Question Video: Finding the Largest Scalar Possible to Multiply a Matrix Given a Restriction Mathematics

Given the matrix 𝐴 = [1, βˆ’16, 5, and 2, βˆ’4, βˆ’3, and βˆ’1, 7, 4], what is the greatest number π‘˜ for which no entry of π‘˜π΄ is greater than 1?

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Video Transcript

Given the matrix 𝐴 is equal to one, negative 16, five, two, negative four, negative three, negative one, seven, four, what is the greatest number π‘˜ for which no entry of π‘˜ times 𝐴 is greater than one?

In this question, we’re given a matrix 𝐴. And we need to determine the greatest number π‘˜ for which no entry of π‘˜ multiplied by 𝐴 is greater than one. There’s a lot to this question, so we’ll break it down. First, we need to know what we mean by π‘˜ multiplied by 𝐴. Remember, π‘˜ is a number and 𝐴 is a matrix, so π‘˜ multiplied by 𝐴 is a scalar multiplied by a matrix. This is scalar multiplication of a matrix. So we need to recall how we do scalar multiplication of a matrix. This means we need to multiply every single entry of our matrix 𝐴 by our scalar π‘˜. So let’s start by finding the expression for π‘˜ multiplied by 𝐴. We need to multiply every entry of 𝐴 by π‘˜.

In matrix 𝐴, in row one column one, the entry is one. So in our matrix π‘˜ multiplied by 𝐴, we need to multiply this value of one by π‘˜. And in row one column two of matrix 𝐴, we have the value of negative 16. So in row one column two of our matrix π‘˜ times 𝐴, we need to multiply this value of negative 16 by π‘˜. And we can carry on this process for the rest of the entries of our matrix. We get the following three-by-three matrix which represents π‘˜ multiplied by 𝐴. All we did is multiply every single entry inside of 𝐴 by our scalar π‘˜.

And of course we can simplify this. All we’ll do is simplify every single entry inside of our matrix. We get the following three-by-three matrix. Now, remember, the question is asking us to find the greatest number π‘˜ such that no entry of this matrix is greater than one. In other words, we need to find the largest value of π‘˜ such that no entry inside of this three-by-three matrix is bigger than one. This means by looking at each entry individually, we could form a system of inequalities. The entry in row one and column one of our matrix is π‘˜. And remember, the question wants us to have no entry of π‘˜ times 𝐴 to be bigger than one. So this entry π‘˜ must be less than or equal to one.

We can form a similar inequality by looking at the entry in row one column two. This entry is negative 16π‘˜. And since it’s not allowed to be bigger than one, it must be less than or equal to one. And by using exactly the same method, we can find seven more inequalities by looking at the other seven entries of our matrix. This gives us the following nine inequalities. All of these inequalities must be true to make no entry of π‘˜ times 𝐴 bigger than one. Solving a system of inequalities is very similar to solving a system of equations.

However, we do need to be careful. Remember, if we divide or multiply both sides of an inequality by a negative number, we need to switch the sign of our inequality. And this can make it quite tricky dealing with inequalities. There’s a few different ways we could solve this. We’ll only go through one of these. Remember, the question wants us to find the greatest number π‘˜ such that this is true. So we’ll start by assuming that π‘˜ is positive. If we find a positive solution for π‘˜, we don’t even need to check for negative solutions of π‘˜ because they’ll be smaller. And we’re any interested in the biggest one.

And this is actually very useful because a negative number multiplied by a positive number is negative. So it’s always going to be less than or equal to one. So all of our inequalities that have a negative coefficient of π‘˜ must be true. Negative 16π‘˜ is negative, negative four π‘˜ is negative, negative three π‘˜ is negative, and negative π‘˜ is negative. So they’re all less than or equal to one. So if π‘˜ is positive, these four inequalities are already true. We just need to make the remaining five inequalities true.

There’s several different ways to do this. The easiest way is to rearrange each inequality for π‘˜. For example, our first inequality is already in terms of π‘˜. We have π‘˜ is less than or equal to one. We can change our third inequality by dividing both sides of the inequality through by five. Doing this, we get that π‘˜ must be less than or equal to one-fifth. And we can do exactly the same thing for our other three positive coefficient inequalities. We just divide through by the coefficient. We get π‘˜ is less than or equal to one-half, π‘˜ is less than or equal to one-seventh, and π‘˜ is less than or equal to one-quarter. And don’t forget our first inequality, π‘˜ must be less than or equal to one.

We need all of these inequalities to be true for our value of π‘˜. But all this is telling us is π‘˜ needs to be positive and π‘˜ needs to be less than or equal to some list of values. For π‘˜ to be less than or equal to all of these values, all we need to do is make π‘˜ less than or equal to the smallest of these values. Of course, the smallest of these positive numbers is one over seven since we’re dividing by the biggest positive number. Therefore, if π‘˜ is less than or equal to one over seven, then we can conclude π‘˜ is less than or equal to one. It’s also less than or equal to one-fifth and one-half and one-quarter because one over seven is smaller than all of these.

But in fact, this is enough to answer our question. We know if π‘˜ is positive, then π‘˜ must also be less than or equal to one-seventh. We want to know the biggest of these values. That will be when π‘˜ is equal to one-seventh. Therefore, we found our answer will be one-seventh. It is worth pointing out again though these are not the only values of π‘˜ which will solve our equation since we assume that π‘˜ will be positive. There are also nonpositive values of π‘˜ which also have this property. For example, π‘˜ is equal to zero also has this property, but we only wanted the largest value of π‘˜ and all of these will be smaller.

Therefore, given the three-by-three matrix 𝐴, we were able to show the greatest number π‘˜ for which no entry of π‘˜π΄ is greater than one was given by one-seventh.

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