Advertisements
Advertisements
Question
Answer the following question:
Find minor and cofactor of elements of the determinant:
`|(-1, 0, 4),(-2, 1, 3),(0, -4, 2)|`
Advertisements
Solution
Here, `|("a"_11, "a"_12, "a"_13),("a"_21, "a"_22, "a"_23),("a"_31, "a"_32, "a"_33)| = |(-1, 0, 4),(-2, 1, 3),(0, -4, 2)|`
M11 = `|(1, 3),(-4, 2)|` = 2 + 12 = 14
∴ C11 = (– 1)1+1 M11 = 1(14) = 14
M12 = `|(-2, 3),(0, 2)|` = – 4 – 0 = – 4
∴ C12 = (– 1)1+2 M12 = (– 1)(– 4) = 4
M13 = `|(-2, 1),(0, -4)|` = 8 – 0 = 8
∴ C13 = (– 1)1+3 M13 = 1(8) = 8
M21 = `|(0, 4),(-4, 2)|` = 0 + 16 =16
∴ C21 = (–1)2+1 M21 = (–1)(16) = – 16
M22 = `|(-1, 4),(0, 2)|` = –2 – 0 = – 2
∴ C22 = (–1)2+2 M22 = 1(–2) = –2
M23 = `|(-1, 0),(0, -4)|` = 4 – 0 = 4
∴ C23 = (–1)2+3 M23 = (–1)(4) = –4
M31 = `|(0, 4),(1, 3)|` = 0 – 4 = – 4
∴ C31 = (–1)3+1 M31 = 1(–4) = –4
M32 = `|(-1, 4),(-2, 3)|` = –3 + 8 = 5
∴ C32 = (–1)3+2 M32 = (–1)(5) = –5
M33 = `|(-1, 0),(-2, 1)|` = –1 + 0 = –1
∴ C33 = (–1)3+3 M33 = 1(–1) = –1
