Matrix Form
Write the equations as
\[AX=C.\]
Here, \(A\) is the coefficient matrix, \(X\) is the matrix containing the unknowns, and \(C\) is the right-hand-side matrix. The augmented matrix is \([A:C]\).
Consistency of Linear Equations
Rank Test, Unique Solutions and Infinite Solutions
Consistency of linear equations is developed through the rank test. Reduce the augmented matrix, compare both ranks with the number of unknowns and solve the resulting echelon equations.
Unit
Matrices
Topic
Consistency of Equations
Question Type
4 Mark Problems
Theory Required for This Problem
Inconsistent System
If
\[R(A)<R([A:C]),\]
the system has no solution.
Unique Solution
If
\[R(A)=R([A:C])=n,\]
the system has a unique solution.
Infinite Solutions
If
\[R(A)=R([A:C])<n,\]
the system has infinitely many solutions. Assign parameters to \(n-r\) unknowns.
Method used: Form the augmented matrix, reduce it to echelon form, compare the two ranks with the number of unknowns, and solve the reduced equations.
Continue through the IOE Engineering Mathematics II Digital Workbook. This consistency of linear equations page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
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4 Marks
Consistency and Solution of Three Equations
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 13 • 4 Marks
Test the consistency of the following system of linear equations and find its solution if it is consistent:
\[\begin{array}{rcl}2x-y+3z&=&8,\\-x+2y+z&=&4,\\3x+y-4z&=&0.\end{array}\]
Solution
Rank test using the augmented matrixLet
\[A=\left[\begin{array}{ccc}2&-1&3\\-1&2&1\\3&1&-4\end{array}\right],\qquad X=\left[\begin{array}{c}x\\y\\z\end{array}\right],\qquad C=\left[\begin{array}{c}8\\4\\0\end{array}\right].\]
The augmented matrix is
\[[A:C]=\left[\begin{array}{ccc|c}2&-1&3&8\\-1&2&1&4\\3&1&-4&0\end{array}\right].\]
Apply
\[\begin{array}{l}R_1\leftrightarrow R_2,\\R_1\rightarrow-R_1.\end{array}\]
Then
\[[A:C]\sim\left[\begin{array}{ccc|c}1&-2&-1&-4\\2&-1&3&8\\3&1&-4&0\end{array}\right].\]
Now apply
\[\begin{array}{l}R_2\rightarrow R_2-2R_1,\\R_3\rightarrow R_3-3R_1.\end{array}\]
Therefore,
\[[A:C]\sim\left[\begin{array}{ccc|c}1&-2&-1&-4\\0&3&5&16\\0&7&-1&12\end{array}\right].\]
Finally, apply
\[R_3\rightarrow3R_3-7R_2.\]The echelon form is
\[[A:C]\sim\left[\begin{array}{ccc|c}1&-2&-1&-4\\0&3&5&16\\0&0&-38&-76\end{array}\right].\]
Number of non-zero rows in \(A=3\). Therefore,
\[R(A)=3.\]
Number of non-zero rows in \([A:C]=3\). Therefore,
\[R([A:C])=3.\]
Also, the number of unknowns is
\[n=3.\]
Hence,
\[R(A)=R([A:C])=n=3.\]
Therefore, the system is consistent and has a unique solution.
From the third row,
\[-38z=-76.\]
\[z=\frac{-76}{-38}.\]
\[z=2.\]
From the second row,
\[3y+5z=16.\]
\[3y+5(2)=16.\]
\[3y=16-10.\]
\[3y=6.\]
\[y=2.\]
From the first row,
\[x-2y-z=-4.\]
\[x-2(2)-2=-4.\]
\[x-6=-4.\]
\[x=2.\]
\[R(A)=R([A:C])=3=n.\]\[x=2,\qquad y=2,\qquad z=2.\]
A Few Important Questions
A Parameter in the Constant Matrix
For what values of \(\lambda\) does the following system have a solution? Solve it in each case.
