1 3
6 2 3
1. Find the solution to ! ' (4 1*
1 4 2
2 7
6∗1+2∗4+3∗2 6∗3+2∗1+3∗7
=! '
1∗1+4∗4+2∗2 1∗3+4∗1+2∗7
6 + 8 + 6 18 + 2 + 21
=! '
1 + 16 + 4 3 + 4 + 14
20 41
=! '
21 21
2 1 2
2. Find the inverse of (1 2 1*
3 2 0
2 1 21 0 0
(1 2 100 1 0*
3 2 00 0 1
R1 – R3
−1 −1 2 1 0 −1
(1 2 100 1 0 *
3 2 00 0 1
R1 – 2xR2
−3 −5 0 1 −2 −1
(1 2 100 1 0*
3 2 00 0 1
R1 + 2.5xR3
4.5 0 0 1 −2 1.5
( 1 2 100 1 0*
3 2 00 0 1
R1 / 4.5
1 0 0 2 −4 3
41 2 159 9 9
0 1 07
3 2 00 0 1
Swap R2 and R3
1 0 0 2 −4 3
43 2 059 9 9
0 0 17
1 2 10 1 0
R2 – 3R1
2 4 3
⎡1 0 0 9 − 9 9⎤
⎢0 2 0; 6 12 ⎥
⎢1 2 1 − 9 9 0⎥
⎣ 0 1 0⎦
R2 / 2
2 4 3
⎡1 0 0 9 − 9 9⎤
⎢ 0 1 0; 3 6 ⎥
⎢1 2 1 − 9 9 0⎥
⎣ 0 1 0⎦
R3 – R1 – 2xR2
2 4 3
⎡ 9 −9 9 ⎤
⎢1 0 0; 3 6 ⎥
⎢0 1 0;− 9 9 0 ⎥
⎢0 0 1 4 1 3⎥
⎣ 9 9 − 9⎦
3 2
3. Find the eigenvalues and eigenvectors of ! '
1 2
|𝑨 − 𝜆𝑰| = C3 − 𝜆 2
C
1 2−𝜆
= (3 − 𝜆)(2 − 𝜆) − 2
= 𝜆! − 5𝜆 + 6 − 2
= 𝜆! − 5𝜆 + 4
= (𝜆 − 1)(𝜆 − 4)
=0
𝜆 = 1 𝑜𝑟 4
For 𝜆 = 1 For 𝜆 = 4
2 2 −1 2
𝑨 − 𝜆𝑰 = ! ' 𝑨 − 𝜆𝑰 = ! '
1 1 1 −2
𝑣" + 𝑣! = 0 𝑣" − 2𝑣! = 0
Setting 𝑣" = 1, 𝑣! = −1 Setting 𝑣! = 1, 𝑣" = 2
1 2
𝑣#$" = ! ' 𝑣#$% = ! '
−1 1