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Linear Algebra, 2020a

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10 Chapter One. <strong>Linear</strong> Systems<br />

̌ 1.19 Use Gauss’s Method to solve each system or conclude ‘many solutions’ or ‘no<br />

solutions’.<br />

(a) 2x + 2y = 5<br />

x − 4y = 0<br />

(b) −x + y = 1<br />

x + y = 2<br />

(c) x − 3y + z = 1<br />

x + y + 2z = 14<br />

(d) −x − y = 1<br />

−3x − 3y = 2<br />

(e) 4y + z = 20<br />

2x − 2y + z = 0<br />

x + z = 5<br />

x + y − z = 10<br />

(f) 2x + z + w = 5<br />

y − w =−1<br />

3x − z − w = 0<br />

4x + y + 2z + w = 9<br />

1.20 Solve each system or conclude ‘many solutions’ or ‘no solutions’. Use Gauss’s<br />

Method.<br />

(a) x + y + z = 5 (b) 3x + z = 7 (c) x + 3y + z = 0<br />

x − y = 0 x − y + 3z = 4 −x − y = 2<br />

y + 2z = 7 x + 2y − 5z =−1 −x + y + 2z = 8<br />

̌ 1.21 We can solve linear systems by methods other than Gauss’s. One often taught<br />

in high school is to solve one of the equations for a variable, then substitute the<br />

resulting expression into other equations. Then we repeat that step until there<br />

is an equation with only one variable. From that we get the first number in the<br />

solution and then we get the rest with back-substitution. This method takes longer<br />

than Gauss’s Method, since it involves more arithmetic operations, and is also<br />

more likely to lead to errors. To illustrate how it can lead to wrong conclusions,<br />

we will use the system<br />

x + 3y = 1<br />

2x + y =−3<br />

2x + 2y = 0<br />

from Example 1.13.<br />

(a) Solve the first equation for x and substitute that expression into the second<br />

equation. Find the resulting y.<br />

(b) Again solve the first equation for x, but this time substitute that expression<br />

into the third equation. Find this y.<br />

What extra step must a user of this method take to avoid erroneously concluding a<br />

system has a solution?<br />

̌ 1.22 For which values of k are there no solutions, many solutions, or a unique<br />

solution to this system?<br />

x − y = 1<br />

3x − 3y = k<br />

1.23 This system is not linear in that it says sin α instead of α<br />

2 sin α − cos β + 3 tan γ = 3<br />

4 sin α + 2 cos β − 2 tan γ = 10<br />

6 sin α − 3 cos β + tan γ = 9<br />

and yet we can apply Gauss’s Method. Do so. Does the system have a solution?<br />

̌ 1.24 What conditions must the constants, the b’s, satisfy so that each of these<br />

systems has a solution? Hint. Apply Gauss’s Method and see what happens to the<br />

right side.

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