Suppose we are given a 3-by-3 matrix,
$$A = \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]$$
The determinant of this 3 x 3 matrix can be calculated using the following formula. By the way, the vertical bars | | don't indicate absolute value, but rather determinant.
$$\det \left( A \right) = a \times \left| {\matrix{ e & f \cr h & i \cr } } \right| - b \times \left| {\matrix{ d & f \cr g & i \cr } } \right| + c \times \left| {\matrix{ d & e \cr g & h \cr } } \right|$$
Observe that the elements of the first row namely \(a\), \(b\), and \(c\) serve as the scalar multipliers. Each of them is multiplied by the determinant of a 2x2 matrix as determined by "deleting" the row and column that passes through it. For instance, the 2x2 matrix associated with scalar multiplier \(a\) is formed by eliminating the entries that are found in the row and column of \(a\). We use the same procedure to find the 2-by-2 matrices associated with \(b\) and \(c\).
2-by-2 matrix for \(a\):
$$\require{cancel}\left[ {\matrix{ a &\cancel{ b} &\cancel{ c} \cr \cancel{d} & e & f \cr \cancel{g} & h & i \cr } } \right]$$
$$\left[ {\matrix{ e & f \cr h & i \cr } } \right]$$
2-by-2 matrix for \(b\):
$$\require{cancel}\left[ {\matrix{ \cancel{a} & b & \cancel{c} \cr d & \cancel{e} & f \cr g & \cancel{h} & i \cr } } \right]$$
$$\left[ {\matrix{ d & f \cr g & i \cr } } \right]$$
2-by-2 matrix for \(c\):
$$\require{cancel}\left[ {\matrix{\cancel{ a} & \cancel{b} & c \cr d & e & \cancel{f} \cr g & h & \cancel{i} \cr } } \right]$$
$$\left[ {\matrix{ d & e \cr g & h \cr } } \right]$$
Finally, the signs of scalar multipliers \(a\), \(b\), and \(c\) alternate in signs. The sign of \(a\) is positive. The sign of \(b\) is negative. And, the sign of c is positive.
Example 1: Find the determinant of the 3-by-3 matrix.
$$X = \left[ {\matrix{ 2 & 5 & 3 \cr { - 2} & { - 1} & 4 \cr 3 & 7 & 5 \cr } } \right]$$
Let's use the formula to set up our solution.
$$\det \left( A \right) = 2\left| {\matrix{ { - 1} & 4 \cr 7 & 5 \cr } } \right| - 5\left| {\matrix{ { - 2} & 4 \cr 3 & 5 \cr } } \right| + 3\left| {\matrix{ { - 2} & { - 1} \cr 3 & 7 \cr } } \right|$$
Finding the determinants of the smaller matrices:
$$\eqalign{ \left| {\matrix{ { - 1} & 4 \cr 7 & 5 \cr } } \right| &= \left( { - 1} \right)\left( 5 \right) - \left( 7 \right)\left( 4 \right) \cr & = - 5 - 28 \cr & = - 33 \cr} $$
$$\eqalign{ \left| {\matrix{ { - 2} & 4 \cr 3 & 5 \cr } } \right| &= \left( { - 2} \right)\left( 5 \right) - \left( 3 \right)\left( 4 \right) \cr & = - 10 - 12 \cr & = - 22 \cr} $$
$$\eqalign{ \left| {\matrix{ { - 2} & { - 1} \cr 3 & 7 \cr } } \right| &= \left( { - 2} \right)\left( 7 \right) - \left( 3 \right)\left( { - 1} \right) \cr & = - 14 + 3 \cr & = - 11 \cr} $$
Putting it all together, we have
$$\eqalign{
\det \left( A \right) &= 2\left( { - 33} \right) - 5\left( { - 22} \right) + 3\left( { - 11} \right) \cr
& = - 66 + 110 - 33 \cr
& = 11 \cr} $$
The final answer is \(11\).
Example 2: Find the determinant of the 3-by-3 matrix.
$$Y = \left[ {\matrix{ { - 1} & { - 3} & 5 \cr 1 & 2 & { - 1} \cr { - 7} & { - 3} & 0 \cr } } \right]$$
I will leave it to you to calculate the determinants of the smaller 2-by-2 matrices.
$$\eqalign{ \ \det \left( Y \right) &= - 1\left( { - 3} \right) - \left( { - 3} \right)\left( { - 7} \right) + 5\left( {11} \right) \cr & = 3 - 21 + 55 \cr & = 37 \cr} $$
The final answer is \(37\).