Find row reduced form of

$A=\left(\begin{array}{lllll} 0 & 3 & 0 & 1 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 4 & 2 & 0 \end{array}\right)$

Solution:

$R_ 3 \leftrightarrow R_4\left(\begin{array}{llll} 0 & 3 & 0 & 1 \\ 0 & 0 & 4 & 1 \\ 0 & 4 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)$

Find row reduced form of $~ A=\left(\begin{array}{ccc}1 & -2 & 3 \\ -4 & 2 & 5\end{array}\right)$

Solution:

$R_2 \rightarrow R_2+4 R_1$

$\left(\begin{array}{ccc}1 & -2 & 3 \\ 0 & -6 & 17\end{array}\right)$

$R_2 \rightarrow-\frac{1}{6} R_2$

$\left(\begin{array}{ccc}1 & -2 & 3 \\ 0 & 1 & \frac{-17}{6}\end{array}\right)$

$R_1 \rightarrow R_2+2 R_2$

$\left[\begin{array}{llr}1 & 0 & 3 - \frac{17}{3} \\0 & 1 & \frac{-17}{6} \end{array}\right]$

$={\left[\begin{array}{ccc}1 & 0 & \frac{-8}{3} \\0 & 1 & \frac{-17}{6} \end{array}\right]}$

The rank of a matrix is defined as the number of non-zero rows in its reduced echelon matrix.

Note : Given a matrix $A$ then the row reduced echelon matrix associated with $A$ is unique.

$A=\left(\begin{array}{llll}0 & 0 & 4 & 1 \\ 0 & 3 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 4 & 2 & 0\end{array}\right)$

$\operatorname{RRE}=\left(\begin{array}{llll}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{array}\right)$

The number of non-zero rows in the RRE of $A$ is 3.

$\therefore \operatorname{Rank}(A)=3$.

$\begin{aligned} A & =\left[\begin{array}{ccc} 1 & -2 & 3 \\ 4 & 2 & 5 \end{array}\right] \\ \operatorname{RRE} & =\left[\begin{array}{ccc} 1 & 0 & \frac{-2}{3} \\ 0 & 1 & \frac{-11}{6} \end{array}\right] \end{aligned}$

The number of non-zero rows in this RRE is 2 .

$\therefore$ Rank of $A$ is 2 .

3. $A=\left[\begin{array}{ll}2 & 3 \\ 4 & 5 \\ 2 & 1\end{array}\right] \quad R_1 \rightarrow \frac{1}{2} R_1$

$\left[\begin{array}{ll}1 & \frac{3}{2} \\ 4 & 5 \\ 2 & 1\end{array}\right]$

This is the row reduced echelon matrix obtained from the given matrix.

The number of non zero rows in this RRE is 2 .

$\therefore$ Rank of the matrix is 2 .

A square matrix $A$ is said to be invertible if a square matrix $B$ with order $B=$ order $A$

such that $A B = B A = I$.

Result:

Suppose $R$ is the RRE matrix obtained from the given matrix $A$. Then $A$ is invertible if and only if $R$ is the identity matrix.

The matrix $B$ is called as the inverse of $A$.

Is the inverse of a matrix unique?

Solution:

Suppose that $A$ is an invertible matrix.

Let, $B$ & $C$ be this inverses, i.e.,

$A$ $B$ & = $I = B A$ $\longrightarrow(1)$

$A$ $C$ & $= I = C A$ $\longrightarrow(2)$

Given any square matrix $A, A \cdot I=A$.

Let $A=\left[a_{i j}\right]$ & $I=\left[b b_{i j}\right],$

$b_{i j}= \begin{cases}1 ,\text{if} ~ ~ i=j \\ 0 ,\text{if} ~ ~ i ≠ j\end{cases}$.

$A I=\left[a_{i j}\right]\left[b_{i j}\right]$

$=\left[\sum_{k=1}^n a_{i k} b_{k j}\right] =\left[a_{i j} b_{j j}\right] =\left[a_{i j}\right]=A . \quad \therefore A I=A \longrightarrow(3)$

$\\ \text { similarly, } A=A \longrightarrow(4) \text {. } \\ \therefore B=B \cdot I \quad \text { (From (3)) } \\ =B \cdot(A C) \text { (From (2)) } \\ =(B A) C \text { (associativity of matrix multiplication)} \\$

$=I.C$ (From (1))

$= C$ (From (A))

Thus, $B = C$.

For any two square matrices $A$ & $B$ of same order, the Invertibility of $A$ & $B$ implies the Invertibility of $A B$ & $(A B)^{-1}=B^{-1} A^{-1}$.

Proof:

$(A B) \cdot\left(B^{-1} A^{-1}\right)=A\left(B B^{-1}\right) \cdot A^{-1}=(A \cdot I) A^{-1}$

$=A \cdot A^{-1}=I$

$\left(B^{-1} A^{-1}\right)(A B)=B^{-1}\left(A^{-1} A\right) B=B^{-1}(I \cdot B)$

$=B^{-1} B=I \cdot$

If $P$ is an elementary row operation on the set of all matrices, then $P$ is invertible as a function i.e., $P$ is both $1-1$ and onto.

Suppose $P_1, P_2, \ldots, P_m$ are the elementary row operations performed on an invertible matrix $A$ as obtain the identity matrix.

Then,$P_m P_{m-1} \cdots P_1(A)=I .$

$\quad A=\left(\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right)$

Making an augumented matrix with the given matrix and identity matrix.

$=\left(\begin{array}{lll|lll} 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 1 & 2 & 0 & 0 & 0 & 1 \end{array}\right)$

Now, applying row operations to it.

$R_2\rightarrow R_2-R_1 \\ R_3 \rightarrow R_3-R_1$

$\left(\begin{array}{lll|ccc}1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & -1 & 1 & 0 \\ 0 & 1 & 2 & -1 & 0 & 1\end{array}\right)$

Thus, $A^{-1}=\left(\begin{array}{ccc}2 & -1 / 2 & -1 / 2 \\ -1 & 1 & 0 \\ 0 & -1 / 2 & 1 / 2\end{array}\right)$.