on the row picture?

268

I'll quit on the row picture before I've successfully found the point where the three planes meet.

269

All I really want to see is that the row picture consists of three planes, and if everything works right,

270

three planes meet in one point, and that's the solution.

271

Now, you can tell I prefer the column picture.

272

Okay?

273

So let me take the column picture.

274

That's x times, so there were two x's in the first equation, minus one x's, I'm just taking, and no x's in the third.

275

It's just the first column of that.

276

And how many y's are there?

277

There's minus one in the first equation, two in the second, and maybe minus three in the third.

278

Just the second column of my matrix.

279

And z times no z's, minus one z's and four z's.

280

And it's those three columns, right, that I have to combine to produce the right-hand side,

281

which is zero, minus one, four.

282

OK.

283

So what have we got on this left-hand side?

284

A linear combination.

285

It's a linear combination now of three vectors, and they happen to be each one is a three-dimensional vector.

286

So we want to know what combination of those three vectors produces that one.

287

Shall I try to draw the column picture then?

288

So since these vectors have three components.

289

So it's some multiple, let me draw in the first column as before, so I, x is two and y is minus one,

290

Maybe there's the first column.

291

Y, the second column has maybe a minus one and a two in the y's and minus three's,

292

somewhere there possibly, column two.

293

And the third column has no zero, minus one, four.

294

So how shall I draw that?

295

So nothing in this was the first component, the second component was a minus one, maybe up here.

296

That's column three, that's the column zero, minus one, and four, this guy.

297

So, again, what's my problem?

298

What this equation is asking me to do is to combine these three vectors with the right combination to produce this one.

299

Well, you can see what the right combination is.

300

Because in this special problem, especially chosen by the lecturer,

301

that right-hand side that I'm trying to get is actually one of these columns.

302

So I know how to get that one.

303

So what's the solution?

304

What combination will work?

305

I just want one of these and none of these.

306

So x should be zero, y should be zero, and z should be one.

307

That's the combination.

308

None of that, none of that, one of those is obviously the right one.

309

So this column three is actually the same as B in this particular problem.

310

I made it work that way just so we would get an answer,

311

zero, zero, one, so somehow that's the point that where those three planes met and I couldn't see it before.

312

Of course, I won't always be able to see it from the column picture, either.

313

It's the next lecture, actually,

314

which is about elimination, which is the systematic way that everybody,

315

every bit of software, too,

316

production, large-scale software would solve the equation.

317

So the lecture that's coming up, if I was to add that to the syllabus, will be about how to find x,

318

y, z in all cases.

319

Can I just think again, though, about the big picture?

320

The big, by the big picture I mean let's keep this same matrix on the left, but imagine that we have a different right-hand side.

321

Oh, let me take a different right-hand side.

322

So I'll change that right-hand side to something that actually is also pretty special.

323

Let me change it to, suppose I, if I add those first two columns that would give me a one and a one and a minus three.

324

There's a very special right-hand side.

325

I just cooked it up by adding this one to this one.

326

Now what's the solution with this new right-hand side?

327

The solution with this new right-hand side is clear.

328

It took, now I took one of these, one of these, and none of those,

329

so actually it just changed around to this when I took this new right-hand side.

330

Okay.

331

So in the row picture, I have three different planes,

332

three new planes meeting now at this point.

333

In the column picture, I have the same three columns, but now I'm combining them to produce this guy,

334

and it turned out that column one plus column two, which would be somewhere there, there is the right column.

335

One of this and one of this would give me the new B.

336

OK.

337

So that's like we squeezed in an extra example.

338

But now think about all Bs, all right-hand sides.

339

Could I get, can I solve these equations for every right-hand side?

340

Can I say that, ask that question?

341

So that's the algebraic question.

342

Can I solve Ax equal b for every b?

343

Let me write that down.

344

Can I solve Ax equal b for every right-hand side b?

345

I mean, is there a solution?

346

And then, if there is, elimination will give me a way to find it.

347

I'm just, I really wanted to ask, is there a solution for every right-hand side?

348

So now, can I put that in different words, in this linear combination words?

349

So in linear combination words, do the linear combinations

350

of the columns fill three-dimensional space.

351

Every b means all the b's in three-dimensional space.

352

So do you see that I'm just asking the same question in different words?

353

Solving AX.

354

Oh, that's very important.

