Determinant of a 3×3 Matrix
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. Note that the vertical bars | | represent the determinant, not absolute value.
$$\det \left( A \right) = a \times \left| {\matrix{ e & f \cr h & i \cr } } \right| – b \times \left| {\matrix{ d & f \cr g & i \cr } } \right| + c \times \left| {\matrix{ d & e \cr g & h \cr } } \right|$$
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 2×2 matrix as determined by “deleting” the row and column that passes through it. For instance, the 2×2 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 \(37\).