Advertisements
Advertisements
Question
By using properties of determinants, prove that `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|` = 0.
Advertisements
Solution
L.H.S. = `|(x + y, y + z, z + x),(z, x, y),(1, 1, 1)|`
Applying R1 → R1 + R2, we get
L.H.S. = `|(x + y + z, x + y + z, x + y + z),(z, x , y),(1, 1, 1)|`
Taking (x + y + z) common from R1, we get
L.H.S. = `(x + y + z)|(1, 1, 1),(z, x, y),(1, 1, 1)|`
= (x + y + z) (0) …[∵ R1 and R3 are identical]
= 0
= R.H.S.
APPEARS IN
RELATED QUESTIONS
Using the properties of determinants, prove the following:
`|[1,x,x+1],[2x,x(x-1),x(x+1)],[3x(1-x),x(x-1)(x-2),x(x+1)(x-1)]|=6x^2(1-x^2)`
If ` f(x)=|[a,-1,0],[ax,a,-1],[ax^2,ax,a]| ` , using properties of determinants find the value of f(2x) − f(x).
By using properties of determinants, show that:
`|(a-b-c, 2a,2a),(2b, b-c-a,2b),(2c,2c, c-a-b)| = (a + b + c)^2`
Using properties of determinants show that
`[[1,1,1+x],[1,1+y,1],[1+z,1,1]] = xyz+ yz +zx+xy.`
Using properties of determinants, prove that \[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix} = a^2 \left( a + x + y + z \right)\] .
Using properties of determinants, prove that
`|[b+c , a ,a ] ,[ b , a+c, b ] ,[c , c, a+b ]|` = 4abc
Using properties of determinants, find the value of x for which
`|(4-"x",4+"x",4+"x"),(4+"x",4-"x",4+"x"),(4+"x",4+"x",4-"x")|= 0`
Without expanding evaluate the following determinant:
`|(2, 3, 4),(5, 6, 8),(6x, 9x, 12x)|`
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)|`
Without expanding determinants show that
`|(1, 3, 6),(6, 1, 4),(3, 7, 12)| + 4|(2, 3, 3),(2, 1, 2),(1, 7, 6)| = 10|(1, 2, 1),(3, 1, 7),(3, 2, 6)|`
Answer the following question:
Evaluate `|(101, 102, 103),(106, 107, 108),(1, 2, 3)|` by using properties
Evaluate: `|(x^2 - x + 1, x - 1),(x + 1, x + 1)|`
Evaluate: `|("a" - "b" - "c", 2"a", 2"a"),(2"b", "b" - "c" - "a", 2"b"),(2"c", 2"c", "c" - "a" - "b")|`
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.
Let 'A' be a square matrix of order 3 × 3, then |KA| is equal to:
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)|`
By using properties of determinant prove that `|(x+y,y+z,z+x),(z,x,y),(1,1,1)|` = 0
By using properties of determinants, prove that
`|(x+y, y+z, z+x),(z, x, y),(1, 1, 1)|` = 0
Without expanding determinant find the value of `|(10, 57, 107),(12, 64, 124),(15, 78, 153)|`
