Advertisements
Advertisements
प्रश्न
Using the factor theorem it is found that a + b, b + c and c + a are three factors of the determinant
The other factor in the value of the determinant is
विकल्प
4
2
a + b + c
none of these
Advertisements
उत्तर
\[\Delta = \begin{vmatrix} - 2a & a + b & a + c\\b + a & - 2b & b + c\\c + a & c + b & - 2c \end{vmatrix}
\text{ Let } a + b = 2C, b + c = 2A\text{ and }c + a = 2B . \]
\[ \Rightarrow a + b + b + c + c + a = 2A + 2B + 2C\]
\[ \Rightarrow 2\left( a + b + c \right) = 2\left( A + B + C \right)\]
\[ \Rightarrow a + b + c = A + B + C\]
Also,
\[a = \left( a + b + c \right) - \left( b + c \right) = \left( A + B + C \right) - 2A = B + C - A\]
Similarly,
\[b = C + A - B, c = A + B - C\]
\[\Delta = \begin{vmatrix} 2A - 2B - 2C & 2C & 2B\\ 2C & 2B - 2C - 2A & 2A\\ 2B & 2A & 2C - 2A - 2B \end{vmatrix} = 8 \times \begin{vmatrix} A - B - C & C & B\\ C & B - C - A & A\\ B & A & C - A - B \end{vmatrix} \left[\text{ taking out 2 common from }R_1 R_2 R_3 \right]\]
\[ = 8 \times \begin{vmatrix} A - B & C + B & B\\ B - A & B - C & A\\ B + A & C - B & C - A - B \end{vmatrix} \left[\text{ Applying }C_1 \to C_1 + C_2 , C_2 \to C_2 + C_3 \right]\]
\[ = 8 \times \begin{vmatrix} \left( A - B \right) & C + B & B\\ 0 & 2B & A + B\\ 2B & 0 & C - B \end{vmatrix} \left[\text{ Applying }R_2 \to R_1 + R_2 , R_3 \to R_2 + R_3 \right]\]
\[ = 8 \times \left\{ \left( A - B \right)\begin{vmatrix} 2B & A + B\\ 0 & C - B \end{vmatrix} + \left( 2B \right) \times \begin{vmatrix} C + B & B\\ 2B & A + B \end{vmatrix} \right\} \left[\text{ Expanding along }C_1 \right]\]
\[ = 16 B\left\{ \left( A - B \right)\left( C - B \right) + \left( C + B \right)\left( A + B \right) - 2 B^2 \right\}\]
\[ = 32 ABC\]
\[ = 32\left( \frac{b + c}{2} \right)\left( \frac{c + a}{2} \right)\left( \frac{a + b}{2} \right)\]
\[ = 4\left( a + b \right)\left( b + c \right)\left( c + a \right)\]
Hence, 4 is the other factor of the determinant .
APPEARS IN
संबंधित प्रश्न
Examine the consistency of the system of equations.
2x − y = 5
x + y = 4
Evaluate the following determinant:
\[\begin{vmatrix}x & - 7 \\ x & 5x + 1\end{vmatrix}\]
Evaluate
\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}^2 .\]
Find the value of x, if
\[\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}\]
For what value of x the matrix A is singular?
\[A = \begin{bmatrix}x - 1 & 1 & 1 \\ 1 & x - 1 & 1 \\ 1 & 1 & x - 1\end{bmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}a + b & 2a + b & 3a + b \\ 2a + b & 3a + b & 4a + b \\ 4a + b & 5a + b & 6a + b\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin^2 23^\circ & \sin^2 67^\circ & \cos180^\circ \\ - \sin^2 67^\circ & - \sin^2 23^\circ & \cos^2 180^\circ \\ \cos180^\circ & \sin^2 23^\circ & \sin^2 67^\circ\end{vmatrix}\]
\[\begin{vmatrix}0 & b^2 a & c^2 a \\ a^2 b & 0 & c^2 b \\ a^2 c & b^2 c & 0\end{vmatrix} = 2 a^3 b^3 c^3\]
Prove that :
Prove that :
3x + y = 19
3x − y = 23
3x + ay = 4
2x + ay = 2, a ≠ 0
3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11
2x + y − 2z = 4
x − 2y + z = − 2
5x − 5y + z = − 2
Solve each of the following system of homogeneous linear equations.
