Homework 4

1. Compute the inverse of the following matrices.

(a) [423−1]\begin{bmatrix} 4 & 2 \\ 3 & -1 \\\end{bmatrix}

Using the fact that [423−1]\begin{bmatrix} 4 & 2 \\ 3 & -1 \\\end{bmatrix} is a 2×22\times2 matrix such that:

det⁡([423−1])=4(−1)−3(2)=−10≠0\det\( \begin{bmatrix} 4 & 2 \\ 3 & -1 \end{bmatrix} \) = 4(-1) - 3(2) = -10 \neq 0

Then:
[423−1]−1=−110[−1−2−34]=[11015310−25]\begin{bmatrix} 4 & 2 \\ 3 & -1 \end{bmatrix}^{-1} = -\frac{1}{10} \begin{bmatrix} -1 & -2 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} \frac{1}{10} & \frac{1}{5} \\[.5em] \frac{3}{10} & -\frac{2}{5} \end{bmatrix}

(b) [1230055002]\begin{bmatrix} 1 & 2 & 30 \\ 0 & 5 & 5 \\ 0 & 0 & 2 \end{bmatrix}

[1230100055010002001]→12R315R2[123010001101500010012]→R1−30R3R2−R3[12010−15010015−120010012]→R1−2R2[1001−25−14010015−120010012]∴[1230055002]−1=[1−25−14015−120012]=110[10−4−14002−5005]\begin{array}{c} \left[ \begin{array}{ccc|c} 1 & 2 & 30 & 1 & 0 & 0 \\ 0 & 5 & 5 & 0 & 1 & 0 \\ 0 & 0 & 2 & 0 & 0 & 1 \end{array} \right] &\xrightarrow[\frac{1}{2}R_3]{\frac{1}{5}R_2} &\left[ \begin{array}{ccc|c} 1 & 2 & 30 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & \frac{1}{5} & 0 \\ 0 & 0 & 1 & 0 & 0 & \frac{1}{2} \end{array} \right] \\ &\xrightarrow[R_1 - 30R_3]{R_2 - R_3} &\left[ \begin{array}{ccc|c} 1 & 2 & 0 & 1 & 0 & -15 \\ 0 & 1 & 0 & 0 & \frac{1}{5} & -\frac{1}{2} \\ 0 & 0 & 1 & 0 & 0 & \frac{1}{2} \end{array} \right] \\ &\xrightarrow{R_1 - 2R_2} &\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & -\frac{2}{5} & -14 \\ 0 & 1 & 0 & 0 & \frac{1}{5} & -\frac{1}{2} \\ 0 & 0 & 1 & 0 & 0 & \frac{1}{2} \end{array} \right] \end{array} \\[2em] \therefore \begin{bmatrix} 1 & 2 & 30 \\ 0 & 5 & 5 \\ 0 & 0 & 2 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -\frac{2}{5} & -14 \\ 0 & \frac{1}{5} & -\frac{1}{2} \\ 0 & 0 & \frac{1}{2} \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 10 & -4 & -140 \\ 0 & 2 & -5 \\ 0 & 0 & 5 \end{bmatrix}

(c) [133316−122]\begin{bmatrix} 1 & 3 & 3 \\ 3 & 1 & 6 \\ -1 & 2 & 2 \end{bmatrix}