\[\begin{array}{rcl}x+y+z&=&1,\\x+2y+4z&=&\lambda,\\x+4y+10z&=&\lambda^2.\end{array}\]
Show Solution
Let
\[A=\left[\begin{array}{ccc}1&1&1\\1&2&4\\1&4&10\end{array}\right],\qquad X=\left[\begin{array}{c}x\\y\\z\end{array}\right],\qquad C=\left[\begin{array}{c}1\\\lambda\\\lambda^2\end{array}\right].\]
The augmented matrix is
\[[A:C]=\left[\begin{array}{ccc|c}1&1&1&1\\1&2&4&\lambda\\1&4&10&\lambda^2\end{array}\right].\]
Apply
\[\begin{array}{l}R_2\rightarrow R_2-R_1,\\R_3\rightarrow R_3-R_1.\end{array}\]
Therefore,
\[[A:C]\sim\left[\begin{array}{ccc|c}1&1&1&1\\0&1&3&\lambda-1\\0&3&9&\lambda^2-1\end{array}\right].\]
Now apply
\[R_3\rightarrow R_3-3R_2.\]
The echelon form is
\[[A:C]\sim\left[\begin{array}{ccc|c}1&1&1&1\\0&1&3&\lambda-1\\0&0&0&\lambda^2-3\lambda+2\end{array}\right].\]
Number of non-zero rows in \(A=2\). Therefore,
\[R(A)=2.\]
For the system to be consistent, the last row of the augmented matrix must be a zero row. Therefore,
\[\lambda^2-3\lambda+2=0.\]
\[(\lambda-1)(\lambda-2)=0.\]
Hence,
\[\lambda=1\]
or
\[\lambda=2.\]
If \(\lambda\neq1,2\), then the last row is non-zero. Hence,
\[R(A)=2<R([A:C])=3.\]
Therefore, the system is inconsistent and has no solution.
Case I: \(\lambda=1\)
Substituting \(\lambda=1\) in the echelon form,
\[[A:C]\sim\left[\begin{array}{ccc|c}1&1&1&1\\0&1&3&0\\0&0&0&0\end{array}\right].\]
Number of non-zero rows in \(A=2\), number of non-zero rows in \([A:C]=2\), and the number of unknowns is \(n=3\). Therefore,
\[R(A)=R([A:C])=2<3=n.\]
Hence, an infinite number of solutions will exist.
The equations are
\[x+y+z=1,\]
\[y+3z=0.\]
Let
\[z=k.\]
Then,
\[y+3k=0.\]
\[y=-3k.\]
Substituting \(y=-3k\) and \(z=k\) in the first equation,
\[x-3k+k=1.\]
\[x-2k=1.\]
\[x=1+2k.\]
\[\lambda=1:\qquad x=1+2k,\qquad y=-3k,\qquad z=k.\]
Case II: \(\lambda=2\)
Substituting \(\lambda=2\) in the echelon form,
\[[A:C]\sim\left[\begin{array}{ccc|c}1&1&1&1\\0&1&3&1\\0&0&0&0\end{array}\right].\]
Number of non-zero rows in \(A=2\), number of non-zero rows in \([A:C]=2\), and the number of unknowns is \(n=3\). Therefore,
\[R(A)=R([A:C])=2<3=n.\]
Hence, an infinite number of solutions will exist.
The equations are
\[x+y+z=1,\]
\[y+3z=1.\]
Let
\[z=k.\]
Then,
\[y+3k=1.\]
\[y=1-3k.\]
Substituting \(y=1-3k\) and \(z=k\) in the first equation,
\[x+(1-3k)+k=1.\]
\[x-2k=0.\]
\[x=2k.\]
\[\lambda=2:\qquad x=2k,\qquad y=1-3k,\qquad z=k.\]
An Inconsistent System
Test the consistency of
\[\begin{array}{rcl}x+y+z&=&4,\\x+y+2z&=&2,\\2x+2y+5z&=&5.\end{array}\]
Show Solution
Let
\[A=\left[\begin{array}{ccc}1&1&1\\1&1&2\\2&2&5\end{array}\right],\qquad X=\left[\begin{array}{c}x\\y\\z\end{array}\right],\qquad C=\left[\begin{array}{c}4\\2\\5\end{array}\right].\]
The augmented matrix is
\[[A:C]=\left[\begin{array}{ccc|c}1&1&1&4\\1&1&2&2\\2&2&5&5\end{array}\right].\]
Apply
\[\begin{array}{l}R_2\rightarrow R_2-R_1,\\R_3\rightarrow R_3-2R_1.\end{array}\]
Therefore,
\[[A:C]\sim\left[\begin{array}{ccc|c}1&1&1&4\\0&0&1&-2\\0&0&3&-3\end{array}\right].\]
Now apply
\[R_3\rightarrow R_3-3R_2.\]
The echelon form is
\[[A:C]\sim\left[\begin{array}{ccc|c}1&1&1&4\\0&0&1&-2\\0&0&0&3\end{array}\right].\]
Number of non-zero rows in \(A=2\). Therefore,
\[R(A)=2.\]
Number of non-zero rows in \([A:C]=3\). Therefore,
\[R([A:C])=3.\]
Hence,
\[R(A)<R([A:C]).\]
Therefore, the system is inconsistent and has no solution.
The system is inconsistent. Therefore, it cannot be solved.