Physics 6A Solutions to Homework Set # Winter 202. Boas, problem 3.2 4. Use an augmented matrix to solve the following equation 2x + 3y z = 2 x + 2y z = 4 4x + 7y 3z = ( We can write this as a 3 x 4 matrix using the coefficient in front of each value of x,y,z as the necessary matrix entry 2 3 2 2 4 4 7 3 We now use our row-reduction methods, and the exact methods used to solve this may vary. To start off, I will subtract 2xRow from Row3 2 3 2 2 4 0 and now I will subtract 2xRow2 from Row 0 0 2 4 0 Then I will add Row3 to Row 0 0 0 2 4 0 Now my first row says that 0 =, we know that this isn t true, so the intersection of these three planes must not exist. 2. Boas, p. 0, problem 3.2 8. Find the rank of the matrix 2 2 2 3 2 4 8 6
We want to find the number of linearly independent rows in the matrix (which is equal to the number of linearly independent columns. First, we find the greatest common denominator of each row (to simplify the numbers we need to deal with later. This leads us to divide the last row by 2. 2 2 2 3 2 4 3 Now let s add row to row2, and subtract 2xrow from row3 and subtract row from row4. This yields 0 2 0 0 0 2 3 3 0 2 3 3 Now we note that the last two rows are the same - so we eliminate the last row (it is not linearly independent of a combination of row and row3. Then let s add row2 to row3. 0 2 0 0 0 0 3 3 0 0 0 0 Now each row has at least one column where it is non-zero, but every other row is 0. That means that the rows must be independent in that direction (i couldn t subtract a combination of other rows to get rid of this column. This means that there are three independent rows, and the rank of the matrix is 3. 3. Boas, problem 3.3 6. Find the determinant of 0 We start with row reduction operations to simplify one of the rows (there are many ways to do this, and there is a lot of row reduction to do. We start by subtracting row2 from, row 3 from row2 to obtain, row4 from row3, and row from row4: 0 0 0 0 0 0 0 0 0 2
this has isolated the last column, so we perform a Laplace development. The residing is in the 4+ position, so we add a negative sign to the calculation of the determinant: 0 0 0 0 now we can add row to row4, which yields: 0 0 0 0 0 2 and can do another Laplace development, with the - in the first column, noting that it is in the + position. This yields: 0 ( 2 0 2 we can now evaluate this 3x3 determinant directly. It is 4, which yields a total answer of 4. 4. Boas, problem 3.6 6. The Pauli spin matrices are defined as: ( ( 0 0 i A =, B = i 0 (, C = 0 There are many little computations here, which are tedious, but each is about one step - so I will try to illustratively show them: A 2 = ( 0 ( 0 = ( 0 0 + + 0 0 + 0 0 0 + = ( 0 The other two behave identically (noting that i * i = - in the computation of B. We will now calculate the commuter of AB (usually written [A, B], which is equal to AB - BA, This can be written as: [A, B] = ( 0 ( 0 i i 0 ( 0 i i 0 ( 0 = ( i ( i 0 0 i (+i ( = 2i 0 Note that this shows both that AB = -BA (because the term inside of the parenthesis stemming from the BA multiplication is the opposite of the AB term - since we subtracted those opposite terms, we got 2 * AB, which is equal to 2iC. All other derivations 3