355

A times x, when I multiply a matrix by a vector, I get a combination of the columns.

356

I'll write that down in a moment.

357

So that you see, but in my column picture, that's really what I'm doing.

358

I'm taking linear combinations of these three columns, and I'm trying to find B.

359

And actually the answer for this matrix will be yes.

360

For this matrix, for this,

361

for this matrix A, for these columns, the answer is yes.

362

This matrix is, this matrix that I chose for an example,

363

is a good matrix, a non-singular matrix, an invertible matrix.

364

Those will be the matrices that we like best.

365

There could be other, and we will see other matrices, where the answer becomes no. Oh,

366

actually, you can see when it would become no. When could, what could go wrong?

367

How could it go wrong that out of these, out of three columns and all their combinations,

368

when would I not be able to produce some b off here?

369

When, when could it go wrong?

370

Do you see that the combinations, suppo- let me say when it goes wrong.

371

If these three columns all lie in the same plane,

372

then their combinations will lie in that same plane.

373

So then we're in trouble.

374

If the three columns of my matrix, if those three vectors happen to lie in the same plane,

375

for example, if column three is just the sum of column one and column two, I would be in trouble.

376

That would be a matrix A where the answer would be no. Because the combinations,

377

if column three is in the same plane as column one and two, I don't get anything new from that.

378

All the combinations are in the plane, and the only right-hand sides B that I could get would be the ones in that plane.

379

So I could solve it for some right-hand sides when B is in in the plane,

380

but most right-hand sides would be out of the plane and unreachable.

381

So that would be a singular case.

382

The matrix would be not invertible.

383

There would not be a solution for every b.

384

The answer would become no for that.

385

OK.

386

I don't know, should we take just a little shot at thinking about nine dimensions?

387

Imagine that we have vectors with nine components.

388

Well, it's going to be hard to visualize those.

389

I don't pretend to do it.

390

But somehow, pretend you do.

391

Pretend we have, if this was nine equations in nine unknowns, then we would have nine columns,

392

and each one would be a vector in nine-dimensional space, and we would be looking at their linear combinations.

393

So we'd be having the linear combinations of nine vectors in nine-dimensional space,

394

and we would be trying to find the combination that hit the correct right-hand side b.

395

And we might also ask the question, can we always do it?

396

Can we get every right-hand side b?

397

And certainly it will depend on those nine columns.

398

Some, sometimes the answer will be yes.

399

If I picked a random matrix, it would be yes, actually.

400

If I used MATLAB and just used the random command, picked out a nine by nine matrix,

401

I guarantee it would be good.

402

It would be non-singular, it would be invertible, all beautiful.

403

But if I choose those columns so that they're not independent,

404

so that they're there, so that the ninth column is the same as the eighth column,

405

then it contributes nothing new and there would be right-hand sides b that I couldn't get.

406

Can you sort of think about nine vectors in nine-dimensional space and take their combinations?

407

That's really the central thought that you get kind of used to in linear algebra,

408

even though you can't really visualize it, you sort of think you can after a while.

409

Those nine columns and all their combinations may very well fill out the whole nine-dimensional space.

410

But if the ninth column happened to be the same as the eighth column and gave nothing new, then probably what it would fill out would be,

411

I hesitate even to say this,

412

it would be a com- a sort of plane,

413

an eight-dimensional plane inside nine-dimensional space.

414

And it's those eight-dimensional planes inside nine-dimensional space that we have to work with eventually.

415

For now, let's stay with the nice case where the matrices work, we can get every right-hand side B,

416

and here we see how to do it with columns.

417

OK.

418

There was one step that I realized, which I realized I was saying in words that I now want to write in letters.

419

Because I'm coming back to the matrix form of the equation, so let me write it here.

420

The matrix form of my equation,

421

of my system, is some matrix A times some vector x equals some right-hand side b.

422

OK.

423

So this is a multiplication, A times x, matrix times vector.

424

And I just want to say, how do you multiply a matrix by a vector?

425

OK, so I'm just going to create a matrix.

426

Let me take two, five, one, three.

427

And let me take a vector, x to be, say, one and two.

428

How do I multiply a matrix by a vector?

429

And just think a little bit about matrix notation and how to do that multiplication.

430

So let me say how I multiply a matrix by a vector.

431

Actually, there are two ways to do it.

432

Let me tell you my favorite way.

433

It's a, it's columns again.

