Question

B=2*A=2*(matrix{2}{3}{1 2 {3~} 4 5 {6~}})=(matrix{2}{3}{2 4 {6~} 8 10 {12~}})
M_3=delim{|}{matrix{3}{3}{1 {-1} {1~} 2 {-2} {2~} 1 1 {-1~}}}{|}=1*(-2)*(-1)+(-1)*2*1+1*2*(-1)-
{}-1*(-2)*1-(-1)*2*(-1)-1*2*(-1)=0~{doubleright}~r(A)=2
det A = delim{|}{A}{|} = delim{|}{matrix{3}{3}{{a_{11}} {a_{12}} {a_{13}} {a_{21}} {a_{22}} {a_{23}} {a_{31}} {a_{32}} {a_{33}}} }{|} = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} +
{}+ a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} - a_{12}a_{21}a_{33} - a_{11}a_{23}a_{32}
кто сможет решить дам 90 баллов

Answer 1

Відповідь:

Відповіль у поясненні.

Покрокове пояснення:

1

B=2A=2(matrix{2}{3}{1 2 {3~} 4 5 {6~}})=(matrix{2}{3}{2 4 {6~} 8 10 {12~}})

Correct.

M_3=delim{|}{matrix{3}{3}{1 {-1} {1~} 2 {-2} {2~} 1 1 {-1~}}}{|}=1*({-2})*({-1})+({-1})*2*1+1*2*({-1})-

{}-1*({-2})*1-({-1})*2*({-1})-1*2*({-1})=0~{doubleright}~r(A)=2

Correct.

det A = delim{|}{A}{|} = delim{|}{matrix{3}{3}{{a_{11}} {a_{12}} {a_{13}} {a_{21}} {a_{22}} {a_{23}} {a_{31}} {a_{32}} {a_{33}}} }{|} = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} +

{}+ a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} - a_{12}a_{21}a_{33} - a_{11}a_{23}a_{32}

The determinant of a 3x3 matrix can be calculated using the following formula:

det A = a11 * a22 * a33 + a12 * a23 * a31 + a13 * a21 * a32 - a13 * a22 * a31 - a12 * a21 * a33 - a11 * a23 * a32

Therefore:

det A = 1 * 4 * -1 + (-1) * 2 * 1 + 1 * 2 * 2 - 1 * 2 * 1 - (-1) * 2 * -1 - 1 * 4 * 2

det A = 0

Since the determinant of A is 0, then A is a singular matrix, which means that it has no inverse.

In other words, there is no matrix B such that AB = BA = I, where I is the identity matrix.