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Ne4ueva [31]
3 years ago
6

A = ???? 4 −2

Mathematics
1 answer:
liraira [26]3 years ago
7 0

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 <u><em>different from zero</em></u>, then the matrix <em>has</em> 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.

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