Advertisements
Advertisements
प्रश्न
Prove that
\[\begin{vmatrix}- bc & b^2 + bc & c^2 + bc \\ a^2 + ac & - ac & c^2 + ac \\ a^2 + ab & b^2 + ab & - ab\end{vmatrix} = \left( ab + bc + ca \right)^3\]
उत्तर
\[∆ = \begin{vmatrix}- bc & b^2 + bc & c^2 + bc \\ a^2 + ac & - ac & c^2 + ac \\ a^2 + ab & b^2 + ab & - ab\end{vmatrix}\]
\[ = \frac{1}{abc}\begin{vmatrix}- abc & a b^2 + abc & a c^2 + abc \\ a^2 b + abc & - abc & c^2 b + abc \\ a^2 c + abc & b^2 c + abc & - abc\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }aR_1 , R_2 \text{ to }bR_2\text{ and }R_3 \text{ to }cR_3\text{ and then dividing by abc }\right]\]
\[ = \frac{abc}{abc}\begin{vmatrix}- bc & ab + ac & ac + ab \\ ab + bc & - ac & cb + ab \\ ac + bc & bc + ac & - ab\end{vmatrix} \left[\text{ Taking out a, b and c common from the three columns }\right]\]
\[\begin{vmatrix}ab + bc + ca & ab + bc + ca & ab + bc + ca \\ ab + bc & - ac & cb + ab \\ ac + bc & bc + ac & - ab\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }R_1 + R_2 + R_3 \right]\]
\[ = (ab + bc + ca)\begin{vmatrix}1 & 1 & 1 \\ ab + bc & - ac & cb + ab \\ ac + bc & bc + ac & - ab\end{vmatrix}\]
\[ = (ab + bc + ca)\begin{vmatrix}0 & 0 & 1 \\ 0 & - (ab + bc + ac) & cb + ab \\ ac + bc + ab & bc + ac + ab & - ab\end{vmatrix} \left[\text{ Applying }C_1 \text{ to }C_1 - C_3 \text{ and }C_2 \text{ to }C_2 - C_3 \right]\]
\[ = (ab + bc + ca)\begin{vmatrix}0 & - (ab + bc + ac) \\ ac + bc + ab & bc + ac + ab\end{vmatrix}\]
\[ = (ab + bc + ca)(ab + bc + ac )^2 \]
\[ = (ab + bc + ca )^3\]
Hence proved.
APPEARS IN
संबंधित प्रश्न
If `|[2x,5],[8,x]|=|[6,-2],[7,3]|`, write the value of x.
Examine the consistency of the system of equations.
2x − y = 5
x + y = 4
Solve system of linear equations, using matrix method.
5x + 2y = 4
7x + 3y = 5
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}1^2 & 2^2 & 3^2 & 4^2 \\ 2^2 & 3^2 & 4^2 & 5^2 \\ 3^2 & 4^2 & 5^2 & 6^2 \\ 4^2 & 5^2 & 6^2 & 7^2\end{vmatrix}\]
Prove that
\[\begin{vmatrix}\frac{a^2 + b^2}{c} & c & c \\ a & \frac{b^2 + c^2}{a} & a \\ b & b & \frac{c^2 + a^2}{b}\end{vmatrix} = 4abc\]
Prove the following identities:
\[\begin{vmatrix}x + \lambda & 2x & 2x \\ 2x & x + \lambda & 2x \\ 2x & 2x & x + \lambda\end{vmatrix} = \left( 5x + \lambda \right) \left( \lambda - x \right)^2\]
Prove the following identity:
\[\begin{vmatrix}2y & y - z - x & 2y \\ 2z & 2z & z - x - y \\ x - y - z & 2x & 2x\end{vmatrix} = \left( x + y + z \right)^3\]
Find the area of the triangle with vertice at the point:
(0, 0), (6, 0) and (4, 3)
Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, −6) and (5, 4).
Prove that :
Prove that :
5x + 7y = − 2
4x + 6y = − 3
x + y + z + 1 = 0
ax + by + cz + d = 0
a2x + b2y + x2z + d2 = 0
3x − y + 2z = 3
2x + y + 3z = 5
x − 2y − z = 1
Write the value of the determinant
\[\begin{bmatrix}2 & 3 & 4 \\ 2x & 3x & 4x \\ 5 & 6 & 8\end{bmatrix} .\]
Find the value of the determinant
\[\begin{bmatrix}101 & 102 & 103 \\ 104 & 105 & 106 \\ 107 & 108 & 109\end{bmatrix}\]
If \[A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}\text{ and B} = \begin{bmatrix}1 & - 4 \\ 3 & - 2\end{bmatrix},\text{ find }|AB|\]
If \[\begin{vmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{vmatrix} = 8\] , write the value of x.
The value of \[\begin{vmatrix}5^2 & 5^3 & 5^4 \\ 5^3 & 5^4 & 5^5 \\ 5^4 & 5^5 & 5^6\end{vmatrix}\]
\[\begin{vmatrix}\log_3 512 & \log_4 3 \\ \log_3 8 & \log_4 9\end{vmatrix} \times \begin{vmatrix}\log_2 3 & \log_8 3 \\ \log_3 4 & \log_3 4\end{vmatrix}\]
The value of the determinant
If x, y, z are different from zero and \[\begin{vmatrix}1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z\end{vmatrix} = 0\] , then the value of x−1 + y−1 + z−1 is
The value of \[\begin{vmatrix}1 & 1 & 1 \\ {}^n C_1 & {}^{n + 2} C_1 & {}^{n + 4} C_1 \\ {}^n C_2 & {}^{n + 2} C_2 & {}^{n + 4} C_2\end{vmatrix}\] is
Solve the following system of equations by matrix method:
3x + 4y − 5 = 0
x − y + 3 = 0
Solve the following system of equations by matrix method:
x + y + z = 3
2x − y + z = − 1
2x + y − 3z = − 9
Show that each one of the following systems of linear equation is inconsistent:
4x − 5y − 2z = 2
5x − 4y + 2z = −2
2x + 2y + 8z = −1
If \[A = \begin{bmatrix}1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1\end{bmatrix}\] , find A−1. Using A−1, solve the system of linear equations x − 2y = 10, 2x − y − z = 8, −2y + z = 7
Use product \[\begin{bmatrix}1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4\end{bmatrix}\begin{bmatrix}- 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2\end{bmatrix}\] to solve the system of equations x + 3z = 9, −x + 2y − 2z = 4, 2x − 3y + 4z = −3.
2x + 3y − z = 0
x − y − 2z = 0
3x + y + 3z = 0
Let \[X = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}, A = \begin{bmatrix}1 & - 1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}3 \\ 1 \\ 4\end{bmatrix}\] . If AX = B, then X is equal to
x + y = 1
x + z = − 6
x − y − 2z = 3
Transform `[(1, 2, 4),(3, -1, 5),(2, 4, 6)]` into an upper triangular matrix by using suitable row transformations
Show that if the determinant ∆ = `|(3, -2, sin3theta),(-7, 8, cos2theta),(-11, 14, 2)|` = 0, then sinθ = 0 or `1/2`.
If a, b, c are non-zeros, then the system of equations (α + a)x + αy + αz = 0, αx + (α + b)y + αz = 0, αx+ αy + (α + c)z = 0 has a non-trivial solution if
What is the nature of the given system of equations
`{:(x + 2y = 2),(2x + 3y = 3):}`
Let A = `[(i, -i),(-i, i)], i = sqrt(-1)`. Then, the system of linear equations `A^8[(x),(y)] = [(8),(64)]` has ______.
