7 1 det 2a 3adja 23det a 3detadj a 23 2

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Unformatted text preview: 4[0, 1, 1] = [0, − 2, 2], that is, 2 2 . 7. (1). det( − 2A− 3adjA) = ( − 2)3(det A)− 3det(adj A) = − 23 2− 3 det(adj A) = − det(adj A) = − det((detA)A− 1) = − (det A)3(det A)− 1 = − 4. 8. (5). Recall that the least squares solution to A X = B can be proven to be a solution for 2 1 1 A⊤ A X = A⊤ B . Here, A⊤A = 3I , so we’re very lucky: X = 3 A⊤B = ( − 3 , 0, − 3 ). 1 0 −2 9. (6). Solving for X , X = A⊤A. To get A invert the given matrix, getting A = 0 1 0 . 01 1 Then...
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