Question

To compute ${A^{ - 1}}$ to a given equation A = \left( {\begin{array}{*{20}{c}} 2&{ - 3} \\\ { - 4}&7 \end{array}} \right)and show that $2{A^{ - 1}} = 9I - A$.

Answer 1

The vector method of a given problem is to transform the inverse vector method and to solve both sides of the equation to prove that the equation is equal, then using the dot product and adjacent method to solve such a form of the equation to determine the given equation.

Formula used: The inverse vector formula,${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)$
I = \left( {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right)
Where,
The adjacent value of the vector method is adj\left( A \right) = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}} \\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right)
The modulus of $A$ is, we using dot product,
$\left| A \right| = {a_{11}}.{a_{22}} - {a_{21}}.{a_{12}}$

Complete step-by-step answer:
Given that,
A = \left( {\begin{array}{*{20}{c}} 2&{ - 3} \\\ { - 4}&7 \end{array}} \right)
Find ${A^{ - 1}}$ and $2{A^{ - 1}} = 9I - A$
We first compute ${A^{ - 1}}$ ,
And the given formula is ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)$
To interchange the given equation,
We get,
$\Rightarrow$ \left( {\begin{array}{*{20}{c}} 7&4 \\\ 3&2 \end{array}} \right)
Then find the value of $\left| A \right|$
$\left| A \right| = {a_{11}}.{a_{22}} - {a_{21}}.{a_{12}}$
And the given value by
${a_{11}} = 7,{a_{12}} = 4,{a_{21}} = 3,{a_{22}} = 2$
Substituting a given equation,
$\Rightarrow$ $\left| A \right| = \left| {14 - 12} \right|$
We get,
$\Rightarrow$ $\left| A \right| = 2$…………..$(1)$
Then we find the adj\left( A \right) = {\left( {\begin{array}{*{20}{c}} {{a_{_{11}}}}&{{a_{12}}} \\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right)^T}
adj = {\left( {\begin{array}{*{20}{c}} 7&4 \\\ 3&2 \end{array}} \right)^T}
To transpose the above equation,
We get,
$\Rightarrow$ adj(A) = \left( {\begin{array}{*{20}{c}} 7&3 \\\ 4&2 \end{array}} \right)……………$(2)$
L.H.S
to substituting both equation $(1)$ and $(2)$,
Simplifying a given equation,
$\Rightarrow$ {A^{ - 1}} = \dfrac{1}{2}\left( {\begin{array}{*{20}{c}} 7&3 \\\ 4&2 \end{array}} \right)
To rearrange a given equation,
Here,
2{A^{ - 1}} = \left( {\begin{array}{*{20}{c}} 7&3 \\\ 4&2 \end{array}} \right)
Left hand side of the equation is 2{A^{ - 1}} = \left( {\begin{array}{*{20}{c}} 7&3 \\\ 4&2 \end{array}} \right)
R.H.S
The equation is given by,
$9I - A$
Substituting the $I$ value in above equation,
We get,
= 9\left( {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 2&{ - 3} \\\ { - 4}&7 \end{array}} \right)
By solving the equation,
We get,
$=$ \left( {\begin{array}{*{20}{c}} 9&0 \\\ 0&9 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 2&{ - 3} \\\ { - 4}&7 \end{array}} \right)
On simplifying a above equation,
$\Rightarrow$ \left( {\begin{array}{*{20}{c}} 7&3 \\\ 4&2 \end{array}} \right)
Here, the right hand side of the equation is given by,
$\Rightarrow$ 9I - A = \left( {\begin{array}{*{20}{c}} 7&3 \\\ 4&2 \end{array}} \right)
Hence this $2{A^{ - 1}} = 9I - A$ proved

Note: The challenge is calculating the matrix value and proving the equation. Matrix-vector multiplication, vector multiplication, or scalar dot vector product, depending on the context. The number of columns in the first matrix must be equal to the number of rows in the second matrix if we want to perform the multiplication among these matrices. I is an identity matrix.