Find cost of each item per kg by matrix method Miscellaneous Examples
Example 19 Use product 
1
1
2
0
2
3
3
2
4
2
0
1
9
2
3
6
1
2
�
�
�
�
�
�
















 to solve the system of equations
x – y + 2z = 1
2y – 3z = 1
3x – 2y + 4z = 2
Solution Consider the product 
1
1
2
2
0
1
0
2
3
9
2
3
3
2
4
6
1
2
–
–
–
–
–
–

 


 


 


 


 

 =
2
9
12
0
2
2
1
3
4
0
18 18
0
4
3
0
6
6
6 18
24
0
4
4
3
6
8
− −
+
−
+
+
−




+
−
+
−
−
+




− −
+
−
+
+
−


 = 
1
0
0
0
1
0
0
0
1








Hence
  
1
1
2
0
2
3
3
2
4
2
0
1
9
2
3
6
1
2
1
–
–
–
–
–
–
–







=








Now, given system of equations can be written, in matrix form, as follows
1
–1
2
0
2
–3
3
–2
4
x
y
z

  

  

  

  
 


 =
1
1
2
 
 
 
 
 
Rationalised 2023-24
DETERMINANTS     99
or
  
  
  
  
  
x
y
z
 =
1
1
1
2
1
0
2
3
1
3
2
4
2
−
−


 


 
−


 


 
−


 
 = 
�
�
�
2
0
1
9
2
3
6
1
2
1
1
2
















=
2
0
2
0
9
2
6
5
6
1
4
3
− +
+


 


 
+
−
=


 


 
+ −


 
Hence
x = 0, y = 5 and z = 3
Miscellaneous Exercises on Chapter 4
1 Prove that the determinant
sin
cos
–sin
–
1
cos
1
x
x
x
θ
θ
θ
θ
 is independent of θ 2