(1.2.1) Left multiplication by an $n\ast n$ matrix $A$ on $n\ast p$ matrices, $AX=Y$, can be computed by operating on the rows of $X$.

(1.2.2) $Y_i=a_{i1}{X}_1+\cdots +a_{in}{X}_n$.

(1.2.3) Left multiplication by an invertible matrix is called a row operation. Some square matrices called elementary matrices are used. There are three types of elementary 2 * 2 matrices:

• $\left|\begin{array}{cc}1& a\\ & 1\end{array}\right|$ or $\left|\begin{array}{cc}1& \\ a& 1\end{array}\right|$.
• $\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|$.
• $\left|\begin{array}{cc}c& \\ & 1\end{array}\right|$ or $\left|\begin{array}{cc}1& \\ & c\end{array}\right|$.

(1.2.4)

• $\left|\begin{array}{ccccc}1& & & & \\ & 1& & a& \\ & & 1& & \\ & & & 1& \\ & & & & 1\end{array}\right|$ or $\left|\begin{array}{ccccc}1& & & & \\ & 1& & & \\ & & 1& & \\ & a& & 1& \\ & & & & 1\end{array}\right|$.
• $\left|\begin{array}{ccccc}1& & & & \\ & 0& & 1& \\ & & 1& & \\ & 1& & 0& \\ & & & & 1\end{array}\right|$.
• $\left|\begin{array}{ccccc}1& & & & \\ & 1& & & \\ & & c& & \\ & & & 1& \\ & & & & 1\end{array}\right|(c\ne 0)$.

(1.2.5)

• with a in the i, j position, add a * (row j) of $X$ to (row i)
• Interchange (row i) and (row j) of $X$
• Multiply (row i) of $X$ by a nonzero scalar c

(1.2.6) Elementary matrices are invertible, and their inverse are alse elementary matrices.

Proof. The inverse of an elementary matrix is the matrix corresponding to the inverse row operation: ‘‘subtract a * (row j) from (row i),’’ ‘‘interchange (row i) and (row j)’’ again, or ‘‘multiply (row i) by $c^{-1}$.’’

(1.2.7) Row reduction: $M^{\prime }=E_k\dots E_2{E}_{1}M$

(1.2.8) $$\begin{gathered} M=\left|\begin{array}{ccccc}1& 1& 2& 1& 5\\ 1& 1& 2& 6& 10\\ 1& 2& 5& 2& 7\end{array}\right|\to \left|\begin{array}{ccccc}1& 1& 2& 1& 5\\ 0& 0& 0& 5& 5\\ 0& 1& 3& 1& 2\end{array}\right| \\ \to \left|\begin{array}{ccccc}1& 1& 2& 1& 5\\ 0& 1& 3& 1& 2\\ 0& 0& 0& 5& 5\end{array}\right|\to \left|\begin{array}{ccccc}1& 0& -1& 0& 3\\ 0& 1& 3& 1& 2\\ 0& 0& 0& 5& 5\end{array}\right| \\ \to \left|\begin{array}{ccccc}1& 0& -1& 0& 3\\ 0& 1& 3& 1& 2\\ 0& 0& 0& 1& 1\end{array}\right|\to \left|\begin{array}{ccccc}1& 0& -1& 0& 3\\ 0& 1& 3& 0& 1\\ 0& 0& 0& 1& 1\end{array}\right|=M^{\prime } \end{gathered}$$.

(1.2.9) $M=\left|\begin{array}{cc}A& B\end{array}\right|=\left|\begin{array}{cccc}{a}_{11}& \dots & a_{1n}& b_1\\ ⋮& \ddots & ⋮& ⋮\\ a_{m1}& \dots & a_{mn}& b_m\end{array}\right|$, $EM=\left|\begin{array}{cc}EA& EB\end{array}\right|$.

(1.2.10) The systems $A^{\prime }X=B^{\prime }$ and $AX=B$ have the same solutions.

Proof. Since $M$ is obtained by a sequence of elementary row operations, there are elementary matrices $E_1,\dots ,E_k$ such that, with $P=E_k\dots E_1,M=E_k\dots E_{1}M=PM$.
The matrix $P$ is invertible, and $M^{\prime }=[A^{\prime }|B^{\prime }]=[PA|PB]$. If X is a solution of the original equation $AX=B$, we multiply by $P$ on the left: $PAX=PB$, which is to say, $A^{\prime }X=B^{\prime }$. So $X$ also solves the new equation. Conversely, if $A^{\prime }X=B^{\prime }$, then $P^{-1}{A}^{\prime }X=P^{-1}{B}^{\prime }$, that is, $AX=B$.

(1.2.11) $\{\begin{array}{l}{x}_1+x_2+2x_3+x_4=5\\ x_1+x_2+2x_3+6x_4=10\\ x_1+2x_2+5x_3+2x_4=7\end{array}\to \{\begin{array}{l}{x}_1-x3=3\\ x_2+3x_3=1\\ x_4=1\end{array}$

(1.2.12)

(a) If (row i) of M is zero, then (row j) is zero for all j > i.

(b) If (row i) isn’t zero, its first nonzero entry is 1. This entry is called a pivot.

(c) The (row (i+1)) isn’t zero, the pivot in (row (i+1)) is to the right of the pivot in (row i).

(d) The entries above a pivot are zero. (The entries below a pivot are zero too, by ©).

(1.2.13) Let $M^{\prime }=[A^{\prime }|B^{\prime }]$ be a block row echelon matrix, where $B^{\prime }$ is a column vector. The system of equations $A^{\prime }X=B^{\prime }$ has a solution if and only if there is no pivot in the last column $B^{\prime }$. In that case, arbitrary values can be assigned to the unknown $x_i$, provided that (column i) does not contain a pivot. When these arbitrary values are assigned, the other unknowns are determined uniquely.