[133100316010−122001]→R3+R1R2+3R3[1331000712013055101]→15R3R2−R3[133100027−11201115015]→R2−R3[133100016−6519501115015]→R3−R2[133100016−6519500−575−1−85]→−15R3[133100016−65195001−72515825]→R2−6R3R1−3R3[1304625−35−24250101225−15−325001−72515825]→R1−3R2[100250−350101225−15−325001−72515825]∴[133316−122]−1=[250−351225−15−325−72515825]=125[100−1512−5−3−758]\def\arraystretch{1.2em} \begin{array}{c} \left[ \begin{array}{ccc|ccc} 1 & 3 & 3 & 1 & 0 & 0 \\ 3 & 1 & 6 & 0 & 1 & 0 \\ -1 & 2 & 2 & 0 & 0 & 1 \end{array} \right] &\xrightarrow[R_3 + R_1]{R_2 + 3R_3} &\left[ \begin{array}{ccc|ccc} 1 & 3 & 3 & 1 & 0 & 0 \\ 0 & 7 & 12 & 0 & 1 & 3 \\ 0 & 5 & 5 & 1 & 0 & 1 \end{array} \right] \\ &\xrightarrow[\frac{1}{5}R_3]{R_2 - R_3} &\left[ \begin{array}{ccc|ccc} 1 & 3 & 3 & 1 & 0 & 0 \\ 0 & 2 & 7 & -1 & 1 & 2 \\ 0 & 1 & 1 & \frac{1}{5} & 0 & \frac{1}{5} \end{array} \right] \\ &\xrightarrow{R_2 - R_3} &\left[ \begin{array}{ccc|ccc} 1 & 3 & 3 & 1 & 0 & 0 \\ 0 & 1 & 6 & -\frac{6}{5} & 1 & \frac{9}{5} \\ 0 & 1 & 1 & \frac{1}{5} & 0 & \frac{1}{5} \end{array} \right] \\ &\xrightarrow{R_3 - R_2} &\left[ \begin{array}{ccc|ccc} 1 & 3 & 3 & 1 & 0 & 0 \\ 0 & 1 & 6 & -\frac{6}{5} & 1 & \frac{9}{5} \\ 0 & 0 & -5 & \frac{7}{5} & -1 & -\frac{8}{5} \end{array} \right] \\ &\xrightarrow{-\frac{1}{5}R_3} &\left[ \begin{array}{ccc|ccc} 1 & 3 & 3 & 1 & 0 & 0 \\ 0 & 1 & 6 & -\frac{6}{5} & 1 & \frac{9}{5} \\ 0 & 0 & 1 & -\frac{7}{25} & \frac{1}{5} & \frac{8}{25} \end{array} \right] \\ &\xrightarrow[R_2 - 6R_3]{R_1 - 3R_3} &\left[ \begin{array}{ccc|ccc} 1 & 3 & 0 & \frac{46}{25} & -\frac{3}{5} & -\frac{24}{25} \\ 0 & 1 & 0 & \frac{12}{25} & -\frac{1}{5} & -\frac{3}{25} \\ 0 & 0 & 1 & -\frac{7}{25} & \frac{1}{5} & \frac{8}{25} \end{array} \right] \\ &\xrightarrow{R_1 - 3R_2} &\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{2}{5} & 0 & -\frac{3}{5} \\ 0 & 1 & 0 & \frac{12}{25} & -\frac{1}{5} & -\frac{3}{25} \\ 0 & 0 & 1 & -\frac{7}{25} & \frac{1}{5} & \frac{8}{25} \end{array} \right] \end{array} \\[2em] \therefore \begin{bmatrix} 1 & 3 & 3 \\ 3 & 1 & 6 \\ -1 & 2 & 2 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{2}{5} & 0 & -\frac{3}{5} \\ \frac{12}{25} & -\frac{1}{5} & -\frac{3}{25} \\ -\frac{7}{25} & \frac{1}{5} & \frac{8}{25} \end{bmatrix} = \frac{1}{25} \begin{bmatrix} 10 & 0 & -15 \\ 12 & -5 & -3 \\ -7 & 5 & 8 \end{bmatrix}

(d) [1333101100110001]\begin{bmatrix} 1 & 3 & 3 & 3 \\ 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}

[13331000101101000011001000010001]→R3−R4R2−R3[13331000100001−100010001−100010001]→R1−3R3[130310−33100001−100010001−100010001]→R1−3R4[130010−30100001−100010001−100010001]→R1−R2[03001−1−20100001−100010001−100010001]→13R2R1↔R2[100001−10010013−13−2300010001−100010001]∴[1333101100110001]−1=[01−1013−13−230001−10001]=13[03−301−1−20003−30003]\begin{array}{c} \left[ \begin{array}{cccc|cccc} 1 & 3 & 3 & 3 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \end{array} \right] &\xrightarrow[R_3 - R_4]{R_2 - R_3} &\left[ \begin{array}{cccc|cccc} 1 & 3 & 3 & 3 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \end{array} \right] \\ &\xrightarrow{R_1 - 3R_3} &\left[ \begin{array}{cccc|cccc} 1 & 3 & 0 & 3 & 1 & 0 & -3 & 3 \\ 1 & 0 & 0 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \end{array} \right] \\ &\xrightarrow{R_1 - 3R_4} &\left[ \begin{array}{cccc|cccc} 1 & 3 & 0 & 0 & 1 & 0 & -3 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \end{array} \right] \\ &\xrightarrow{R_1 - R_2} &\left[ \begin{array}{cccc|cccc} 0 & 3 & 0 & 0 & 1 & -1 & -2 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \end{array} \right] \\ &\xrightarrow[\frac{1}{3}R_2]{R_1 \leftrightarrow R_2} &\left[ \begin{array}{cccc|cccc} 1 & 0 & 0 & 0 & 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & 0 & \frac{1}{3} & -\frac{1}{3} & -\frac{2}{3} & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \end{array} \right] \end{array} \\[2em] \therefore \begin{bmatrix} 1 & 3 & 3 & 3 \\ 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 0 & 1 & -1 & 0 \\ \frac{1}{3} & -\frac{1}{3} & -\frac{2}{3} & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 0 & 3 & -3 & 0 \\ 1 & -1 & -2 & 0 \\ 0 & 0 & 3 & -3 \\ 0 & 0 & 0 & 3 \end{bmatrix}

