Advertisements
Advertisements
प्रश्न
Show that x = 2 is a root of the equation
Advertisements
उत्तर
\[\text{ Let }∆ = \begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 3 & 2x & x + 2\end{vmatrix}\]
\[ = \begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 3 - x & 2x + 6 & x + 3\end{vmatrix} \left[\text{ Applying }R_3\text{ to }R_3 - R_1 \right]\]
\[ = \left( x + 3 \right)\begin{vmatrix}x & - 6 & - 1 \\ 2 & - 3x & x - 3 \\ - 1 & 2 & 1\end{vmatrix} \]
\[ = \left( x + 3 \right)\begin{vmatrix}x - 2 & 3x - 6 & - x + 2 \\ 2 & - 3x & x - 3 \\ - 1 & 2 & 1\end{vmatrix} \left[\text{ Applying } R_1 \text{ to }R_1 - R_2 \right]\]
\[ = \left( x + 3 \right)\left( x - 2 \right)\begin{vmatrix}1 & 3 & - 1 \\ 2 & - 3x & x - 3 \\ - 1 & 2 & 1\end{vmatrix} \]
\[ = \left( x + 3 \right)\left( x - 2 \right)\begin{vmatrix}1 & 3 & 0 \\ 2 & - 3x & x - 1 \\ - 1 & 2 & 0\end{vmatrix} \left[\text{ Applying }C_3 \text{ to }C_3 + C_1 \right]\]
\[ = \left( x + 3 \right)\left( x - 2 \right)\left( x - 1 \right)\begin{vmatrix}1 & 3 & 0 \\ 2 & - 3x & 1 \\ - 1 & 2 & 0\end{vmatrix} \]
\[ = \left( x + 3 \right)\left( x - 2 \right)\left( x - 1 \right)\left\{ - 1\begin{vmatrix}1 & 3 \\ - 1 & 2\end{vmatrix} \right\} \left[ \text{ Expanding along }C_3 \right]\]
\[ = - 5\left( x + 3 \right)\left( x - 2 \right)\left( x - 1 \right)\]
\[x = 2, - 3, 1\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following determinant:
\[\begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}\]
Evaluate
\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}^2 .\]
If \[A = \begin{bmatrix}2 & 5 \\ 2 & 1\end{bmatrix} \text{ and } B = \begin{bmatrix}4 & - 3 \\ 2 & 5\end{bmatrix}\] , verify that |AB| = |A| |B|.
If A \[\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}\] , then show that |3 A| = 27 |A|.
Find the value of x, if
\[\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}\]
Find the value of x, if
\[\begin{vmatrix}3 & x \\ x & 1\end{vmatrix} = \begin{vmatrix}3 & 2 \\ 4 & 1\end{vmatrix}\]
Evaluate the following determinant:
\[\begin{vmatrix}1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}a & b & c \\ a + 2x & b + 2y & c + 2z \\ x & y & z\end{vmatrix}\]
\[\begin{vmatrix}b^2 + c^2 & ab & ac \\ ba & c^2 + a^2 & bc \\ ca & cb & a^2 + b^2\end{vmatrix} = 4 a^2 b^2 c^2\]
Prove the following identities:
\[\begin{vmatrix}x + \lambda & 2x & 2x \\ 2x & x + \lambda & 2x \\ 2x & 2x & x + \lambda\end{vmatrix} = \left( 5x + \lambda \right) \left( \lambda - x \right)^2\]
Solve the following determinant equation:
Solve the following determinant equation:
Prove that :
2x − y = 17
3x + 5y = 6
x+ y = 5
y + z = 3
x + z = 4
Find the real values of λ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions
\[2 \lambda x - 2y + 3z = 0\]
\[ x + \lambda y + 2z = 0\]
\[ 2x + \lambda z = 0\]
If a, b, c are non-zero real numbers and if the system of equations
(a − 1) x = y + z
(b − 1) y = z + x
(c − 1) z = x + y
has a non-trivial solution, then prove that ab + bc + ca = abc.
Find the value of the determinant
\[\begin{bmatrix}101 & 102 & 103 \\ 104 & 105 & 106 \\ 107 & 108 & 109\end{bmatrix}\]
Find the value of x from the following : \[\begin{vmatrix}x & 4 \\ 2 & 2x\end{vmatrix} = 0\]
If \[∆_1 = \begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{vmatrix}, ∆_2 = \begin{vmatrix}1 & bc & a \\ 1 & ca & b \\ 1 & ab & c\end{vmatrix},\text{ then }\]}
Let \[A = \begin{bmatrix}1 & \sin \theta & 1 \\ - \sin \theta & 1 & \sin \theta \\ - 1 & - \sin \theta & 1\end{bmatrix},\text{ where 0 }\leq \theta \leq 2\pi . \text{ Then,}\]
If \[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix}\] , then x =
Let \[f\left( x \right) = \begin{vmatrix}\cos x & x & 1 \\ 2\sin x & x & 2x \\ \sin x & x & x\end{vmatrix}\] \[\lim_{x \to 0} \frac{f\left( x \right)}{x^2}\] is equal to
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 + 2z = 8
2y − 3z = 3
x − 2y + 6z = −2
Show that the following systems of linear equations is consistent and also find their solutions:
x − y + z = 3
2x + y − z = 2
−x −2y + 2z = 1
The prices of three commodities P, Q and R are Rs x, y and z per unit respectively. A purchases 4 units of R and sells 3 units of P and 5 units of Q. B purchases 3 units of Q and sells 2 units of P and 1 unit of R. Cpurchases 1 unit of P and sells 4 units of Q and 6 units of R. In the process A, B and C earn Rs 6000, Rs 5000 and Rs 13000 respectively. If selling the units is positive earning and buying the units is negative earnings, find the price per unit of three commodities by using matrix method.
2x + 3y − z = 0
x − y − 2z = 0
3x + y + 3z = 0
If \[\begin{bmatrix}1 & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ - 1 \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}\] , find x, y and z.
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
The number of solutions of the system of equations:
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
Write the value of `|(a-b, b- c, c-a),(b-c, c-a, a-b),(c-a, a-b, b-c)|`
Solve the following system of equations by using inversion method
x + y = 1, y + z = `5/3`, z + x = `4/3`
The cost of 4 dozen pencils, 3 dozen pens and 2 dozen erasers is ₹ 60. The cost of 2 dozen pencils, 4 dozen pens and 6 dozen erasers is ₹ 90. Whereas the cost of 6 dozen pencils, 2 dozen pens and 3 dozen erasers is ₹ 70. Find the cost of each item per dozen by using matrices
If the system of equations x + ky - z = 0, 3x - ky - z = 0 & x - 3y + z = 0 has non-zero solution, then k is equal to ____________.
Let A = `[(1,sin α,1),(-sin α,1,sin α),(-1,-sin α,1)]`, where 0 ≤ α ≤ 2π, then:
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
If `|(x + a, beta, y),(a, x + beta, y),(a, beta, x + y)|` = 0, then 'x' is equal to
