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Let $A=\left[\begin{array}{ll}{1} & {2} \\ {1} & {3} \\ {1} & {5}\end{array}\right] .$ Construct a $2 \times 3$ matrix $C$ (by trial and error) using only $1,-1,$ and 0 as entries, such that $C A=I_{2}$ Compute $A C$ and note that $A C \neq I_{3} .$

$\neq I _ { 3 }$

Algebra

Chapter 2

Matrix Algebra

Section 2

The Inverse of a Matrix

Introduction to Matrices

Campbell University

Oregon State University

Harvey Mudd College

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in this example were provided with the Matrix A. And our goal here is to find a new matrix C such that when we take see times a we obtain that two by two identity matrix. First, let's analyze what sigh c must be the matrix A is of size three by two So we have a three by two matrix Here The result is a two by two matrix I to And so let's begin right down the dimensions for C. We have to have a three here only because that's the only way this and this are equal to allow for this matrix multiplication to be well defined. Now the two, as we see here in here give come from the two here and they will come from a two if we place it here. This allows the outer dimensions here in here to result in the two by two for I to So what we've shown so far is that C must be a two by three matrix. Now, this is a tricky example because what we're occurs to Dio is too right out to sea as a two by three matrix using only end elements one negative one and zero such that this equation above will hold. So we're encouraged to use just trial and error. And I found that if you take 11 and negative one negative 11 and zero, this should work. Let's verify this by actually multiplying. See times A so C is copying negative 11 11 and negative 10 for the columns A has calm 111 then 235 The result is of size to buy two. So now you are ready to begin multiplying for this entry here, take row one and multiply its corresponding entries from column one will have one times one plus one times one minus one times one so altogether. This results in a one. Then for this entry which will go here, take the same calm one. Multiply through by row. To this time we have one times one or negative one times one rather plus one times one plus zero times one. And this one results in a value of Let's see, here we get zero altogether. So far, we're in good shape. That's the first column of thy Danny matrix that we're trying to land on when multiplying. Now take the similar strategy to get this entry. We stay on this column and multiply. The corresponding entries in this row will have to plus three minus five, resulting in a zero. Now, our hope of hopes is that we get a one here, So let's check that out. This time we're taking these entries multiplying by the second column we see here and we obtain negative two plus three plus zero times five, which is zero, and this does result in a positive one. So this matrix c displayed here provides us with an answer for obtaining the two by two identity matrix. Notice that it almost seems as though the matrix A here is in vertebral because we have C times A equals the identity matrix. However, if we were to reverse the multiplication and consider a time see, then the Matrix A is of size three by two. Matrix C is of size to buy two and that means a product. It's not going to be an identity matrix necessarily, but it will be of size three by two, so it cannot possibly be an identity matrix. So despite the fact we did land on a convincing equation, A is not in vertebral

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