How To Divide Matrices With Pictures

Note that [A] * [B]-1 and [B]-1 * [A] are not the same problem. You may need to solve both to find all possible solutions. For example, instead of (13263913)÷(7423){\displaystyle {\begin{pmatrix}13&26\39&13\end{pmatrix}}\div {\begin{pmatrix}7&4\2&3\end{pmatrix}}}, write (13263913)∗(7423)−1{\displaystyle {\begin{pmatrix}13&26\39&13\end{pmatrix}}{\begin{pmatrix}7&4\2&3\end{pmatrix}}^{-1}}. You may also need to calculate (7423)−1∗(13263913){\displaystyle {\begin{pmatrix}7&4\2&3\end{pmatrix}}^{-1}{\begin{pmatrix}13&26\39&13\end{pmatrix}}}, which may have a different answer.

The term “divisor matrix” is a little loose, since this is not technically a division problem. For [A] * [B]-1, this refers to matrix [B]. In our example problem, this is (7423){\displaystyle {\begin{pmatrix}7&4\2&3\end{pmatrix}}}. A matrix that has an inverse is called “invertible” or “non-singular. " Matrices without an inverse are “singular. "

For example, if [A] is a 4 x 3 matrix (4 rows, 3 columns) and [B] is a 2 x 2 matrix (2 rows, 2 columns), there is no solution. [A] * [B]-1 does not work since 3 ≠ 2, and [B]-1 * [A] does not work since 2 ≠ 4. Note that the inverse [B]-1 always has the same number of rows and columns as the original matrix [B]. There’s no need to calculate the inverse to complete this step. In our example problem, both matrices are 2 x 2s, so they can be multiplied in either order.

2 x 2 matrix: The determinant of the matrix (abcd){\displaystyle {\begin{pmatrix}a&b\c&d\end{pmatrix}}} is ad - bc. [7] X Research source In other words, take the product of the main diagonal (top left to bottom right), then subtract the product of the anti-diagonal (top right to bottom left). For example, the matrix (7423){\displaystyle {\begin{pmatrix}7&4\2&3\end{pmatrix}}} has the determinant (7)(3) - (4)(2) = 21 - 8 = 13. This is nonzero, so it is possible to find the inverse.

3 x 3 matrix: Choose any element and cross out the row and column it belongs to. Find the determinant of the remaining 2 x 2 matrix, multiply by the chosen element, and refer to a matrix sign chart to determine the sign. Repeat this for the other two elements in the same row or column as the first one you chose, then sum all three determinants. Read this article for step-by-step instructions and tips to speed this up. Larger matrices: Using a graphing calculator or software is recommended. The method is similar to the 3 x 3 matrix method, but is tedious by hand. [9] X Research source For example, to find the determinant of a 4 x 4 matrix, you need to find the determinants of four 3 x 3 matrices.

(7423){\displaystyle {\begin{pmatrix}7&4\2&3\end{pmatrix}}} → (3427){\displaystyle {\begin{pmatrix}3&4\2&7\end{pmatrix}}} Note: Most people use calculators to find the inverse of a 3 x 3 matrix or larger. If you’d like to calculate it by hand, refer to the end of this section.

(3427){\displaystyle {\begin{pmatrix}3&4\2&7\end{pmatrix}}} → (3−4−27){\displaystyle {\begin{pmatrix}3&-4\-2&7\end{pmatrix}}}

In our example, the determinant is 13. The reciprocal of this is 113{\displaystyle {\frac {1}{13}}}.

113∗(3−4−27){\displaystyle {\frac {1}{13}}*{\begin{pmatrix}3&-4\-2&7\end{pmatrix}}}=(313−413−213713){\displaystyle {\begin{pmatrix}{\frac {3}{13}}&{\frac {-4}{13}}\{\frac {-2}{13}}&{\frac {7}{13}}\end{pmatrix}}}

For the example problem, multiply (7423)∗(313−413−213713)=(1001){\displaystyle {\begin{pmatrix}7&4\2&3\end{pmatrix}}*{\begin{pmatrix}{\frac {3}{13}}&{\frac {-4}{13}}\{\frac {-2}{13}}&{\frac {7}{13}}\end{pmatrix}}={\begin{pmatrix}1&0\0&1\end{pmatrix}}}. Here’s a refresher on how to multiply matrices. Note: Matrix multiplication is not commutative: the order of the factors matters. However, when multiplying a matrix by its inverse, both options will result in the identity matrix. [14] X Research source

Adjoin the identity matrix I to the right side of your matrix. For example, [B] → [B | I ]. The identity matrix has “1” elements along the main diagonal, and “0” elements in all other positions. Perform row operations to reduce the matrix until the left side is in row-echelon form, then continue reducing until the left side is the identity matrix. Once the operation is complete, your matrix will be in the form [I | B-1]. In other words, the right side will be the inverse of the original matrix.

[A] * [B]-1 is the solution x for the problem x[B] = [A]. [B]-1 * [A] is the solution x for the problem [B]x = [A]. If this is part of an equation, make sure you are performing the same operation on both sides. If [A] = [C], then [B]-1[A] does not equal [C][B]-1, because the [B]-1 is on the left side of [A] but the right side of [C]. [18] X Research source

Returning to our original example, both (13263913){\displaystyle {\begin{pmatrix}13&26\39&13\end{pmatrix}}} and (313−413−213713){\displaystyle {\begin{pmatrix}{\frac {3}{13}}&{\frac {-4}{13}}\{\frac {-2}{13}}&{\frac {7}{13}}\end{pmatrix}}} are 2 x 2 matrices so the dimensions of the answer are also 2 x 2. To take a more complicated example, if [A] is a 4 x 3 matrix and [B]-1 is a 3 x 3 matrix, then the matrix [A] * [B]-1 has dimensions 4 x 3.

To find row 1, column 1 of [A][B]-1, find the dot product of [A] row 1 and [B]-1 column 1. That is, for a 2 x 2 matrix, calculate a1,1∗b1,1+a1,2∗b2,1{\displaystyle a_{1,1}b_{1,1}+a_{1,2}b_{2,1}}. In our example (13263913)∗(313−413−213713){\displaystyle {\begin{pmatrix}13&26\39&13\end{pmatrix}}{\begin{pmatrix}{\frac {3}{13}}&{\frac {-4}{13}}\{\frac {-2}{13}}&{\frac {7}{13}}\end{pmatrix}}}, row 1 column 1 of our answer is:(13∗313)+(26∗−213){\displaystyle (13{\frac {3}{13}})+(26*{\frac {-2}{13}})}=3+−4{\displaystyle =3+-4}=−1{\displaystyle =-1}

(13263913)∗(313−413−213713)=(−1107−5){\displaystyle {\begin{pmatrix}13&26\39&13\end{pmatrix}}{\begin{pmatrix}{\frac {3}{13}}&{\frac {-4}{13}}\{\frac {-2}{13}}&{\frac {7}{13}}\end{pmatrix}}={\begin{pmatrix}-1&10\7&-5\end{pmatrix}}} If you need to find the other solution, (313−413−213713)∗(13263913)=(−92193){\displaystyle {\begin{pmatrix}{\frac {3}{13}}&{\frac {-4}{13}}\{\frac {-2}{13}}&{\frac {7}{13}}\end{pmatrix}}{\begin{pmatrix}13&26\39&13\end{pmatrix}}={\begin{pmatrix}-9&2\19&3\end{pmatrix}}}