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Question: Find \({(AB)^{ - 1}},\) if \(A = \left[ \begin{array}{*{20}{c}} 1&{{\text{ }}2}&{{\text{ }}...

Find (AB)−1,{(AB)^{ - 1}}, if A = \left[ \begin{array}{*{20}{c}} 1&{{\text{ }}2}&{{\text{ }}3} \end{array} \\\ \begin{array}{*{20}{c}} 1&{ - 2}&{ - 3} \end{array} \\\ \right], B = \left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} 1 \\\ 1 \\\ 1 \end{array}}&{\begin{array}{*{20}{c}} { - 1} \\\ 2 \\\ { - 2} \end{array}} \end{array}} \right]

Explanation

Solution

Here first of all we will find the product of the given two matrices. Then the adjoint of the matrix and the determinant of the product of the matrix and at last will place the values in the standard formula. (AB)−1=adj(AB)∣AB∣{(AB)^{ - 1}} = \dfrac{{adj(AB)}}{{\left| {AB} \right|}}

Complete step by step solution:
First of all multiply the two given matrices.
AB = \left[ \begin{array}{*{20}{c}} 1&{{\text{ }}2}&{{\text{ }}3} \end{array} \\\ \begin{array}{*{20}{c}} 1&{ - 2}&{ - 3} \end{array} \\\ \right]\left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} 1 \\\ 1 \\\ 1 \end{array}}&{\begin{array}{*{20}{c}} { - 1} \\\ {{\text{ }}2} \\\ { - 2} \end{array}} \end{array}} \right]
Now, apply the properties of the multiplication of matrices by row into column.
AB = \left[ {\begin{array}{*{20}{c}} {1(1) + 2(1) + 3(1)}&{1( - 1) + 2(2) + 3( - 2)} \\\ {1(1) - 2(1) - 3(1)}&{1( - 1) - 2(2) - 3( - 2)} \end{array}} \right]
Open the brackets and simplify the above expressions –
AB = \left[ {\begin{array}{*{20}{c}} {1 + 2 + 3}&{ - 1 + 4 - 6} \\\ {1 - 2 - 3}&{ - 1 - 4 + 6} \end{array}} \right]
Always remember when you are having two negative terms, you have to do addition and sign of minus. When you are having one positive term and one negative term then do subtraction and give sign of bigger numbers.
AB = \left[ {\begin{array}{*{20}{c}} 6&{ - 3} \\\ { - 4}&1 \end{array}} \right] ..... (I)
Now, the adjoint of the matrix with 2×22 \times 2 order, is the transpose of the co-factor of the matrix.
adjAB = {\left[ {\begin{array}{*{20}{c}} 1&4 \\\ 3&6 \end{array}} \right]^{'}}
In transpose we change rows into columns and columns into rows.
adjAB = \left[ {\begin{array}{*{20}{c}} 1&3 \\\ 4&6 \end{array}} \right] .... (II)
Take determinant of the matrix from the equation (I)
\left| {AB} \right| = \left| {\begin{array}{*{20}{c}} 6&{ - 3} \\\ { - 4}&1 \end{array}} \right|
Expand the determinant –
∣AB∣=∣6−(−4)(−3)∣ ∣AB∣=∣6−12∣ ∣AB∣=∣−6∣ ∣AB∣=6  ..... (III)  \left| {AB} \right| = \left| {6 - ( - 4)( - 3)} \right| \\\ \left| {AB} \right| = \left| {6 - 12} \right| \\\ \left| {AB} \right| = \left| { - 6} \right| \\\ \left| {AB} \right| = 6\,{\text{ }}.....{\text{ (III)}} \\\
Now place the values in the standard formula from the equation (II) and (III)
(AB)−1=adj(AB)∣AB∣{(AB)^{ - 1}} = \dfrac{{adj(AB)}}{{\left| {AB} \right|}}
\Rightarrow {(AB)^{ - 1}} = \dfrac{{\left[ {\begin{array}{*{20}{c}} 1&3 \\\ 4&6 \end{array}} \right]}}{6}
The above equation can be re-written as –
\Rightarrow {(AB)^{ - 1}} = \dfrac{1}{6}\left[ {\begin{array}{*{20}{c}} 1&3 \\\ 4&6 \end{array}} \right] is the required answer.

Note: Always remember that the product of two matrices is only possible when the inner dimensions are equal, means that the number of columns of the first matrix is equal to the number of the rows of the second matrix. If the product of the matrix does not exist then the inverse of the product of the two matrices also does not exist. Also, the determinant of the matrix should be not equal to zero to get its inverse.