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:

$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ 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.

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4. Matrix multiplication by a unit matrix

Theory: