Question

If$\mathit{A}$is an$\mathit{n}\mathbf{\times }\mathit{m}$ matrix, prove that ${\mathit{I}}_{\mathbf{n}}\mathbf{\cdot }\mathit{A}\mathbf{=}\mathit{A}$and$\mathit{A}\mathbf{\cdot }{\mathit{I}}_{\mathbf{m}}\mathbf{=}\mathit{A}$ .

Short answer

The required identity $A\mathbf{+}Z\mathbf{=}A$ has been proved.

Step-by-step solution

Find the product of the matrices A and In

As it is given that$A$is an$n\mathbf{\times }m$matrix with elements indexed$a_{ij}$that is;

$\begin{array}{c}A\mathbf{=}\left(a_{ij}\right)\\ \mathbf{=}\left[\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{m1}& \cdots & a_{mn}\end{array}\right]\end{array}$.

Let$I_n$be$n\mathbf{\times }m$zero matrix, that is;

.$I_n\mathbf{=}\left[\begin{array}{ccc}1& \cdots & 0\\ ⋮& \ddots & ⋮\\ 0& \cdots & 1\end{array}\right]$

Then, the product of the matrices$I_n$andA can be written as;

$\begin{array}{c}{I}_n\cdot A\mathbf{=}\left[\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{m1}& \cdots & a_{mn}\end{array}\right]\left[\begin{array}{ccc}1& \cdots & 0\\ ⋮& \ddots & ⋮\\ 0& \cdots & 1\end{array}\right]\\ \mathbf{=}\left[\begin{array}{ccc}{a}_{11}\mathbf{+}0\mathbf{+}0\mathbf{+}\dots \mathbf{+}0& \cdots & 0\mathbf{+}0\mathbf{+}0\mathbf{+}\dots \mathbf{+}{a}_{1n}\\ ⋮& \ddots & ⋮\\ a_{m1}\mathbf{+}0\mathbf{+}0\mathbf{+}\dots \mathbf{+}0& \cdots & 0\mathbf{+}0\mathbf{+}0\mathbf{+}\dots \mathbf{+}{a}_{mn}\end{array}\right]\\ \mathbf{=}\left[\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{m1}& \cdots & a_{mn}\end{array}\right]\\ \mathbf{=}A\end{array}$

Hence, it is proved that the property of identity matrix is $I_n\cdot A\mathbf{=}A$.

Find the product of the matrices A and  Im 

As it is given that$A$is an$n\mathbf{\times }m$matrix with elements indexed$a_{ij}$that is;

$\begin{array}{c}A\mathbf{=}\left(a_{ij}\right)\\ \mathbf{=}\left[\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{m1}& \cdots & a_{mn}\end{array}\right]\end{array}$

.

Let$I_n$be$n\mathbf{\times }m$zero matrix, that is;

.$I_m\mathbf{=}\left[\begin{array}{ccc}1& \cdots & 0\\ ⋮& \ddots & ⋮\\ 0& \cdots & 1\end{array}\right]$

Then, the product of the matrices A and$I_m$can be written as;

$\begin{array}{c}A\cdot I_m\mathbf{=}\left[\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{m1}& \cdots & a_{mn}\end{array}\right]\left[\begin{array}{ccc}1& \cdots & 0\\ ⋮& \ddots & ⋮\\ 0& \cdots & 1\end{array}\right]\\ \mathbf{=}\left[\begin{array}{ccc}{a}_{11}\mathbf{+}0\mathbf{+}0\mathbf{+}\dots \mathbf{+}0& \cdots & 0\mathbf{+}0\mathbf{+}0\mathbf{+}\dots \mathbf{+}{a}_{1n}\\ ⋮& \ddots & ⋮\\ a_{m1}\mathbf{+}0\mathbf{+}0\mathbf{+}\dots \mathbf{+}0& \cdots & 0\mathbf{+}0\mathbf{+}0\mathbf{+}\dots \mathbf{+}{a}_{mn}\end{array}\right]\\ \mathbf{=}\left[\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{m1}& \cdots & a_{mn}\end{array}\right]\\ \mathbf{=}A\end{array}$

Hence, it is proved that the property of identity matrix is $A\cdot I_m\mathbf{=}A$$A\cdot I_m\mathbf{=}A$.