434

It's a column at a time.

435

For me, that matrix mo, this matrix multiplication says I take one of that column and two of that column and add.

436

So this is, in my, the way I would think of it, is one of the first column and two of the second column,

437

and let's just see what we get.

438

So in the first component I'm getting a two and a ten, I'm getting a twelve there.

439

In the second component I'm getting a one and a six, I'm getting a seven.

440

So that, that matrix times that vector is 12-7.

441

Now, you could do that another way.

442

You could do it a row at a time.

443

You could do, and you would get this 12, and actually I pretty much did it here, this way.

444

Two, I could take that row times my vector.

445

This is, this is all the time, this is the idea of a dot product.

446

This vector times this vector, two times one plus five times two is the twelve.

447

This vector times this vector, one times one plus three times two is the seven.

448

So I can do it by rows, and in each row, each row times my x is what I'll later call a dot product.

449

But I also like to see it by columns.

450

I see this as a linear combination of the columns.

451

So here's my point.

452

A times x is a combination of the columns of A.

453

That's how I hope you'll think of A times x when we need it.

454

Right now we've got, with small ones we can always do it in different ways, but later think of it that way.

455

Okay.

456

So that's the picture for a two-by-two system.

457

And if the right-hand side b happened to be 12-7, then of course the correct solution would be one-two.

458

Okay.

459

So let me come back next time to a systematic way,

460

using elimination, to find the solution, if there is one, to any,

461

a system of any size, and find out because the system, because eliminate, if elimination fails,

462

find out when there isn't a solution.

463

OK, thanks.

464

Thank you.

📺 Mesmo canal

✨ Vídeo recomendado

91 frases

Linear combinations, span, and basis vectors | Chapter 2, Essence of linear algebra

Cambridge Exam PrepUniversity StudentsWorking Adults

301 frases

Lecture 01. Introduction to the Course - Philosophy of the Natural Sciences (Univ of New Orleans)

TOEFL PrepCambridge Exam PrepUniversity Students

244 frases

1. Introduction to statistics - Lecture 1

IELTS PrepTOEFL PrepUniversity Students

128 frases

Vectors and 2D Motion: Crash Course Physics #4

TOEFL PrepCambridge Exam PrepUniversity Students

24 frases

English vocabulary in use Advanced Unit 2

IELTS PrepTOEFL PrepCambridge Exam Prep

181 frases

The Limit (do not use L'Hospital rule)

Cambridge Exam PrepUniversity StudentsWorking Adults

506 frases

Analyzing Statements of Cash Flows 1 (2025 CFA® Level I Exam – FSA – Learning Module 4)

Cambridge Exam PrepBusiness EnglishUniversity Students

552 frases

The Professor Who Taught People How To Think (1962)

IELTS PrepTOEFL PrepUniversity Students

102 frases

Orbital Mechanics - Newton's Laws

TOEFL PrepCambridge Exam PrepUniversity Students

124 frases

bassham ch3pt1

IELTS PrepTOEFL PrepCambridge Exam Prep

311 frases

Eco 155: Principles of Macroeconomics Class 1

IELTS PrepTOEFL PrepCambridge Exam Prep

Vocabulário e dicas de fala para esta lição

Este vídeo tem 464 frases e 5054 palavras para praticar shadowing. A fala dura 39:39. O falante mantém um ritmo estável de cerca de 127 palavras por minuto, confortável para o shadowing. Apenas 80% das palavras estão entre as 3.000 mais comuns do inglês, por isso o vocabulário é exigente.

Vocabulário principal deste vídeo

15 palavras do vídeo que vale a pena aprender, com pronúncia e significado:

Palavra	Pronúncia	Significado
column substantivo	/ˈkɒləm/	coluna
equation substantivo	/ɪˈkweɪ.ʒən/	equação
minus	/ˈmaɪ.nəs/	menos
vector substantivo	/ˈvɛktɚ/	vector, vetor
linear adjetivo	/ˈlɪn.i.əɹ/	linear
solve verbo	/sɒlv/	resolver, solucionar
component substantivo	/kəmˈpoʊ.nənt/	componente
dimensional adjetivo	/daɪˈmɛn.ʃə.nəl/	dimensional
lecture substantivo	/ˈlɛk.t͡ʃə/	conferência, palestra
algebra substantivo	/ˈæl.d͡ʒɪ.bɹə/	álgebra
combine verbo	/kəmˈbaɪn/	combinar
dimension substantivo	/daɪˈmɛn.ʃən/	dimensão
satisfy verbo	/ˈsætɪsfaɪ/	satisfazer
pretend verbo	/pɹəˈtɛnd/	fingir
singular adjetivo	/ˈsɪŋ.ɡjə.lɚ/	único, simples

