Question

A = ???? 4 −2

Answer 1

Answer:

1. $A^-^1=\left[\begin{array}{cc}\frac{3}{10}&\frac{1}{5}\\\frac{1}{10}&\frac{2}{5} \end{array}\right]$

2. $|B|=0$

Step-by-step explanation:

We have the matrix:

$A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]$

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 $A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$

$A^-^1=\frac{1}{|A|} \left[\begin{array}{cc}d&-b\\-c&a\end{array}\right]$

And,

$|A|$ is the determinant of the matrix, the determinant has to be different from zero.

If $|A|=0$ then the matrix doesn't have inverse.

$|A|=ad-bc$

Then,

$A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]$

$a=4, b=-2, c=-1, d=3$

First we are going to calculate the determinant:

$|A|=ad-bc\\|A|=4.3-(-2).(-1)=12-2\\|A|=10$

The determinant is different from zero, then the matrix has inverse.

Then the inverse of A is:

$A^-^1=\frac{1}{|A|} \left[\begin{array}{cc}d&-b\\-c&a\end{array}\right]$

$A^-^1=\frac{1}{10} \left[\begin{array}{cc}3&-(-2)\\-(-1)&4\end{array}\right]\\\\\\A^-^1=\frac{1}{10} \left[\begin{array}{cc}3&2\\1&4\end{array}\right]\\\\\\A^-^1=\left[\begin{array}{cc}\frac{3}{10}&\frac{2}{10}\\\frac{1}{10}&\frac{4}{10} \end{array}\right]\\\\\\A^-^1=\left[\begin{array}{cc}\frac{3}{10}&\frac{1}{5}\\\frac{1}{10}&\frac{2}{5} \end{array}\right]$

2. We have the matrix,

$B=\left[\begin{array}{cc}6&3\\4&2\end{array}\right]$

$a=6, b=3, c=4,d=2$

We have to calculate the determinant:

$|B|=ad-bc\\|B|=6.2-3.4=12-12\\|B|=0$

We said that a matrix can have an inverse only if its determinant is nonzero.

In this case $|B|=0$ then, the matrix B doesn't have inverse.