Advertisements
Advertisements
Question
Evaluate: `|(x + 4, x, x),(x, x + 4, x),(x, x, x + 4)|`
Solution
We have, `|(x + 4, x, x),(x, x + 4, x),(x, x, x + 4)|`
= `|(3x + 4, x + 4, x + 4),(x, x + 4, x),(x, x, x + 4)|` .....[Applying R1 → R1 + R2 + R3]
= `(3x + 4)|(1, 1, 1),(x, x + 4, x),(x, x, x + 4)|`
= `(3x + 4) |(0, 0, 1),(-4, 4, x),(0, -4, x + 4)|` ...[Applying C1 → C1 – C2, C2 → C2 – C3]
= 16(3x + 4)
APPEARS IN
RELATED QUESTIONS
Using properties of determinants, prove that
`|((x+y)^2,zx,zy),(zx,(z+y)^2,xy),(zy,xy,(z+x)^2)|=2xyz(x+y+z)^3`
Using the property of determinants and without expanding, prove that:
`|(1, bc, a(b+c)),(1, ca, b(c+a)),(1, ab, c(a+b))| = 0`
By using properties of determinants, show that:
`|(-a^2, ab, ac),(ba, -b^2, bc),(ca,cb, -c^2)| = 4a^2b^2c^2`
Without expanding the determinant, prove that
`|(a, a^2,bc),(b,b^2, ca),(c, c^2,ab)| = |(1, a^2, a^3),(1, b^2, b^3),(1, c^2, c^3)|`
Using properties of determinants, prove that `|(1,1,1+3x),(1+3y, 1,1),(1,1+3z,1)| = 9(3xyz + xy + yz+ zx)`
Using properties of determinants, prove the following :
Prove the following using properties of determinants :
\[\begin{vmatrix}a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b\end{vmatrix} = 2\left( a + b + c \right) {}^3\]
Using properties of determinants, prove that:
`|(a,b,b+c),(c,a,c+a),(b,c,a+b)|` = (a+b+c)(a-c)2
Without expanding determinants, find the value of `|(2014, 2017, 1),(2020, 2023, 1),(2023, 2026, 1)|`
Without expanding determinants, prove that `|("a"_1, "b"_1, "c"_1),("a"_2, "b"_2, "c"_2),("a"_3, "b"_3, "c"_3)| = |("b"_1, "c"_1, "a"_1),("b"_2, "c"_2, "a"_2),("b"_3, "c"_3, "a"_3)| = |("c"_1, "a"_1, "b"_1),("c"_2, "a"_2, "b"_2),("c"_3, "a"_3, "b"_3)|`
Solve the following equation:
`|(x + 2, x + 6, x - 1),(x + 6, x - 1, x + 2),(x - 1, x + 2, x + 6)|` = 0
If `|("x"^"k", "x"^("k" + 2), "x"^("k" + 3)),("y"^"k", "y"^("k" + 2), "y"^("k" + 3)),("z"^"k", "z"^("k" + 2), "z"^("k" + 3))|` = (x - y) (y - z) (z - x)`(1/"x"+ 1/"y" + 1/"z") ` then
Answer the following question:
Evaluate `|(101, 102, 103),(106, 107, 108),(1, 2, 3)|` by using properties
Answer the following question:
By using properties of determinant prove that `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|` = 0
Answer the following question:
Without expanding determinant show that
`|(x"a", y"b", z"c"),("a"^2, "b"^2, "c"^2),(1, 1, 1)| = |(x, y, z),("a", "b", "c"),("bc", "ca", "ab")|`
Answer the following question:
Without expanding determinant show that
`|(l, "m", "n"),("e", "d", "f"),("u", "v", "w")| = |("n", "f", "w"),(l, "e", "u"),("m", "d", "v")|`
Evaluate: `|(0, xy^2, xz^2),(x^2y, 0, yz^2),(x^2z, zy^2, 0)|`
Evaluate: `|("a" - "b" - "c", 2"a", 2"a"),(2"b", "b" - "c" - "a", 2"b"),(2"c", 2"c", "c" - "a" - "b")|`
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.
The determinant `|("b"^2 - "ab", "b" - "c", "bc" - "ac"),("ab" - "a"^2, "a" - "b", "b"^2 - "ab"),("bc" - "ac", "c" - "a", "ab" - "a"^2)|` equals ______.
If the determinant `|(x + "a", "p" + "u", "l" + "f"),("y" + "b", "q" + "v", "m" + "g"),("z" + "c", "r" + "w", "n" + "h")|` splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.
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
If `|(α, 3, 4),(1, 2, 1),(1, 4, 1)|` = 0, then the value of α is ______.
The value of the determinant `|(6, 0, -1),(2, 1, 4),(1, 1, 3)|` is ______.
By using properties of determinant prove that
`|(x+ y,y+z, z+x ),(z, x,y),(1,1,1)|` = 0
Without expanding determinants, 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)|`
Without expanding determinant find the value of `|(10, 57, 107),(12, 64, 124),(15, 78, 153)|`
