Skill: Calculate the determinant of a 3 by 3 matrix using cofactor expansion

Explanation and Free Practice Resources

For a 3×33\times3 matrix, cofactor expansion expresses the determinant as a sum of entries multiplied by the signed determinants of the corresponding 2×22\times2 submatrices; the alternating signs ensure that expansion along any row or column gives the same value. The determinant represents signed volume scaling and indicates whether the matrix is invertible; determinants of larger matrices and more abstract generalizations are beyond this scope.

Detailed Explanation: Calculate the determinant of a 3 by 3 matrix using cofactor expansion

To find a 3×33\times3 determinant by cofactor expansion, choose a row or column and expand across it. For each entry, multiply it by the determinant of the 2×22\times2 matrix left after deleting that entry’s row and column. The signs alternate in a checkerboard pattern:

(+−+−+−+−+)\begin{pmatrix} +&-&+\\ -&+&-\\ +&-&+ \end{pmatrix}

For a 2×22\times2 matrix, use ∣abcd∣=ad−bc\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc.

For example, find

det⁡(2130−14520).\det\begin{pmatrix} 2&1&3\\ 0&-1&4\\ 5&2&0 \end{pmatrix}.

Expand along the first row, whose signs are +,−,++,-,+:

det⁡=2∣−1420∣−1∣0450∣+3∣0−152∣=2((−1)(0)−(4)(2))−((0)(0)−(4)(5))+3((0)(2)−(−1)(5))=2(−8)−(−20)+3(5)=19.\begin{aligned} \det &=2\begin{vmatrix}-1&4\\2&0\end{vmatrix} -1\begin{vmatrix}0&4\\5&0\end{vmatrix} +3\begin{vmatrix}0&-1\\5&2\end{vmatrix}\\ &=2\bigl((-1)(0)-(4)(2)\bigr) -\bigl((0)(0)-(4)(5)\bigr) +3\bigl((0)(2)-(-1)(5)\bigr)\\ &=2(-8)-(-20)+3(5)\\ &=19. \end{aligned}

So the determinant is 1919. Remember to apply the alternating signs as well as the 2×22\times2 determinant rule.

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Matrices - Determinant (3x3) Cofactor Expansion - Matrix to Answer


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