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Unless otherwise specified, assume that all matrices in these exercises are $n \times n .$ Determine which of the matrices in Exercises $1-10$ are invertible. Use as few calculations as possible. Justify your answers.$$\left[\begin{array}{cccc}{1} & {3} & {7} & {4} \\ {0} & {5} & {9} & {6} \\ {0} & {0} & {2} & {8} \\ {0} & {0} & {0} & {10}\end{array}\right]$$

The $4 \times 4$ Matrix has four pivot positions so by the inverse matrix transform rule the matrix $\mathrm{A}$ is invertible.

Algebra

Chapter 2

Matrix Algebra

Section 3

Characterizations of Invertible Matrices

Introduction to Matrices

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Lectures

01:32

In mathematics, the absolu…

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Unless otherwise specified…

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For the following exercise…

we have four by poor matrix, which is say, Hey, is first row is 137 or second row is 0596 The third rule still was able to eight. The last room there was there was zero and Tech. Okay, So to find out a determinant again, we need to expend it along some row or some column. So in this case, I would recommend to do the first column because it has, um, three entries of zeros and this collector is one. So he's very easy to calculate, so I'll just do that. Expend a We pressed into the by now determined and Ave and expanded along the first column. So with the purse entry is one times since today this. Ah, first column and first row. So we have connected to power up to. But that's the next squared, naked squared and crossing. Now the first column Members row. So we won. We left is determined off three by three matrix 596 Bye 96 0 to 8 and 00 10. All right, so negative. One squared. It's one. So it's just a determinant of it's three by three matrix and 00 10. All right, so for it is three by three missions. We again, we need to expend it. Um, so right now, I would expect I would recommend to expand along the first column again because it has three as two zeros. You can also expend alone the third. The third row on it also has two zeroes here. So that's the same where I would just do this by expending along the first column, so that would be determined. Sorry. Um, there's this five and two connective one. Next, we once what next? One squared. So that's one. I'll just Ah, and we're not write down one here, so he's just a determined, uh, two by two matrix, which to AIDS. 00 10 280 10 minus. Because we have now the second term off the first column. So that is, uh, two plus one. So that's, uh, three. So that's 9 to 1 neck to onto the power of three. So he's We have ah, minus here, and we're crossing now. The second ruin. First column. So that is 96 Do you attend term end? Uh, 960 by the way, since we have zero as a defector. So we don't You don't need to care about that determined here. So should be zero. And the last term should also be zero because we have a zero point last year burning us entry on the first call him. So the only thing we need to care about is the first term, as five times determined, determined updates. Matrix two times 10 is 20 and eight times zero is zero. So that's 20. So that would be 100 which is a non zero. So a is non singular, which means a convertible singular, singular. All right, and we're not.

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