(e) [cos⁡θ−sin⁡θsin⁡θcos⁡θ]\begin{bmatrix}\cos \theta & -\sin \theta \\\sin \theta & \cos \theta \end{bmatrix}

[cos⁡θ−sin⁡θsin⁡θcos⁡θ]\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}

Using the fact that [cos⁡θ−sin⁡θsin⁡θcos⁡θ]\begin{bmatrix}\cos \theta & -\sin \theta \\\sin \theta & \cos \theta \end{bmatrix} is a 2×22\times2 matrix such that:

det⁡([cos⁡θ−sin⁡θsin⁡θcos⁡θ])=cos⁡2θ−(−sin⁡θsin⁡θ)=cos⁡2θ+sin⁡2θ=1≠0\det\( \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \) = \cos^2\theta - (-\sin\theta\sin\theta) = \cos^2\theta + \sin^2\theta = 1 \neq 0

Then:
[cos⁡θ−sin⁡θsin⁡θcos⁡θ]−1=11[cos⁡θsin⁡θ−sin⁡θcos⁡θ]=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}^{-1} = \frac{1}{1} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}

2. Let T:R3→R3T:\R^3\to\R^3 be a linear transformation that maps the standard vector e1\mathbf{e}_1, e2\mathbf{e}_2 and e3\mathbf{e}_3 to [110]\begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}, [100]\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix} and [111]\begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}. Is TT invertible? Explain.

T:R3→R3T(x⃗)=[111101001][x1x2x3]T: \R^3 \to \R^3 \\ T(\vec{x}) = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}

Let A=[111101001]A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}. Then, T−1T^{-1} exists if A−1A^{-1} exists.

[111101001]→R2−R3R1−R2[010100001]→R1↔R2[100010001]\begin{array}{c} \left[ \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{array} \right] &\xrightarrow[R_2 - R_3]{R_1 - R_2} &\left[ \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right] &\xrightarrow{R_1 \leftrightarrow R_2} &\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \end{array}

AA has a full rank. Therefore, TT is invertible.

3.

(a) Find invertible matrix AA, BB such that A+BA + B is not invertible.

An n×nn\times n matrix MM is invertible if and only if rank⁡(M)=n\operatorname{rank}(M) = n.

Let AA and BB be 1×11\times1 matrices.

A=[1]  ⟹  rank⁡(A)=1  ⟹  A−1=[1]B=[−1]  ⟹  rank⁡(B)=1  ⟹  B−1=[−1]\begin{array}{c} A = \begin{bmatrix} 1 \end{bmatrix} &\implies &\operatorname{rank}(A) = 1 &\implies &A^{-1} = \begin{bmatrix} 1 \end{bmatrix} \\ B = \begin{bmatrix} -1 \end{bmatrix} &\implies &\operatorname{rank}(B) = 1 &\implies &B^{-1} = \begin{bmatrix} -1 \end{bmatrix} \end{array}

Hence, AA and BB are invertible.

A+B=[1]+[−1]=[0]  ⟹  rank⁡(A+B)=0≠1\begin{array}{c} A + B = \begin{bmatrix} 1 \end{bmatrix} + \begin{bmatrix} -1 \end{bmatrix} = \begin{bmatrix} 0 \end{bmatrix} &\implies &\operatorname{rank}(A+B) = 0 \neq 1 \end{array}

Hence, A+BA+B is not invertible.

(b) Find non-invertible matrix AA, BB such that A+BA + B is invertible.

