Question

Find the inverse of the matrix given below

3&{ - 5} \\\ { - 1}&2 \end{array}} \right|$$

Answer 1

Inverse of the matrix is the relation of the adjoint and determinant value of the matrix, which is represented as${A^{ - 1}} = \dfrac{{adj(A)}}{{\det (A)}}$ where the determinant and adjoint are calculated individually and are kept in the formula. The given matrix is a $2 \times 2$matrix that contains four elements in it.

Complete step-by-step answer:
A square matrix is the one that has the same number of rows as well as columns. The standard $2 \times 2$ square matrix is\left| {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right|.
Compare the given matrix A = \left| {\begin{array}{*{20}{c}} 3&{ - 5} \\\ { - 1}&2 \end{array}} \right| with the standard $2 \times 2$ matrix\left| {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right|.
By equating, we get the value as${\text{a = 3, b = - 5, c = - 1, d = 2}}$.
A determinant of a matrix is the scalar value of a matrix that can be computed from elements of a matrix in $\left( {ad - bc} \right)$ form. The matrix which has its determinant value as 0 is known as Singular or Non-invertible matrix.
First, we find the determinant value of the matrix,
Let $\det ({\text{A}})$represents the determinant value of the matrix A = \left| {\begin{array}{*{20}{c}} 3&{ - 5} \\\ { - 1}&2 \end{array}} \right|

$$\det A = \left[ {\left( {a \times d} \right) - \left( {b \times c} \right)} \right] \\\ = \left[ {\left( {3 \times 2} \right) - \left( {\left( { - 1} \right) \times \left( { - 5} \right)} \right)} \right] \\\ = \left[ {6 - 5} \right] \\\ = 1 \\\$$

Now the next step for the inverse of the matrix is to find the adjoint value of the matrix

a&b; \\\ c&d; \end{array}} \right|$$ In general, adjoint of the matrix is the transpose of the cofactor of the matrix. To find the adjoint of a $2 \times 2$ square matrix, swap the positions of a and d, and put a negative sign in front of b and c in the matrix $$adjA = {\left| {\begin{array}{*{20}{c}} d&{ - b} \\\ { - c}&a; \end{array}} \right|^{}}$$ We get $$a = 2,b = 5,c = 1,d = 3$$ So we can write adjoint of the matrix $$A = \left| {\begin{array}{*{20}{c}} 3&{ - 5} \\\ { - 1}&2 \end{array}} \right|$$ as, $$adjA = {\left| {\begin{array}{*{20}{c}} 2&5 \\\ 1&3 \end{array}} \right|^{}}$$ Finally, substitute $$adjA = {\left| {\begin{array}{*{20}{c}} 2&5 \\\ 1&3 \end{array}} \right|^{}}$$ and $\det (A) = 1$ in the formula $${A^{ - 1}} = \dfrac{{adj(A)}}{{\det (A)}}$$ to determine the inverse of the matrix.

{A^{ - 1}} = \dfrac{{adj(A)}}{{\det (A)}} \\
= \dfrac{1}{{\left| 1 \right|}}\left| {\begin{array}{{20}{c}}
2&5 \\
1&3
\end{array}} \right| \\
= \left| {\begin{array}{{20}{c}}
{\dfrac{2}{1}}&{\dfrac{5}{1}} \\
{\dfrac{1}{1}}&{\dfrac{3}{1}}
\end{array}} \right| \\
= \left| {\begin{array}{*{20}{c}}
2&5 \\
1&3
\end{array}} \right| \\

Hence, the inverse of the matrix $$A = \left| {\begin{array}{*{20}{c}} 3&{ - 5} \\\ { - 1}&2 \end{array}} \right|$$ is $${A^{ - 1}} = \left| {\begin{array}{*{20}{c}} 2&5 \\\ 1&3 \end{array}} \right|$$ **Note:** Inverse of a matrix A is applicable only when it satisfies the condition $$A \times {A^{ - 1}} = {A^{ - 1}} \times A = I$$ where $$I$$ represents the singular matrix.