Linear Algebra, 5th Edition

Section 3.2

1. 1.

   1. (a) F

   2. (b) F

   3. (c) T

   4. (d) T

   5. (e) F

   6. (f) T

   7. (g) T

   8. (h) T

   9. (i) T

2. 2.

   1. (a) 2

   2. (c) 2

   3. (e) 3

   4. (g) 1

3. 4. (a) $\left(\begin{array}{llll}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 0\end{array}\right);$ the rank is 2.

4. 5.

   1. (a) The rank is 2, and the inverse is $\left(\begin{array}{rr}-1& 2\\ 1& -1\end{array}\right)$.

   2. (c) The rank is 2, and so no inverse exists.

   3. (e) The rank is 3, and the inverse is $\left(\begin{array}{rrr}{\displaystyle \frac{1}{6}}& -{\displaystyle \frac{1}{3}}& {\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{2}}& 0& -{\displaystyle \frac{1}{2}}\\ -{\displaystyle \frac{1}{6}}& {\displaystyle \frac{1}{3}}& {\displaystyle \frac{1}{2}}\end{array}\right)$.

   4. (g) The rank is 4, and the inverse is $\left(\begin{array}{rrrr}-51& 15& 7& 12\\ 31& -9& -4& -7\\ -10& 3& 1& 2\\ -3& 1& 1& 1\end{array}\right)$.

5. 6.

   1. (a) ${\text{T}}^{-1}(a{x}^2+bx+c)=-a{x}^2-(4a+b)x-(10a+2b+c)$

   2. (c) ${\text{T}}^{-1}(a,b,c)=\left({\displaystyle \frac{1}{6}}a-{\displaystyle \frac{1}{3}}b+{\displaystyle \frac{1}{2}}c,{\displaystyle \frac{1}{2}}a-{\displaystyle \frac{1}{2}}c,-{\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{3}}b+{\displaystyle \frac{1}{2}}c\right)$

   3. (e) ${\text{T}}^{-1}(a,b,c)=\left({\displaystyle \frac{1}{2}}a-b+{\displaystyle \frac{1}{2}}c\right)x^2+\left(-{\displaystyle \frac{1}{2}}a+{\displaystyle \frac{1}{2}}c\right)x+b$

6. 7. $\left(\begin{array}{rrr}1& 0& 0\\ 0& 1& 0\\ 1& 0& 1\end{array}\right)\left(\begin{array}{rrr}1& 0& 0\\ 1& 1& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{rrr}1& 0& 0\\ 0& -2& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{rrr}1& 2& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{rrr}1& 0& 0\\ 0& 1& 0\\ 0& -1& 1\end{array}\right)\left(\begin{array}{rrr}1& 0& 1\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$

7. 20.

   1. (a) $\left(\begin{array}{rrrrr}1& 3& 0& 0& 0\\ -2& 1& 0& 0& 0\\ 1& 0& 0& 0& 0\\ 0& -2& 0& 0& 0\\ 0& 1& 0& 0& 0\end{array}\right)$