Identity Matrix - Linear Algebra Step by Step

Show the 3x3 identity matrix I and demonstrate that multiplying any matrix A on the right by I leaves A unchanged: A · I = A. Each entry of the result picks out the matching entry of A because the dot product of a row of A with a column of I (a standard basis vector) selects exactly one element.

identity matrix The n×n identity matrix I has ones on the diagonal and zeros elsewhere. Multiplying any matrix A (with n columns) on the right by I_n gives A: A · I_n = A.

column selection Column j of I_n is the j-th standard basis vector e_j — all zeros except a 1 in row j. The dot product of row i of A with e_j is A[i,j], so the product A · I simply copies each entry of A.

Example

Multiply by the identity matrix and fill the copied result entries.

highlighted = computed this step

Step 1 — Set up

Start with the given matrix data.

A=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix},\quad B=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix},\quad C=\begin{bmatrix}\square&\square&\square\\\square&\square&\square\end{bmatrix}

Step 2 — Top-Left entry

Fill top-left entry: 1*1 = 1.

C=\begin{bmatrix}\hl{1}&\square&\square\\\square&\square&\square\end{bmatrix},\quad 1\cdot 1=1

Step 3 — Top-Right entry

Fill top-right entry: 2*1 = 2.

C=\begin{bmatrix}1&\hl{2}&\square\\\square&\square&\square\end{bmatrix},\quad 2\cdot 1=2

Step 4 — Top-Third entry

Fill top-third entry: 3*1 = 3.

C=\begin{bmatrix}1&2&\hl{3}\\\square&\square&\square\end{bmatrix},\quad 3\cdot 1=3

Step 5 — Bottom-Left entry

Fill bottom-left entry: 4*1 = 4.

C=\begin{bmatrix}1&2&3\\\hl{4}&\square&\square\end{bmatrix},\quad 4\cdot 1=4

Step 6 — Bottom-Right entry

Fill bottom-right entry: 5*1 = 5.

C=\begin{bmatrix}1&2&3\\4&\hl{5}&\square\end{bmatrix},\quad 5\cdot 1=5

Step 7 — Bottom-Third entry

Fill bottom-third entry: 6*1 = 6.

C=\begin{bmatrix}1&2&3\\4&5&\hl{6}\end{bmatrix},\quad 6\cdot 1=6

Step 8 — Result

Read the completed AI matrix.

C=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix},\quad \text{AI complete}