There is no significant method for matrix division. We cannot divide two matrices. But, we can multiply one matrix by the inverse of the other in order to divide the matrices. To solve the system of linear equations, we can use division property of matrices.

A matrix may be divided by a scalar and also by a matrix. Here, we will learn how to divide matrices with the help of few examples.

Solved Examples

Question 1: If A = $\begin{bmatrix}
6 &0 \\
-4 &-4 \\
2 &6
\end{bmatrix}$ and B = $\begin{bmatrix}
8 &-2 \\
7 &6 \\
-9 &3
\end{bmatrix}$ are the two given matrices, find the matrix X such that 3A + 2X = 4B
Solution:
We have 3A + 2X = 4B
2X = 4B - 3A
2X = 4 $\begin{bmatrix}
8 &-2 \\
7 &6 \\
-9 &3
\end{bmatrix}$ - 3 $\begin{bmatrix}
6 &0 \\
-4 &-4 \\
2 &6
\end{bmatrix}$

2X = $\begin{bmatrix}
32 &-8 \\
-28 &24 \\
-36 &12
\end{bmatrix}$ - $\begin{bmatrix}
18 &0 \\
-12 &-12 \\
6 &18
\end{bmatrix}$

X = $\frac{1}{2}$ $\begin{bmatrix}
14 &-8 \\
-16 &36\\
-42 &-6
\end{bmatrix}$

X = $\begin{bmatrix}
7 &-4 \\
-8 &13 \\
21 &-3
\end{bmatrix}$
Then, X = $\begin{bmatrix}
7 &-4 \\
-8 &13 \\
21 &-3
\end{bmatrix}$

Question 2: If A = $\begin{bmatrix}
6 &2 \\
3 &2
\end{bmatrix}$ and B = $\begin{bmatrix}
6 &-4 \\
2 &1
\end{bmatrix}$, then calculate B/A
Solution:
Let C = B/A = B.A-1

For this, first we have to calculate A-1.

We know that, A-1 = $\frac{adj A}{\left | A \right |}$

For adj A, first we have to find the cofactor matrix of A and then the transpose of that cofactor matrix.

Cofactor matrix of A = $\begin{bmatrix}
2 &-3\\
-2 &6
\end{bmatrix}$ = C

Now, $C^T$ = $\begin{bmatrix}
2 &-2 \\
-3 &6
\end{bmatrix}$

Again, $\left | A \right |$ = 12 - 6 = 6.

Then, A-1 = $\frac{1}{6}$ $\begin{bmatrix}
2 &-2 \\
-3 &6
\end{bmatrix}$

For $\frac{B}{A}$ = D = B x A-1
= B x $\frac{adj A}{\left | A \right |}$

D = $\begin{bmatrix}
6 &-4 \\
2 &1
\end{bmatrix}$ . $\frac{1}{6}$ $\begin{bmatrix}
2 &-2 \\
-3 &6
\end{bmatrix}$

= $\frac{1}{6}$. $\begin{bmatrix}
24&-36 \\
1&2
\end{bmatrix}$

D = $\begin{bmatrix}
4&-6 \\
\frac{1}{6}& \frac{1}{3}
\end{bmatrix}$