Let AA and BB be 2×22\times2 matrices.

A=[1000]  ⟹  rank⁡(A)=1≠2B=[0001]  ⟹  rank⁡(B)=1≠2\begin{array}{c} A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} &\implies &\operatorname{rank}(A) = 1 \neq 2 \\ B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} &\implies &\operatorname{rank}(B) = 1 \neq 2 \end{array}

Hence, AA and BB are not invertible.

A+B=[1000]+[0001]=[1001]  ⟹  rank⁡(A+B)=2  ⟹  (A+B)−1=[1001]\\ \begin{array}{ccl} A + B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} &\implies &\operatorname{rank}(A+B) = 2 \\ &\implies &(A+B)^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{array}

Hence, A+BA+B is invertible.

4. For which values of the constant aa, bb is the following matrix not invertible? [abbaabaaa]\begin{bmatrix} a & b & b \\ a & a & b \\ a & a & a \end{bmatrix} (Hint: find its row echelon form)

[abbaabaaa]→R3−R2R2−R1[abb0a−b000a−b]\begin{array}{c} \begin{bmatrix} a & b & b \\ a & a & b \\ a & a & a \end{bmatrix} &\xrightarrow[R_3 - R_2]{R_2 - R_1} &\begin{bmatrix} a & b & b \\ 0 & a-b & 0 \\ 0 & 0 & a-b \end{bmatrix} \end{array}

The matrix is not invertible if a=ba = b or a=0a=0.

5. Determine if the following statement true or false.

(a) If a matrix AA has a completely zero row, then it is not invertible.

True. If an n×nn\times n matrix AA contains a zero row, then rank⁡(A)<n\operatorname{rank}(A) < n.

(b) Upper triangular matrices are always invertible.

False. If there is a zero on the major diagonal, then they will not have a full rank.

(c) If AA is invertible, then Ax=0Ax = 0 may have non-trivial solution.

False. rref⁡(A)\operatorname{rref}(A) would be the identity matrix and therefore Ax⃗=0⃗  ⟺  x⃗=0⃗A\vec{x} = \vec{0} \iff \vec{x} = \vec{0}.

(d) If ABAB is invertible, then AA is invertible.

False. Let AA be an m×nm\times n matrix and BB be an n×mn\times m for nonzero m≠nm\neq n.

AA and BB are not invertible because they are not square matrices. ABAB is an m×mm\times m square matrix and could be invertible if rank⁡(AB)=n\operatorname{rank}(AB) = n.

For example: let A=[11]A = \begin{bmatrix}1 & 1\end{bmatrix} and B=[11]B =\begin{bmatrix}1 \\ 1\end{bmatrix}. Then, AB=[2]  ⟹  (AB)−1=[12]AB = \begin{bmatrix}2\end{bmatrix} \implies (AB)^{-1} = \begin{bmatrix}\displaystyle\frac{1}{2}\end{bmatrix}.

6. For the matrices AA, BB are invertible, Is the following true? If it is true, verify it. If it is false, give an example to explain why it is false.

(i) (A2)−1=(A−1)2(A^2)^{-1} = (A^{-1})^2

True.

(A2)−1=(AA)−1=A−1A−1=(A−1)2\begin{align*} (A^2)^{-1} &= (AA)^{-1} \\ &= A^{-1} A^{-1} \\ &= (A^{-1})^2 \end{align*}

(ii) (A+B)−1=A−1+B−1(A+B)^{-1} = A^{-1} + B^{-1}

False. Let A=B=[1]A = B = \begin{bmatrix} 1 \end{bmatrix}. Then:

(A+B)−1=[2]−1=[12]A−1+B−1=[1]+[1]=[2]∴(A+B)−1≠A−1+B−1(A+B)^{-1} = \begin{bmatrix} 2 \end{bmatrix}^{-1} = \begin{bmatrix} \displaystyle\frac{1}{2} \end{bmatrix} \\ A^{-1} + B^{-1} = \begin{bmatrix} 1 \end{bmatrix} + \begin{bmatrix} 1 \end{bmatrix} = \begin{bmatrix} 2 \end{bmatrix} \\[1em] \therefore (A+B)^{-1} \neq A^{-1} + B^{-1}

7.

(i) Let AA be an n×nn \times n matrix. Expand (I−A)(I+A+A2)(I - A)(I + A + A^2).