(1.2.14) Every system $AX=0$ of $m$ homogeneous equations in $n$ unknowns, with $m<n$, has a solution $X$ in which some $x_i$ is nonzero.

Proof. Row reduction of the block matrix $[A|0]$ yields a matrix $[A^{\prime }|0]$ in which $A^{\prime }$ is in row echelon form. The equation $A^{\prime }X=0$ has the same solutions as $AX=0$. The number, say $r$, of pivots of $A^{\prime }$ is at most equal to the number $m$ of rows, so it is less than $n$. The proposition tells us that we may assign arbitrary values to $n-r$ variables xi.

(1.2.15) A square row echelon matrix $M$ is either the identity matrix $I$, or else its bottom row is zero.

Proof. Say that M is an $n\ast n$ row echelon matrix. Since there are $n$ columns, there are at most $n$ pivots, and if there are $n$ of them, there has to be one in each column. In this case, $M=I$. If there are fewer than $n$ pivots, then some row is zero, and the bottom row is zero too.

(1.2.16) Let $A$ be a square matrix. The following conditions are equivalent:

(a) $A$ can be reduced to the identity by a sequence of elementary row operations.

(b) $A$ is a product of elementary matrices.
(c) $A$ is invertible.

Proof. We prove the theorem by proving the implications (a) ⇒ (b) ⇒ © ⇒ (a). Suppose that $A$ can be reduced to the identity by row operations, say $E_k\dots E_{1}A=I$. Multiplying both sides of this equation on the left by $E_1^{-1}\dots E_k^{-1}$, we obtain $A=E_1^{-1}\dots E_k^{-1}$. Since the inverse of an elementary matrix is elementary, (b) holds, and therefore (a) implies (b). Because a product of invertible matrices is invertible, (b) implies ©. Finally, we prove the implication © ⇒ (a). If $A$ is invertible, so is the end result $A^{\prime }$ of its row reduction. Since an invertible matrix cannot have a row of zeros, Lemma 1.2.15 shows that $A^{\prime }$ is the identity.

(1.2.17) Let $A$ be an invertible matrix. To compute its inverse, one may apply elementary row operations $E_1,\dots ,E_k$ to $A$, reducing it to the identity matrix. The same sequence of operations, when applied to the identity matrix $I$, yields $A^{-1}$.

(1.2.18) We invert the matrix $A=\left|\begin{array}{cc}1& 5\\ 2& 6\end{array}\right|$. To do this, we form the 2 * 4 block matrix $\left|\begin{array}{cccc}1& 5& 1& 0\\ 2& 6& 0& 1\end{array}\right|$.We perform row operations to reduce $A$ to the identity, carrying the right side along, and thereby end up with $A^{-1}$ on the right.

(1.2.19) $\left|\begin{array}{cc}A& I\end{array}\right|=\left|\begin{array}{cccc}1& 5& 1& 0\\ 2& 6& 0& 1\end{array}\right|\to \left|\begin{array}{cccc}1& 5& 1& 0\\ 0& -4& -2& 1\end{array}\right|\to \left|\begin{array}{cccc}1& 5& 1& 0\\ 0& 1& \frac{1}{2}& -\frac{1}{4}\end{array}\right|\to \left|\begin{array}{cccc}1& 0& -\frac{3}{2}& \frac{5}{4}\\ 0& 1& \frac{1}{2}& -\frac{1}{4}\end{array}\right|=\left|\begin{array}{cc}I& A^{-1}\end{array}\right|$

(1.2.20) Let $A$ be a square matrix that has either a left inverse or a right inverse, a matrix B such that either $BA=I$ or $AB=I$. Then $A$ is invertible, and $B$ is its inverse.

Proof. Suppose that $AB=I$. We perform row reduction on $A$. Say that $A^{\prime }=PA$, where $P=E_k\dots E_1$ is the product of the corresponding elementary matrices, and $A^{\prime }$ is a row echelon matrix. Then $A^{\prime }B=PAB=P$. Because $P$ is invertible, its bottom row isn’t zero. Then the bottom row of $A^{\prime }$ can’t be zero either. Therefore $A^{\prime }$ is the identity matrix (1.2.15), and so $P$ is a left inverse of $A$. Then $A$ has both a left inverse and a right inverse, so it is invertible and $B$ is its inverse.
If $BA=I$, we interchange the roles of $A$ and $B$ in the above reasoning. We find that $B$ is invertible and that its inverse is $A$. Then $A$ is invertible, and its inverse is $B$.

(1.2.21) Square systems: The following conditions on a square matrix $A$ are equivalent:

(a) $A$ is invertible.

(b) The system of equations $AX=B$ has a unique solution for every column of $B$.

(c) The system of homogeneous euqations $AX=0$ has only the trival solution $X=0$.

Proof. Given the system $AX=B$, we reduce the augmented matrix $[A|B]$ to row echelon form $[A^{\prime }|B^{\prime }]$. The system $A^{\prime }X=B^{\prime }$ has the same solutions. If $A$ is invertible, then $A^{\prime }$ is the identity matrix, so the unique solution is $X=B^{\prime }$. This shows that (a) ⇒ (b). If an $n\ast n$ matrix $A$ is not invertible, then $A^{\prime }$ has a row of zeros. One of the equations making up the system $A^{\prime }X=0$ is the trivial equation. So there are fewer than $n$ pivots. The homogeneous system $A^{\prime }X=0$ has a nontrivial solution (1.2.13), and so does $AX=0$ (1.2.14). This shows that if (a) fails, then © also fails, hence that © ⇒ (a). Finally, it is obvious that (b) ⇒ ©.