Question

Find the inverse of the matrix\left( {\begin{array}{*{20}{c}} { - 3}&2 \\\ 5&{ - 3} \end{array}} \right). Hence find the matrix P satisfying the matrix equation:
P\left( {\begin{array}{*{20}{c}} { - 3}&2 \\\ 5&{ - 3} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&2 \\\ 2&{ - 1} \end{array}} \right)

Answer 1

Hint : 1. When a matrix A is multiplied with its inverse A-1, it gives out identity matrix as a result.
i.e. $A{A^{ - 1}} = I$
2. Inverse of a matrix is defined as the adjoint of the matrix divided by its determinant.
i.e. ${A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}$
3. If we are given thatA = \left( {\begin{array}{*{20}{c}} a&b; \\\ d&c; \end{array}} \right)
Then, adjoint of a matrix is defined as adjA = \left( {\begin{array}{*{20}{c}} c&{ - b} \\\ d&a; \end{array}} \right)

Complete step-by-step answer :
We are given that A = \left( {\begin{array}{*{20}{c}} { - 3}&2 \\\ 5&{ - 3} \end{array}} \right)
1. Then by using the hint adjA = \left( {\begin{array}{*{20}{c}} { - 3}&{ - 2} \\\ { - 5}&{ - 3} \end{array}} \right)
2. Determinant of matrix A can be found out as,
$\begin{gathered} \left| A \right| = \left( {( - 3) \times ( - 3)} \right) - \left( {5 \times 2} \right) \\\ \left| A \right| = 9 - 10 \\\ \left| A \right| = - 1 \\\ \end{gathered}$
3. Now for finding the inverse of the matrix, then
${A^{ - 1}} = \dfrac{{adjA}}{{\left| A \right|}}$
Putting the values of $adjA$and $\left| A \right|$ in the above equation, we get
\begin{gathered} {A^{ - 1}} = \dfrac{{\left( {\begin{array}{*{20}{c}} { - 3}&{ - 2} \\\ { - 5}&{ - 3} \end{array}} \right)}}{{ - 1}} \\\ {A^{ - 1}} = \left( {\begin{array}{*{20}{c}} 3&2 \\\ 5&3 \end{array}} \right) \\\ \end{gathered}
The inverse of the matrix A is{A^{ - 1}} = \left( {\begin{array}{*{20}{c}} 3&2 \\\ 5&3 \end{array}} \right).
4. Now for finding out the value of P such that
P\left( {\begin{array}{*{20}{c}} { - 3}&2 \\\ 5&{ - 3} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&2 \\\ 2&{ - 1} \end{array}} \right) ...... (1)
Since we know that
A = \left( {\begin{array}{*{20}{c}} { - 3}&2 \\\ 5&{ - 3} \end{array}} \right)
Let, B = \left( {\begin{array}{*{20}{c}} 1&2 \\\ 2&{ - 1} \end{array}} \right)
So Eqn (1) can also be written as
$PA = B$
Multiplying by ${A^{ - 1}}$both sides we get
$PA{A^{ - 1}} = B{A^{ - 1}}$
Since we know that when a matrix is multiplied it with its inverse than it results in identity matrix
i.e. $A{A^{ - 1}} = I$
$PI = B{A^{ - 1}}$
$P = B{A^{ - 1}}$
Since, now we know the values of ${A^{ - 1}}$and$B$. Putting in the above equation, we get
\begin{gathered} P = \left( {\begin{array}{*{20}{c}} 1&2 \\\ 2&{ - 1} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 3&2 \\\ 5&3 \end{array}} \right) \\\ P = \left( {\begin{array}{*{20}{c}} {(1 \times 3) + (2 \times 5)}&{(1 \times 2) + (2 \times 3)} \\\ {(2 \times 3) + ( - 1 \times 5)}&{(2 \times 2) + ( - 1 \times 3)} \end{array}} \right) \\\ P = \left( {\begin{array}{*{20}{c}} {3 + 10}&{2 + 6} \\\ {6 + ( - 5)}&{4 + ( - 3)} \end{array}} \right) \\\ P = \left( {\begin{array}{*{20}{c}} {13}&8 \\\ {6 - 5}&{4 - 3} \end{array}} \right) \\\ P = \left( {\begin{array}{*{20}{c}} {13}&8 \\\ 1&1 \end{array}} \right) \\\ \end{gathered}
Therefore, the value of P is:
P = \left( {\begin{array}{*{20}{c}} {13}&8 \\\ 1&1 \end{array}} \right)

Note : When an identity matrix $I$is multiplied to any matrix then the resultant matrix is same as the multiplied matrix
i.e. $A{I^{ - 1}} = A$