(I−A)(I+A+A2)=I(I−A)+A(I−A)+A2(I−A)=I2 −AI+AI−A2+A2I−A3A2I=A2=I2−A3=I−A3\begin{align*} (I - A)(I + A + A^2) &= I(I-A) + A(I-A) + A^2(I-A) \\ &= I^2 \cancel{\thinspace- AI + AI - A^2 + A^2I }- A^3 \qquad\qquad\boxed{A^2I = A^2} \\ &= I^2 - A^3 \\ &= I - A^3 \end{align*}

(ii) Suppose that A3=OA^3 = O, the zero matrix. Use (i), find (I−A)−1(I - A)^{-1} in terms of AA.

(I−A)(I+A+A2)=I  ⟺  (I−A)−1(I+A+A2)=I  ⟺  (I−A)(I+A+A2)−1=I∴(I−A)−1=(I+A+A2)\begin{align*} (I - A)(I + A + A^2) = I &\iff (I - A)^{-1}(I + A + A^2) = I \\ &\iff (I - A)(I + A + A^2)^{-1} = I \end{align*} \\[1em] \therefore (I-A)^{-1} = (I+A+A^2)

(iii) (Bonus 1 point) If Ak=OA^k = O, find (I−A)−1(I - A)^{-1}.

From (i) and (ii), we use

(I−A)(I+A+A2)(I-A)(I+A+A^2)

to find (I−A)−1(I-A)^{-1} for A3=OA^3=O. Then, for Ak=OA^k = O, we can use:

(I−A)(I+A+A2+⋯+Ak−1)(I-A)(I + A + A^2 + \cdots + A^{k-1})

We note that the second term is in a geometric form, something that I definitely remembered from Calculus II without Dr. Lai pointing it out.

So, let the identity matrix II be the coefficient and the invertible n×nn\times n matrix AA be the common factor. As such, we have:

(I−A)∑n=0k−1IAn=(I−A)(IA0+IA1+IA2+⋯+IAk−1)=(I−A)(I+IA+IA2+⋯+IAk−1)=(I−A)(I+A+A2+⋯+Ak−1)=I(I+A+A2+⋯+Ak−1)−A(I+A+A2+⋯+Ak−1)=(I+A+A2+⋯+Ak−1)−(A+A2+⋯+Ak−1+Ak)=I+(A+A2+⋯+Ak−1)−(A+A2+⋯+Ak−1+Ak)=I−Ak\begin{align*} (I-A)\sum_{n=0}^{k-1} IA^n &= (I-A)(IA^0 + IA^1 + IA^2 + \cdots + IA^{k-1}) \\ &= (I-A)(I + IA + IA^2 + \cdots + IA^{k-1}) \\ &= (I-A)(I + A + A^2 + \cdots + A^{k-1}) \\ &= I(I + A + A^2 + \cdots + A^{k-1}) - A(I + A + A^2 + \cdots + A^{k-1}) \\ &= (I + A + A^2 + \cdots + A^{k-1}) - (A + A^2 + \cdots + A^{k-1} + A^k) \\ &= I + (\cancel{A + A^2 + \cdots + A^{k-1}}) - (\cancel{A + A^2 + \cdots + A^{k-1}} + A^k) \\ &= I - A^k \end{align*}

Then, if Ak=OA^k = O,
(I−A)∑n=0k−1IAn=I(I-A)\sum_{n=0}^{k-1} IA^n = I

which implies
(I−A)−1=∑n=0k−1IAn=I+A+A2+⋯+Ak−1.(I-A)^{-1} = \sum_{n=0}^{k-1} IA^n = I + A + A^2 + \cdots + A^{k-1}.

Homework 4
  1. Compute the inverse of the following matrices.
  1. Let T:R3→R3T:\R^3\to\R^3 be a linear transformation that maps the standard vector e1\mathbf{e}_1, e2\mathbf{e}_2 and e3\mathbf{e}_3 to [110]\begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}, [100]\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix} and [111]\begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}. Is TT invertible? Explain.
  1. For which values of the constant aa, bb is the following matrix not invertible? [abbaabaaa]\begin{bmatrix} a & b & b \\ a & a & b \\ a & a & a \end{bmatrix} (Hint: find its row echelon form)
  1. Determine if the following statement true or false.
  1. For the matrices AA, BB are invertible, Is the following true? If it is true, verify it. If it is false, give an example to explain why it is false.
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