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Determinant & EROs

Swapping Two Rows

If A swap two rows BA \xrightarrow{\text{ swap two rows }} B, then det(B)=det(A)\det{B} = -\det{A},
Example
det[1234]=det[3412]{\rm det} \begin{bmatrix} 1&2\\3&4 \end{bmatrix} = -{\rm det} \begin{bmatrix} 3&4\\1&2 \end{bmatrix}
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Multiplying a Row by a Non-Zero Constant

If A multiply a row by k BA \xrightarrow{\text{ multiply a row by } k\ } B, then det(B)=kdet(A)\det{B} = k \cdot \det{A}.
Example
det[1268]=2det[1234]{\rm det} \begin{bmatrix} 1&2\\6&8 \end{bmatrix} = 2 \cdot {\rm det} \begin{bmatrix} 1&2\\3&4 \end{bmatrix}
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Adding a Multiple of One Row to Another Row

If A add a multiple of one row to another row BA \xrightarrow{\text{ add a multiple of one row to another row }} B, then det(B)=det(A)\det{B}=\det{A}
Example
det[1100]=det[1111]{\rm det} \begin{bmatrix} 1&1\\0&0 \end{bmatrix} = {\rm det} \begin{bmatrix} 1&1\\-1&-1 \end{bmatrix}

Wize Tip
These determinant properties hold even if the operations are performed on the columns instead of the rows.

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Example: Determinants & EROs

The following EROs were performed on AA to get to BB:
A3R2A1R3: R3+R1A2R1R2BA\xrightarrow{-3R_2}A_1\xrightarrow{R_3:\ R_3+R_1}A_2\xrightarrow{R_1\leftrightarrow R_2}B
If det(A)=5\det A = 5, find det(B)\det B.
Let's work backward, starting from BB.
Swapping rows means the determinant is negated (multiply by -1): det(B)=det(A2)\det B =-\det{A_2}
Next, we added a multiple of one row to another which does not change the determinant: det(A2)=det(A1)\det{A_2} = \det{A_1}
Then we multiplied a row by -3, so the determinant is multiplied by -3: det(A1)=(3)det(A)\det{A_1} = (-3)\cdot\det A
Putting it all together:

det(B)=det(A2)swapped rows=det(A1)added multiple of row=(3)det(A)multiplied row by 3=(3)(5)=15\det B \quad = \quad \underbrace{-\det{A_2}}_\text{swapped rows} \quad= -\underbrace{\det{A_1}}_\text{added multiple of row} \quad= -\underbrace{(-3)\cdot \det{A}}_{\text{multiplied row by }-3} \quad= -(-3)(5)=\boxed{15}
A=[abcpqrxyz],B=[4x4y4zp+xq+yr+zabc]A= \begin{bmatrix} a&b&c\\ p&q&r\\ x&y&z \end{bmatrix}, \quad B= \begin{bmatrix} 4x&4y&4z\\ p+x&q+y&r+z\\ a&b&c \end{bmatrix}

If det(A)=3\det A=3, find det(B)\det B.
Extra Practice