(which should be shown in the homework are completely identical, and in the interest of saving paper, will not be shown here.. Boas problem 3.6 8. Show (by multiplying matrices that the following equation represents an ellipse: ( x y ( 7 7 3 ( x y = 30 Remember Matrix multiplication is associative, so we can start with either the first and second, or second and third terms. Choosing the latter, we get ( ( x 7y x y = 30 7x + 3y the second multiplication yields: which gives the equation x 2 7xy + 7xy + 3y 2 = 30 3 x2 + y 2 = 0 which is the equation for an ellipse. 6. Boas problem 3.6 2. Solve the following system of equations by finding the inverse of the coefficient matrix: x + 2z = 8 2x y = x + y + z = 4 (2 We can quickly convert this to a 3x3 matrix of coefficients which looks like: 2 2 now we can find the inverse of this matrix from it s cofactors, which gives us: C = 2 3 2 2 4 4
we then transpose this cofactor matrix, and also can find that the determinant of our original matrix is, so that gives us an inverse: 2 2 2 4 3 now we can multiply this matrix by the vector of answers to our linear equations above: 2 2 8 x 2 4 = y 3 4 z which yields: x = -0/ = -2, y = / =, z= 2/ =. 7. Boas problem 3.7 34. Determine whether the following matrix is a reflection or a rotation, and determine either the reflecting plane or the axis of rotation: 0 0 0 0 The determinant of this matrix is -, so it is a reflection, using the equation Lr = -r, we have 0 x x 0 0 y = y 0 z z which yields equations x=-x, -y = -z, -z=-. Thus x=0, and we can arbitrarily pick y=, which means z=. This gives us a vector (0,, - so we have an equation for the plane of reflection x=0, y+z=0, which gives y=-z. 7. Boas problem 3.8 8. Solve the following system of equations by row-reducing the matrix: 2x + 3z = 0 4x + 2y + z = 0 x y + 2z = 0 (3 we set up a 3x4 matrix for our coefficients:
2 0 3 0 4 2 0 2 0 now we can subtract 2xrow from row2, and then divide both rows by 2 to yield. 0 0 0. 0 2 0 then we can subtract row from row3, and then add row 2 to row3, to obtain:. 0 0 0. 0 0 0 0 0 which eliminates the third row - so we can write the resulting equations as x = - 3 2 z and y = 2z. 8. Boas problem 3.8 24. Find the values of λ for which we have non-trivial solutions, and for each λ solve the equations: (6 λx + 3y = 0 3x (2 + λy = 0 (4 we can place these coefficients in a matrix as usual ( 6 λ 3 0 3 2 λ 0 we can multiply the 2nd row by (6-λ/3 to obtain ( 6 λ 3 0 6 λ ( 2 λ(6 λ/3 0 subtracting the first line from the second yields: ( 6 λ 3 0 0 ( 2 λ(6 λ/3 3 0 which gives us an equation in λ of: which yields: ( 2 λ(6 λ 9 = 0 λ 2 4λ 2 = 0 6
(λ 7(λ + 3 = 0 so λ = {-3, 7}. Plugging the first into the first equation we get 9x + 3y = 0 which yields x = - y, and the second yields -x + 3y = 0 which gives x = 3y. 3 8. Boas problem 3.9 4. Find the transpose, inverse, complex conjugate, and transpose conjugate, and find AA for: 0 2i A = i 2 0 3 0 0 we get the transpose by flipping (x,y to (y,x which yields: 0 i 3 A T = 2i 2 0 0 the complex conjugate is given by flipping i to -i, and yields: 0 2i A = i 2 0 3 0 0 the transpose conjugate is given by changing i to -i in A T and yields 0 i 3 (A T = 2i 2 0 0 the inverse is given by taking the transpose of the cofactor matrix, so first we find the cofactor matrix as 0 0 6 C = 0 3 6i 2 i 2 we note the determinant of A is 6, and then we take the transpose of the co-factor matrix to obtain A = 0 2 0 3 i 6 6 6i 2 we then calculate AA as 0 2i i 2 0 6 3 0 0 so this all checks out. 0 0 2 0 3 i 6 6i 2 7 = 0 0 0 0
8. Boas problem 3.9 23. Show that the following matrices are hermitian, regardless of whether A is hermitian or not: AA, A + A, i(a - A We first note as identities that A = A, and that (AB = B A. Thus the Hermitian conjugate of the first yields: (AA = A A = AA which checks out, for the second, we note that matrices are commutative under addition, so (A + A = A + A = A + A = A + A finally, for the last we remember that (ca = c A, so we have [i(a A ] = ia ( ia = ia + ia = i(a A and thus this also holds. 8