Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

How do you solve 5x5y+10z=11, 10x+5y5z=1 and 15x15y10z=1 using matrices?

Answer
VerifiedVerified
449.7k+ views
like imagedislike image
Hint: Use the matrix equation AX=B to represent the given system of equations. You need A1 to solve this equation. Use A1=1|A|AdjA to find A1 and solve the equation after substituting the values.

Complete step by step solution:
The given system of equations has three unknown quantities and three linear equations. So, we can find the particular solution of the system, if it exists. Now, we will use matrices to solve this system.
The given system of linear equations is,
5x5y+10z=11
10x+5y5z=1
15x15y10z=1
We can represent the system in the matrix form as AX=B.
Here, A=[55101055151510], X=[xyz] and B=[1111].
Now, we get X=A1B from the matrix equation, if the matrix A1 exists.
To confirm the existence of A1 we check whether |A|0.
We have the formula to find |A| given as, a11(a22a33a23a32)a21(a12a33a13a32)+a31(a12a23a13a22) where aij is the element at the ith row and jth column of the matrix A.
So, we get |A|=5(5075)(5)(100+75)+10(15075).
|A|=5(125)(5)(25)+10(225)
|A|=6251252250
|A|=3000
Since, |A|0, we conclude that A1 exists.
Now we will use the formula A1=1|A|AdjA to get A1. But firstly, we need to find AdjA.
A=[55101055151510]
The adjoint of matrix A is the transpose of its cofactor matrix whose elements are the cofactors of each element in A.
Cofactor of a11=|551510|=5(10)(5)(15)=125
Cofactor of a12=|1051510|=(10(10)(5)(15))=25
Cofactor of a13=|1051515|=10(15)(5)(15)=225
Cofactor of a21=|5101510|=((10)(5)(10)(15))=200
Cofactor of a22=|5101510|=5(10)(10)(15)=200
Cofactor of a23=|551515|=((15)(5)(5)(15))=0
Cofactor of a31=|51055|=(5)(5)(10)(5)=25
Cofactor of a32=|510105|=((5)(5)(10)(10))=125
Cofactor ofa33=|55105|=(5)(5)(10)(5)=75.
So, the cofactor matrix is [1252522520020002512575].
Now adjoint is the transpose of the co-factor matrix. So, AdjA=[1252002525200125225075].
Now, using A1=1|A|AdjA, we have A1=[124115112011201151243400140].
So, using the equation X=A1B we get X=[124115112011201151243400140][1111]

X=[1124+1151120+11120+115+1243340+140]
X=[48120241203240]
X=[251545]
Hence, the solution of the 5x5y+10z=11, 10x+5y5z=1 and 15x15y10z=1 is x=25, y=15 and z=45.

Note:
It is important to be careful while doing calculations in these questions. As, it would be tedious and time-consuming to re-check the calculations, once done. Also, remember that if the coefficient matrix doesn’t have an inverse, then the system either is inconsistent (does not have a solution) or has infinitely many solutions.

Latest Vedantu courses for you
Grade 11 Science PCM | CBSE | SCHOOL | English
CBSE (2025-26)
calendar iconAcademic year 2025-26
language iconENGLISH
book iconUnlimited access till final school exam
tick
School Full course for CBSE students
PhysicsPhysics
ChemistryChemistry
MathsMaths
₹41,848 per year
Select and buy