Answer: (a) The coefficient matrix is
\(\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}.\)
Its determinant is
\(\left|\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right|=(5\cdot 9-6\cdot 8)-2(4\cdot 9-6\cdot 7)+3(4\cdot 8-5\cdot 7)=-3+12-9=0.\)
So the coefficient matrix is singular, and the system does not have a unique solution.
(b) From the first equation,
\(z=\dfrac{1-x-2y}{3}.\)
Substituting into the second gives
\(4x+5y+2(1-x-2y)=1\Rightarrow 2x+y=-1\Rightarrow y=-2x-1.\)
Substituting the same expression for \(z\) into the third gives
\(7x+8y+3(1-x-2y)=1\Rightarrow 4x+2y=-2\Rightarrow y=-2x-1.\)
All three equations reduce to the same single linear relation, so the system is consistent and has infinitely many solutions.
Geometrically, the three equations represent three planes intersecting in one common line.
(a) A system of three linear equations in three unknowns has a unique solution only when the determinant of its coefficient matrix is non-zero.
Here the coefficient matrix is
\(A=\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}.\)
Now
\(\det(A)=\left|\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right|\)
\(=(5\cdot 9-6\cdot 8)-2(4\cdot 9-6\cdot 7)+3(4\cdot 8-5\cdot 7)\)
\(=-3+12-9=0.\)
So \(A\) is singular, hence the system does not have a unique solution.
(b) To show the system is consistent, use all three equations and reduce them to one relation in two unknowns.
From
\(x+2y+3z=1\)
we get
\(z=\dfrac{1-x-2y}{3}.\)
Substitute this into the second equation:
\(4x+5y+6z=1\)
\(4x+5y+6\left(\dfrac{1-x-2y}{3}\right)=1\)
\(4x+5y+2(1-x-2y)=1\)
\(2x+y+2=1\)
\(2x+y=-1\)
\(y=-2x-1.\)
Now substitute \(z=\dfrac{1-x-2y}{3}\) into the third equation:
\(7x+8y+9z=1\)
\(7x+8y+9\left(\dfrac{1-x-2y}{3}\right)=1\)
\(7x+8y+3(1-x-2y)=1\)
\(4x+2y+3=1\)
\(4x+2y=-2\)
\(2x+y=-1\)
\(y=-2x-1.\)
So all three equations reduce to the same single linear relation. Therefore the system is consistent, and since there is only one independent relation between the variables, there are infinitely many solutions.
For example, taking \(x=t\), we get
\(y=-2t-1,\qquad z=\dfrac{1-t-2(-2t-1)}{3}=t+1.\)
Thus the solution set is a line.
Geometrically, each equation represents a plane, and the three planes intersect in a common line.
