M=(a11a12a13a21a22a23a31a32a33)=(153247462){\displaystyle M={\begin{pmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}\end{pmatrix}}={\begin{pmatrix}1&5&3\2&4&7\4&6&2\end{pmatrix}}}
Let’s choose the first row of our example matrix A. Circle the 1 5 3. In general terms, circle a11 a12 a13.
In our example, our reference row is 1 5 3. The first element is in row 1 and column 1. Cross out all of row 1 and column 1. Write the remaining elements as a 2 x 2 matrix: 1 5 3 2 4 7 4 6 2
In our example, the determinant of the matrix (4762){\displaystyle {\begin{pmatrix}4&7\6&2\end{pmatrix}}} = 4 * 2 - 7 * 6 = -34. This determinant is called the minor of the element we chose in our original matrix. [5] X Research source In this case, we just found the minor of a11.
In our example, the determinant of the matrix (4762){\displaystyle {\begin{pmatrix}4&7\6&2\end{pmatrix}}} = 4 * 2 - 7 * 6 = -34. This determinant is called the minor of the element we chose in our original matrix. [5] X Research source In this case, we just found the minor of a11.
In our example, we selected a11, which had a value of 1. Multiply this by -34 (the determinant of the 2x2) to get 1*-34 = -34.
- +- + -+ - + Since we chose a11, marked with a +, we multiply the number by +1. (In other words, leave it alone. ) The answer is still -34. Alternatively, you can find the sign with the formula (-1)i+j, where i and j are the element’s row and column. [7] X Research source
- +- + -+ - + Since we chose a11, marked with a +, we multiply the number by +1. (In other words, leave it alone. ) The answer is still -34. Alternatively, you can find the sign with the formula (-1)i+j, where i and j are the element’s row and column. [7] X Research source
Cross out the row and column of that element. In our case, select element a12 (with a value of 5). Cross out row one (1 5 3) and column two (546){\displaystyle {\begin{pmatrix}5\4\6\end{pmatrix}}}. Treat the remaining elements as a 2x2 matrix. In our example, the matrix is (2742){\displaystyle {\begin{pmatrix}2&7\4&2\end{pmatrix}}} Find the determinant of this 2x2 matrix. Use the ad - bc formula. (22 - 74 = -24) Multiply by the chosen element of the 3x3 matrix. -24 * 5 = -120 Determine whether to multiply by -1. Use the sign chart or the (-1)ij formula. We chose element a12, which is - on the sign chart. We must change the sign of our answer: (-1)*(-120) = 120.
Cross out row 1 and column 3 to get (2446){\displaystyle {\begin{pmatrix}2&4\4&6\end{pmatrix}}} Its determinant is 26 - 44 = -4. Multiply by element a13: -4 * 3 = -12. Element a13 is + on the sign chart, so the answer is -12.
In our example the determinant is -34 + 120 + -12 = 74.
Let’s say you pick row 2, with elements a21, a22, and a23. To solve this problem, we’ll be looking at three different 2x2 matrices. Let’s call them A21, A22, and A23. The determinant of the 3x3 matrix is a21|A21| - a22|A22| + a23|A23|. If terms a22 and a23 are both 0, our formula becomes a21|A21| - 0*|A22| + 0*|A23| = a21|A21| - 0 + 0 = a21|A21|. Now we only have to calculate the cofactor of a single element.
For example, say you have a 3 x 3 matrix: (9−1231075−2){\displaystyle {\begin{pmatrix}9&-1&2\3&1&0\7&5&-2\end{pmatrix}}} In order to cancel out the 9 in position a11, we can multiply the second row by -3 and add the result to the first. The new first row is [9 -1 2] + [-9 -3 0] = [0 -4 2]. The new matrix is (0−4231075−2){\displaystyle {\begin{pmatrix}0&-4&2\3&1&0\7&5&-2\end{pmatrix}}} Try to use the same trick with columns to turn a12 into a 0 as well.
Upper triangular matrix: All the non-zero elements are on or above the main diagonal. Everything below is a zero. Lower triangular matrix: All the non-zero elements are on or below the main diagonal. Diagonal matrix: All the non-zero elements are on the main diagonal. (A subset of the above. ) You can use the method of minors or the elementary row operations to find the inverse of a 3 x 3 matrix. [11] X Research source If you use the latter method to find the inverse of a matrix A, begin by setting up the formula [A | I]. Where I is the 3 x 3 identity matrix. [12] X Research source Then, use elementary row operations to reduce the left-hand side of the formula to I. The resulting formula will be [I | A-1], where A-1 is the inverse of A. [13] X Research source
