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Exercises $17-20$ refer to the matrices $A$ and $B$ below. Make appropriate calculations that justify your answers and mention an appropriate theorem.$$A=\left[\begin{array}{rrrr}{1} & {3} & {0} & {3} \\ {-1} & {-1} & {-1} & {1} \\ {0} & {-4} & {2} & {-8} \\ {2} & {0} & {3} & {-1}\end{array}\right] \quad B=\left[\begin{array}{rrrr}{1} & {3} & {-2} & {2} \\ {0} & {1} & {1} & {-5} \\ {1} & {2} & {-3} & {7} \\ {-2} & {-8} & {2} & {-1}\end{array}\right]$$Do the columns of $B$ span $\mathbb{R}^{4} ?$ Does the equation $B \mathbf{x}=\mathbf{y}$ have a solution for each $\mathbf{y}$ in $\mathbb{R}^{4} ?$

only three rows contain pivot positionsColumns of $\mathrm{B}$ do not $\operatorname{span} \mathbb{R}^{4}$the equation $\mathrm{Bx}=y$ does not have a solution for each $\mathrm{y}$ in $\mathbb{R}^{4}$Check theorem 4

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 4

The Matrix Equation Ax D b

Introduction to Matrices

Campbell University

McMaster University

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Lectures

01:32

In mathematics, the absolu…

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02:09

Exercises $17-20$ refer to…

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In Exercises $17-26,$let

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In Exercises $17-22,$ use …

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For the following exercise…

in this video, we're start out with the Matrix B that's provided here as a four by four matrix. And our goal is to determine first where the pivot positions would be. To that end, let's begin row reduction on this matrix. We have our first pivot here and we need to eliminate the entries one and negative to found below. So first, let's copy down the 1st 2 rows, they're 13 negative two and two, then for row to it. 01 one and negative five Now to eliminate this century multiply row one by negative one added to row three will obtain zero negative one negative one and a five. Next. To eliminate this entry, we have to this time multiply row one by two and add it to row. For this time, we obtained zero negative Too negative too. And three. So our new pivot position to consider is here, and we're going for just echelon form so we can analyze the pivots. That means we need Teoh. Just eliminate thes two entries below, but none above. So let me once again start by copying the 1st 2 rows which are 13 negative, two and two Then row two is 01 one and negative five. Now to eliminate this take row to and added to row three, the result is going to be zero 00 and zero. So we have a zero doubt row. Then, to eliminate this entry here, our role operation will be to multiply row two by two and added to row for we'll obtain 00 zero and then finally we have negative 10 plus three for a negative seven in our last row operation. Let's interchange growths three and four. So I'm gonna first start by copying 13 negative two and two. Then the next non zero is 011 negative five. And here's where we make our interchange. The next non zero is 000 negative seven and place the zeroed out to row down below. Now we're in echelon form and we find that there is a pivot here, here and here. Now, with the results that we have obtained above, we know that be is of size four by four. I'm going to make the number of rows special by highlighting as blue and there are we found three pivots. So there is not for pivots to correspond, to pivot in every row. This gives us to immediate conclusions. We can say that the columns of be do not span are four and the reason being would be since we require four pivots in order span a four dimensional space and we're short on pivots. So the columns do not spend the target space are four. And this also means that if we pick in a equation to consider be X equals y then this system is not consistent for all vectors. Why in are for again in order for this equation to always be consistent we required four pivots, but we're short on pivots and this is a big theme in linear algebra that were studying at this level, where really knowing where the pivots are can tell us a lot of fundamental information about the Matrix as well as matrix equations like these

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In mathematics, the absolute value or modulus |x| of a real number x is its …

Exercises $17-20$ refer to the matrices $A$ and $B$ below. Make appropriate …

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