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Consider the (3 X 3) matrix A with rows defined as Row 1: [k,-2,-6] Row 2: [7,1,-2] Row 3: [-2,0,-1]If det(A) = 1,then k =...

Question

Consider the (3 X 3) matrix A with rows defined as Row 1: [k,-2,-6] Row 2: [7,1,-2] Row 3: [-2,0,-1]If det(A) = 1,then k =

Consider the (3 X 3) matrix A with rows defined as Row 1: [k,-2,-6] Row 2: [7,1,-2] Row 3: [-2,0,-1] If det(A) = 1,then k =



Answers

Given the matrix $\mathbf{K}$ $$\mathbf{K}=\left[\begin{array}{rr|r}5 & 2 & 1 \\1 & -4 & 3\end{array}\right]$$ write the matrix obtained by interchanging rows 1 and 2

Okay. Yeah. All right. We're gonna find the determinant of matrix A. By expanding this first column. So we're determined in the bay it's going to include three. It's going to include two. It's going to include one but we want to then look at their co factors. So the co factors are going to include a negative one for each of these but they're expert is going to be different. Three is in the first row, first column. So one plus one is two for the X men. Two is in the second row, first column. That's going to be three as the X moment. Then one is gonna have an expert before from there we want they're minor which is the two by two matrix of items that are not in the same row or column. So three is in the top row, it's in the first column. So 2 to 31 is going to be the two by two matrix there. Here we get 2331 as our matrix. and then one is going to give us 2322 sort of determinant of a is going to equal three times one times two minus six plus two times negative, one times two minus nine plus one times one times four minus six. So the determinant for this major because it's gonna be three times negative, four minus two times negative seven plus negative two. So we end up with the determinant Equalling zero, which means the inverse does not exist.

All right. They want us to find the determine of a. By expanding on this first column. So to do that we take the three things that are in the first column. We have one, zero and zero. Yeah. And now we need their CO factors in this case we need negative one and then the row plus the column that it's in one is in the first row, first column. So we have an expert to we want things that are not in the same row or column. That's these four things. Two negative three negative 13 And now the good part about picking column one is that I'm gonna get zero for the next two terms. We're gonna practice the setup. We have four negative seven negative 13 for this second zero. And then I heard those zero in the second row, zero in the third row. We get that as our two by two matrix. But again we're taking zero times the co factor zero times the co factor we get zero and zero. And that means our first one is going to give us all the good. So two times three is six minus negative, one times negative three, which is positive. Three are determinant, is three. Second part of this question. Ask us is it in vertebral while three does not equal zero, So it is in fact convertible and you're done.

Here. We want to find a determine if it exists. Because it's a square matrix are determined, does exist. So we want the first diagonal minus sec title. So we multiply three over to buy negative to over three, three over to buy negative to over three in this equals negative one castle. It three cancel the to Then we have negative one minus the second angle. So one times negative. What? That's negative. So here we get negative one plus one. Let me The determinant is zero.

Uh huh. We have to use the determinant to determine the convertibility of matrix A given by roads and every 3010 to 506 and 0303 As was shown here on the right to answer this question, we need to know first how to take the determinant and secondly what the relationship between determinants and convertibility is for matrices. So on the left here, the determinant with three by three matrix with rows A B C D E F G H I is eight times the I minus F H minus B. 10 D. I'm in sf G plus C 10 D. H minus E G. And a matrix A is inevitable if and only if is determining A does not equal zero. So to solve this problem, we're going to use the formula on the left to find the determinant of a. Then we're going to check whether or not the determinant is equal to zero to determine the convertibility of A. So first the determining A is simply negative 3 10 0 minutes, zero minus 0, 10 15 minutes, 48 +10 minutes zero, which sums up to zero. Since our determining A is zero, this matrix is not convertible A is not inevitable.


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