Answer:
1. [tex]A^-^1=\left[\begin{array}{cc}\frac{3}{10}&\frac{1}{5}\\\frac{1}{10}&\frac{2}{5} \end{array}\right][/tex]
2. [tex]|B|=0[/tex]
Step-by-step explanation:
We have the matrix:
[tex]A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right][/tex]
It's a 2×2 matrix (This means that the matrix has two rows and two columns).
1. We have to find the inverse of A.
For a 2×2 matrix the inverse is:
If you have [tex]A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right][/tex]
[tex]A^-^1=\frac{1}{|A|} \left[\begin{array}{cc}d&-b\\-c&a\end{array}\right][/tex]
And,
[tex]|A|[/tex] is the determinant of the matrix, the determinant has to be different from zero.
If [tex]|A|=0[/tex] then the matrix doesn't have inverse.
[tex]|A|=ad-bc[/tex]
Then,
[tex]A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right][/tex]
[tex]a=4, b=-2, c=-1, d=3[/tex]
First we are going to calculate the determinant:
[tex]|A|=ad-bc\\|A|=4.3-(-2).(-1)=12-2\\|A|=10[/tex]
The determinant is different from zero, then the matrix has inverse.
Then the inverse of A is:
[tex]A^-^1=\frac{1}{|A|} \left[\begin{array}{cc}d&-b\\-c&a\end{array}\right][/tex]
[tex]A^-^1=\frac{1}{10} \left[\begin{array}{cc}3&-(-2)\\-(-1)&4\end{array}\right]\\\\\\A^-^1=\frac{1}{10} \left[\begin{array}{cc}3&2\\1&4\end{array}\right]\\\\\\A^-^1=\left[\begin{array}{cc}\frac{3}{10}&\frac{2}{10}\\\frac{1}{10}&\frac{4}{10} \end{array}\right]\\\\\\A^-^1=\left[\begin{array}{cc}\frac{3}{10}&\frac{1}{5}\\\frac{1}{10}&\frac{2}{5} \end{array}\right][/tex]
2. We have the matrix,
[tex]B=\left[\begin{array}{cc}6&3\\4&2\end{array}\right][/tex]
[tex]a=6, b=3, c=4,d=2[/tex]
We have to calculate the determinant:
[tex]|B|=ad-bc\\|B|=6.2-3.4=12-12\\|B|=0[/tex]
We said that a matrix can have an inverse only if its determinant is nonzero.
In this case [tex]|B|=0[/tex] then, the matrix B doesn't have inverse.
