Matrix multiplication by a unit matrix — lesson. Mathematics State Board, Class 10.

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Consider \(A\) is a square matrix of order \(n×n\) and \(I\) is the unit matrix of the 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:

\begin{array}{l} AI = [\begin{array}{l} 1 & 2 \\ 3 & 4 \end{array}] [\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array}] \\ AI = [\begin{array}{l} (1 \times 1) + (2 \times 0) & (1 \times 0) + (2 \times 1) \\ (3 \times 1) + (4 \times 0) & (3 \times 0) + (4 \times 1) \end{array}] \end{array}

\(AI = \begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}\)……(1)

Let's find \(IA\) matrix.

\begin{array}{l} IA = [\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array}] [\begin{array}{l} 1 & 2 \\ 3 & 4 \end{array}] \\ IA = [\begin{array}{l} (1 \times 1) + (0 \times 3) & (1 \times 2) + (0 \times 4) \\ (0 \times 1) + (1 \times 3) & (0 \times 2) + (1 \times 4) \end{array}] \end{array}

\(IA = \begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}\)……(2)

From \((1)\) and \((2)\) we get that \((1) = (2) = A\).

Hence, \(AI = IA = A\) proved.

Did you find an error?

4. Matrix multiplication by a unit matrix