UPSKILL MATH PLUS

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Theory:

Consider \(A\) is a square matrix of order \(n×n\) and \(I\) is the unit matrix of same order then \(AI = IA = A\).
 
Here, the \(2 × 2\) order of \(I\) matrix is \(\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}\)
 
Then, the \(3 × 2\) order of \(I\) matrix will be \(\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}\)
 
Now let's prove \(AI = IA = A\) using \(2 × 2\) order matrix.
Example:
If \(A = \begin{bmatrix}
1 & 2 \\ 
3 & 4
\end{bmatrix}\) then we have:
 
AI=12341001AI=(1×1)+(2×0)(1×0)+(2×1)(3×1)+(4×0)(3×0)+(4×1)
 
\(AI = \begin{bmatrix}
1 & 2 \\ 
3 & 4
\end{bmatrix}\)……(1)
 
Let's find \(IA\) matrix.
 
IA=10011234IA=(1×1)+(0×3)(1×2)+(0×4)(0×1)+(1×3)(0×2)+(1×4)
 
\(IA = \begin{bmatrix}
1 & 2 \\ 
3 & 4
\end{bmatrix}\)……(2)
 
From \((1)\) and \((2)\) we get that \((1) = (2) = A\).
 
Hence, \(AI = IA = A\) proved.