Advertisements
Advertisements
Question
Using properties of determinants, show that `|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|` = 4abc.
Solution
L.H.S. = `|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|`
Applying C1 → C1 – (C2 + C3), we get
L.H.S. = `|(0, "a", "b"),(-2"c", "a" + "c", "c"),(-2"c", "c", "b" + "c")|`
Taking (– 2) common from C1, we get
L.H.S. = ` - 2|(0, "a", "b"),("c", "a" + "c", "c"),("c", "c", "b" + "c")|`
Applying C2 → C2 – C1 and C3 → C3 – C1, we get
L.H.S. = `-2|(0, "a", "b"),("c", "a", 0),("c", 0, "b")|`
= - 2[0(ab – 0) – a(bc – 0) + b(0 – ac)]
= – 2(0 – abc – abc)
= – 2(– 2abc)
= 4abc
= R.H.S.
APPEARS IN
RELATED QUESTIONS
By using properties of determinants, show that:
`|(1,1,1),(a,b,c),(a^3, b^3,c^3)|` = (a-b)(b-c)(c-a)(a+b+c)
Evaluate `|(x, y, x+y),(y, x+y, x),(x+y, x, y)|`
Using properties of determinants, prove that \[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix} = a^2 \left( a + x + y + z \right)\] .
Without expanding the determinants, show that `|("b" + "c", "bc", "b"^2"c"^2),("c" + "a", "ca", "c"^2"a"^2),("a" + "b", "ab", "a"^2"b"^2)|` = 0
Without expanding the determinants, show that `|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|` = 0
Prove that `|(x + y, y + z, z + x),(z + x, x + y, y + z),(y + z, z + x, x + y)| = 2|(x, y, z),(z, x, y),(y, z, x)|`
Answer the following question:
If `|("a", 1, 1),(1, "b", 1),(1, 1, "c")|` = 0 then show that `1/(1 - "a") + 1/(1 - "b") + 1/(1 - "c")` = 1
Evaluate: `|("a" - "b" - "c", 2"a", 2"a"),(2"b", "b" - "c" - "a", 2"b"),(2"c", 2"c", "c" - "a" - "b")|`
Prove that: `|(y + z, z, y),(z, z + x, x),(y, x, x + y)|` = 4xyz
If `[(4 - x, 4 + x, 4 + x),(4 + x, 4 - x, 4 + x),(4 + x, 4 + x, 4 - x)]` = 0, then find values of x.
If the value of a third order determinant is 12, then the value of the determinant formed by replacing each element by its co-factor will be 144.
`|(x + 1, x + 2, x + "a"),(x + 2, x + 3, x + "b"),(x + 3, x + 4, x + "c")|` = 0, where a, b, c are in A.P.
In a third order matrix B, bij denotes the element in the ith row and jth column. If
bij = 0 for i = j
= 1 for > j
= – 1 for i < j
Then the matrix is
`f : {1, 2, 3) -> {4, 5}` is not a function, if it is defined by which of the following?
Which of the following is correct?
If A, B and C are the angles of a triangle ABC, then `|(sin2"A", sin"C", sin"B"),(sin"C", sin2"B", sin"A"),(sin"B", sin"A", sin2"C")|` = ______.
The value of the determinant `|(6, 0, -1),(2, 1, 4),(1, 1, 3)|` is ______.
Without expanding determinant find the value of `|(10,57,107),(12,64,124),(15,78,153)|`
Without expanding the determinant, find the value of `|(10,57,107),(12,64,124),(15,78,153)|`
