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$\mathbf{A}=\left[ \begin{array}{rr}{3} & {-2} \\ {0} & {3}\end{array}\right]$
All right, let's clear denominators by multiplying every term by a minus three. Cancel out there as will they here. All right, so a minus three times to we have to use distributive property. So that will be two a minus six on then plus the three and plus the A right there. All right. Combine like terms on the left. Negative six plus three is negative. Three. Let's attract to a from both sides a minus two A's negative a and then multiply both sides by negative one. We get that a is three. However, this is the first time this has occurred. When you look up here, these denominators if we make a three three minus three would be zero and we'd be dividing by zero, which we cannot dio. So there is no solution to this equation because they cannot be three
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I am going to solve the following equation. I have one over a equal to 1/3 minus 2/3. A first thing I need to look for is my lowest common denominator, and that is three. A. If I can multiply everything by three, Ai will eliminate all of my denominators. So three a on the right side and times three a on the left, so 38 times one over a just reduces to or simplifies to become three. But I have three a times 1/3 minus three a times to over three a, and we simplify the right side. Ah, the threes cancel out here and I'm left with a minus and then three A's both cancel out because they're in the numerator and denominator. So a minus two. But add to teach side and five is equal to a Now I'm going to take and I'm going to substitute five for both of the A's in my original equation. So on the left side I have 1/5 equals 1/3 minus 2/3 times five. Now I will work to simplify the right hand side and see if we come up with an answer of 1/5. So I've got 1/3 minus 2/15. I need to have common denominator denominators. Three goes into 15 5 times. So if I multiply this by five over five, then I will have 5/15 minus 2/15 and I will get 3/15 and then three goes into both the numerator and the denominator. So I'm going to divide them both by three. And I get 1/5. So yes, this is a valid, um, solution. The answer is five A equals five.
According to the potion we have given question is Gerry's a Q minus three. A scwill is equal to 40. A first awful. The rearrange This equation There is a Q minus. Three is well minus for P A is equal to real. And then we take common. In this equation we have a square minus three minus 40 is equal to zero. And then the expanded the question. We have these will minus it a plus fight a minus 40 physical real on with the common. In this equation we have a A minus eight. And in this side of the day, common fight we have a minus eight. The factors comes out in this equation. Dairies a A plus fight. A minus eight is equal to zero. We know that in this caution we used the principle off zeros product. They say that all the factories they will do zero there is a good 20 A plus five is equal to zero a minus. It is equal to zero. We find out the value of a there is a is equal to minus five is equal to eight and is equal to zero. We can say that the value of a comes out. Zero minus five and eight