Phrasal verbs que você vai ouvir

Palavra	Pronúncia	Significado
fill out verbo	/fɪl ˈaʊt/	preencher
find out verbo		descobrir
come back verbo		voltar
end up verbo		terminar, acabar
go up verbo		subir
turn out verbo		terminar
write down verbo		anotar

Pronúncia para ficar de olho

O falante usa 91 contrações e formas reduzidas, como I'm, I'll, don't. Diga-as na forma curta, do jeito que você ouve.

Os sons de “th”: eighth /eɪtθ/, ninth /naɪnθ/
Os sons de “sh” e “zh”: equation /ɪˈkweɪ.ʒən/, combinations /kɑmbɪˈneɪʃənz/, dimensional /daɪˈmɛn.ʃə.nəl/, dimension /daɪˈmɛn.ʃən/, elimination /ɪlɪmɪˈneɪʃ(ə)n/
Palavras longas — acerte a sílaba tônica: combinations /kɑmbɪˈneɪʃənz/, dimensional /daɪˈmɛn.ʃə.nəl/, elimination /ɪlɪmɪˈneɪʃ(ə)n/, fundamental /ˌfʌn.dəˈmɛn.təl/, horizontal /ˌhoɹəˈzɑn.təl/

Como praticar com este vídeo

Ouça o vídeo inteiro uma vez sem falar e anote as palavras que você não conhece.
Faça shadowing frase por frase na velocidade normal, repetindo cada uma até o seu ritmo ficar igual ao do falante.
Grave a sua voz e compare com o original, prestando atenção a palavras como column, equation, minus.

Gramática neste vídeo

As estruturas que o falante mais usa, com as palavras exatas do vídeo:

Estrutura	No vídeo
Frases condicionais if + oração, will/would + verbo — uma condição e seu resultado	if there is, elimination will give
Present perfect have/has + particípio passado — uma ação passada que ainda importa agora	you've seen · I've done · I've put
Orações relativas who / which + oração — informação extra sobre uma pessoa ou coisa	where, which will · algebra, which is · point, which is
Voz passiva be + particípio passado — o foco está no que acontece, não em quem faz	was arranged · is called

O que é a Técnica de Shadowing?

Shadowing é uma técnica de aprendizado de idiomas com base científica, originalmente desenvolvida para o treinamento de intérpretes profissionais. O método é simples, mas poderoso: você ouve áudio em inglês nativo e repete imediatamente em voz alta — como uma sombra seguindo o falante com 1-2 segundos de atraso. Pesquisas mostram melhora significativa na precisão da pronúncia, entonação, ritmo, sons conectados, compreensão auditiva e fluência na fala.

Técnica de shadowing: leia o guia completo passo a passo →

Prática de Shadowing: Lec 1 | MIT 18.06 Linear Algebra, Spring 2005 - Aprenda a falar inglês com vídeo

11. Introduction to Machine Learning

Independence, Basis, and Dimension

Big Picture: Derivatives

Lec 1: Introduction to Principles of Microeconomics and Supply & Demand

Linear combinations, span, and basis vectors | Chapter 2, Essence of linear algebra

Lecture 01. Introduction to the Course - Philosophy of the Natural Sciences (Univ of New Orleans)

1. Introduction to statistics - Lecture 1

Vectors and 2D Motion: Crash Course Physics #4

English vocabulary in use Advanced Unit 2

The Limit (do not use L'Hospital rule)

Analyzing Statements of Cash Flows 1 (2025 CFA® Level I Exam – FSA – Learning Module 4)

The Professor Who Taught People How To Think (1962)

Orbital Mechanics - Newton's Laws

bassham ch3pt1

Eco 155: Principles of Macroeconomics Class 1

Vocabulário e dicas de fala para esta lição

Vocabulário principal deste vídeo

Phrasal verbs que você vai ouvir

Pronúncia para ficar de olho

Como praticar com este vídeo

Gramática neste vídeo

O que é a Técnica de Shadowing?