2x + 3y + 4z = 0
x + y + z = 0
2x − y + 3z = 0
Find the real values of λ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions
\[2 \lambda x - 2y + 3z = 0\]
\[ x + \lambda y + 2z = 0\]
\[ 2x + \lambda z = 0\]
State whether the matrix
\[\begin{bmatrix}2 & 3 \\ 6 & 4\end{bmatrix}\] is singular or non-singular.
Evaluate \[\begin{vmatrix}4785 & 4787 \\ 4789 & 4791\end{vmatrix}\]
If \[A = \left[ a_{ij} \right]\] is a 3 × 3 diagonal matrix such that a11 = 1, a22 = 2 a33 = 3, then find |A|.
If \[A = \begin{bmatrix}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{bmatrix}\]. Write the cofactor of the element a32.
The determinant \[\begin{vmatrix}b^2 - ab & b - c & bc - ac \\ ab - a^2 & a - b & b^2 - ab \\ bc - ca & c - a & ab - a^2\end{vmatrix}\]
If \[\begin{vmatrix}a & p & x \\ b & q & y \\ c & r & z\end{vmatrix} = 16\] , then the value of \[\begin{vmatrix}p + x & a + x & a + p \\ q + y & b + y & b + q \\ r + z & c + z & c + r\end{vmatrix}\] is
Solve the following system of equations by matrix method:
x + y + z = 6
x + 2z = 7
3x + y + z = 12
Solve the following system of equations by matrix method:
If A = `[(1, 2, 0), (-2, -1, -2), (0, -1, 1)]`, find A−1. Using A−1, solve the system of linear equations x − 2y = 10, 2x − y − z = 8, −2y + z = 7.
A total amount of ₹7000 is deposited in three different saving bank accounts with annual interest rates 5%, 8% and \[8\frac{1}{2}\] % respectively. The total annual interest from these three accounts is ₹550. Equal amounts have been deposited in the 5% and 8% saving accounts. Find the amount deposited in each of the three accounts, with the help of matrices.
The system of equations:
x + y + z = 5
x + 2y + 3z = 9
x + 3y + λz = µ
has a unique solution, if
(a) λ = 5, µ = 13
(b) λ ≠ 5
(c) λ = 5, µ ≠ 13
(d) µ ≠ 13
If \[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\] ,find A–1 and hence solve the system of equations x – 2y = 10, 2x + y + 3z = 8 and –2y + z = 7.
Show that \[\begin{vmatrix}y + z & x & y \\ z + x & z & x \\ x + y & y & z\end{vmatrix} = \left( x + y + z \right) \left( x - z \right)^2\]
x + y = 1
x + z = − 6
x − y − 2z = 3
System of equations x + y = 2, 2x + 2y = 3 has ______
If `alpha, beta, gamma` are in A.P., then `abs (("x" - 3, "x" - 4, "x" - alpha),("x" - 2, "x" - 3, "x" - beta),("x" - 1, "x" - 2, "x" - gamma)) =` ____________.
`abs ((1, "a"^2 + "bc", "a"^3),(1, "b"^2 + "ca", "b"^3),(1, "c"^2 + "ab", "c"^3))`
Solve the following system of equations x − y + z = 4, x − 2y + 2z = 9 and 2x + y + 3z = 1.
For what value of p, is the system of equations:
p3x + (p + 1)3y = (p + 2)3
px + (p + 1)y = p + 2
x + y = 1
consistent?
If the system of linear equations x + 2ay + az = 0; x + 3by + bz = 0; x + 4cy + cz = 0 has a non-zero solution, then a, b, c ______.
