Let \(A=(a_{ij})\) be an \(n\times m\) matrix over a field, let \[ X=(x_1,\ldots,x_m)^t, \qquad B=(b_1,\ldots,b_n)^t, \] and consider the system \[ \begin{aligned} a_{11}x_1+\cdots+a_{1m}x_m&=b_1,\\ &\vdots\\ a_{n1}x_1+\cdots+a_{nm}x_m&=b_n. \end{aligned} \]
Show that the system has a simultaneous solution if and only if the matrix equation \(AX=B\) has a solution.
If \(A_1,B_1\) are obtained from \(A,B\) by performing the same sequence of elementary row operations, show that \(X\) solves \(AX=B\) if and only if it solves \(A_1X=B_1\).
Let \[ C=[A\mid B] \] be the \(n\times(m+1)\) augmented matrix. Show that \(AX=B\) has a solution if and only if \[ \operatorname{rank}A=\operatorname{rank}C. \] When a solution exists, show that it is unique if and only if \[ \operatorname{rank}A=m. \]
If \(B=0\), show that the homogeneous system has a nontrivial solution if and only if \[ \operatorname{rank}A<m. \] In particular, show that if \(n<m\), then the homogeneous system has a nontrivial solution.
Solution
1. The displayed scalar system is exactly the coordinate form of \(AX=B\).
The \(i\)th entry of the product \(AX\) is \[ (AX)_i=\sum_{j=1}^m a_{ij}x_j. \] Therefore \[ AX=B \] holds exactly when, for every \(1\le i\le n\), \[ \sum_{j=1}^m a_{ij}x_j=b_i. \] These are precisely the \(n\) displayed equations. Hence a vector \(X\) is a simultaneous solution of the scalar system if and only if it solves the matrix equation.
2. Simultaneous elementary row operations do not change the solution set.
A sequence of elementary row operations is left multiplication by an invertible matrix \(E\), a product of elementary matrices. Thus \[ A_1=EA, \qquad B_1=EB. \] If \(AX=B\), then multiplying by \(E\) gives \[ A_1X=EAX=EB=B_1. \] Conversely, if \(A_1X=B_1\), then \[ EAX=EB. \] Multiplying by \(E^{-1}\) yields \(AX=B\). Therefore the two systems have exactly the same solutions.
3. Consistency is equivalent to equality of the two ranks.
The equation \[ AX=B \] has a solution exactly when the column \(B\) belongs to the column space of \(A\). Appending a column already in the column space does not increase its dimension, while appending a column outside it increases that dimension by one. Hence \[ B\in\operatorname{Col}(A) \iff \operatorname{rank}[A\mid B]=\operatorname{rank}A. \] Since \(C=[A\mid B]\), this is exactly \[ \boxed{AX=B\text{ is solvable}\iff\operatorname{rank}A=\operatorname{rank}C.} \]
4. A consistent system has a unique solution exactly when \(\operatorname{rank}A=m\).
Assume \(AX=B\) is consistent and choose one solution \(X_0\). Then \(X\) is any other solution if and only if \[ A(X-X_0)=0. \] Thus the full solution set is the affine space \[ X_0+\ker A. \] It consists of exactly one point if and only if \[ \ker A=0. \] By rank-nullity for the linear map \[ A:F^m\longrightarrow F^n, \] one has \[ m=\dim\ker A+\operatorname{rank}A. \] Hence \(\ker A=0\) if and only if \(\operatorname{rank}A=m\). Therefore, for a consistent system, \[ \boxed{\text{the solution is unique}\iff\operatorname{rank}A=m.} \]
5. The homogeneous system has a nontrivial solution exactly when \(\operatorname{rank}A<m\).
For \(B=0\), the solution set is precisely \(\ker A\). It contains a nonzero vector if and only if \[ \dim\ker A>0. \] Rank-nullity gives \[ \dim\ker A=m-\operatorname{rank}A, \] so \[ \ker A\ne0 \iff \operatorname{rank}A<m. \]
6. More unknowns than equations force a nontrivial homogeneous solution.
Since \(A\) has \(n\) rows, \[ \operatorname{rank}A\le n. \] If \(n<m\), then \[ \operatorname{rank}A\le n<m, \] so step 5 gives a nontrivial solution of \(AX=0\).