Inverse
Exercise Compute the inverse of the following matrices. Then use your result to solve the following systems.
1)
2x_{1} 
+ 3x_{2} 
= 
2 
, 
2x_{1} 
+ 3x_{2} 
= 
1 
7x_{1} 
+ 9x_{2} 
= 
1 
7x_{1} 
+ 9x_{2} 
= 
0 
Answer The inverse of the matrix is
(2×9  7×3)^{1}[ 
9 
3 
] = [ 
3 
1 
] 
7 
2 
7/3 
2/3 
Since
[ 
2 
3 
]^{1}[ 
2 
] = [ 
3 
1 
] [ 
2 
] = [ 
7 
] 
7 
9 
1 
7/3 
2/3 
1 
16/3 
the solution of the first system is x_{1} = 7, x_{2} = 4. Similarly, the solution of the second system is
[ 
x_{1} 
] = [ 
2 
3 
]^{1}[ 
1 
] = [ 
3 
] 
x_{2} 
7 
9 
0 
7/3 
2)
x_{1} 

+ 2x_{3} 
= 
1 
, 
x_{1} 

+ 2x_{3} 
= 
0 
2x_{1} 
 x_{2} 
+ 3x_{3} 
= 
0 
2x_{1} 
 x_{2} 
+ 3x_{3} 
= 
2 
4x_{1} 
 x_{2} 
+ 8x_{3} 
= 
1 
4x_{1} 
 x_{2} 
+ 8x_{3} 
= 
1 
Answer Row operations change
[ 
1 
0 
2 
1 
0 
0 
] 
2 
1 
3 
0 
1 
0 
4 
1 
8 
0 
0 
1 
into
[ 
1 
0 
0 
5 
2 
2 
] 
0 
1 
0 
4 
0 
1 
0 
0 
1 
2 
1 
1 
Thus the inverse of the matrix is
[ 
5 
2 
2 
] 
4 
0 
1 
2 
1 
1 
and the solution to the two systems are respectively
[ 
x_{1} 
] = [ 
5 
2 
2 
] [ 
1 
] = [ 
7 
] 
x_{2} 
4 
0 
1 
0 
5 
x_{3} 
2 
1 
1 
1 
3 
[ 
x_{1} 
] = [ 
5 
2 
2 
] [ 
0 
] = [ 
2 
] 
x_{2} 
4 
0 
1 
2 
1 
x_{3} 
2 
1 
1 
1 
1 
3)
[ 
2 
0 
2 
1 
] 
1 
1 
1 
0 
0 
1 
2 
1 
1 
1 
2 
1 
2x_{1} 

+ 2x_{3} 
+ x_{4} 
= 
1 
x_{1} 
+ x_{2} 
 x_{3} 

= 
0 

x_{2} 
 2x_{3} 
 x_{4} 
= 
1 
x_{1} 
 x_{2} 
+ 2x_{3} 
+ x_{4} 
= 
0 
Answer The inverse of the matrix is
[ 
0 
0 
1 
1 
] 
1 
0 
1 
2 
1 
1 
0 
1 
1 
2 
2 
0 
The solution of the system is
[ 
x_{1} 
] = [ 
0 
0 
1 
1 
] [ 
1 
] = [ 
1 
] 
x_{2} 
1 
0 
1 
2 
0 
0 
x_{3} 
1 
1 
0 
1 
1 
1 
x_{4} 
1 
2 
2 
0 
0 
3 