हिंदी

Prove that : ∣ ∣ ∣ ∣ a + B B + C C + a B + C C + a A + B C + a A + B B + C ∣ ∣ ∣ ∣ = 2 ∣ ∣ ∣ ∣ a B C B C a C a B ∣ ∣ ∣ - Mathematics

Advertisements
Advertisements

प्रश्न

Prove that :

\[\begin{vmatrix}a + b & b + c & c + a \\ b + c & c + a & a + b \\ c + a & a + b & b + c\end{vmatrix} = 2\begin{vmatrix}a & b & c \\ b & c & a \\ c & a & b\end{vmatrix}\]

 

Advertisements

उत्तर

\[\text{ Let }\Delta = \begin{vmatrix} a + b & b + c & c + a \\b + c & c + a & a + b\\c + a & a + b & b + c \end{vmatrix}\] 

Using the property of determinants that if each element of a row or column is expressed as the sum of two or more quantities, the determinant is expressed as the sum of two or more determinants, we get
\[\Delta = \begin{vmatrix} a & b & c \\b & c & a\\c & a & b \end{vmatrix} + \begin{vmatrix} b & c & a \\c & a & b\\a & b & c \end{vmatrix} \] 
\[ = \begin{vmatrix} a & b & c \\b & c & a\\c & a & b \end{vmatrix} + \left( - 1 \right)\begin{vmatrix} a & c & b \\b & a & c\\c & b & a \end{vmatrix}\left[\text{ Applying }C_1 \leftrightarrow C_3\text{ in second determinant to get negative value of the deteminant }\right]\] 
\[ = \begin{vmatrix} a & b & c \\b & c & a\\c & a & b \end{vmatrix} | + \left( - 1 \right)\left( - 1 \right) \begin{vmatrix} a & b & c \\b & c & a\\c & a & b \end{vmatrix} \left[\text{ Applying }C_2 \leftrightarrow C_3 \right]\] 
\[ = 2 \begin{vmatrix} a & b & c \\b & c & a\\c & a & b \end{vmatrix} = RHS\] 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Determinants - Exercise 6.2 [पृष्ठ ५८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 6 Determinants
Exercise 6.2 | Q 13 | पृष्ठ ५८

संबंधित प्रश्न

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method.


Show that

\[\begin{vmatrix}\sin 10^\circ & - \cos 10^\circ \\ \sin 80^\circ & \cos 80^\circ\end{vmatrix} = 1\]


Evaluate

\[\begin{vmatrix}2 & 3 & - 5 \\ 7 & 1 & - 2 \\ - 3 & 4 & 1\end{vmatrix}\] by two methods.

 

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|.

 

Evaluate the following determinant:

\[\begin{vmatrix}a & h & g \\ h & b & f \\ g & f & c\end{vmatrix}\]


Evaluate the following determinant:

\[\begin{vmatrix}1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9\end{vmatrix}\]


Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}1 & a & a^2 - bc \\ 1 & b & b^2 - ac \\ 1 & c & c^2 - ab\end{vmatrix}\]


Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}\sin\alpha & \cos\alpha & \cos(\alpha + \delta) \\ \sin\beta & \cos\beta & \cos(\beta + \delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma + \delta)\end{vmatrix}\]


Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}\sin^2 A & \cot A & 1 \\ \sin^2 B & \cot B & 1 \\ \sin^2 C & \cot C & 1\end{vmatrix}, where A, B, C \text{ are the angles of }∆ ABC .\]


Prove the following identity:

\[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix} = a^2 \left( a + x + y + z \right)\]

 


Show that
`|(x-3,x-4,x-alpha),(x-2,x-3,x-beta),(x-1,x-2,x-gamma)|=0`, where α, β, γ are in A.P.

 


Using determinants show that the following points are collinear:

(2, 3), (−1, −2) and (5, 8)


Using determinants prove that the points (ab), (a', b') and (a − a', b − b') are collinear if ab' = a'b.

 

Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3) and (−5, −3). Are the given points collinear?


Prove that :

\[\begin{vmatrix}1 & b + c & b^2 + c^2 \\ 1 & c + a & c^2 + a^2 \\ 1 & a + b & a^2 + b^2\end{vmatrix} = \left( a - b \right) \left( b - c \right) \left( c - a \right)\]

 


Prove that :

\[\begin{vmatrix}\left( a + 1 \right) \left( a + 2 \right) & a + 2 & 1 \\ \left( a + 2 \right) \left( a + 3 \right) & a + 3 & 1 \\ \left( a + 3 \right) \left( a + 4 \right) & a + 4 & 1\end{vmatrix} = - 2\]

 


3x + ay = 4
2x + ay = 2, a ≠ 0


Given: x + 2y = 1
            3x + y = 4


x − y + 3z = 6
x + 3y − 3z = − 4
5x + 3y + 3z = 10


If \[A = \begin{bmatrix}0 & i \\ i & 1\end{bmatrix}\text{  and }B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\] , find the value of |A| + |B|.


If A and B are non-singular matrices of the same order, write whether AB is singular or non-singular.


For what value of x is the matrix  \[\begin{bmatrix}6 - x & 4 \\ 3 - x & 1\end{bmatrix}\]  singular?


If \[D_k = \begin{vmatrix}1 & n & n \\ 2k & n^2 + n + 2 & n^2 + n \\ 2k - 1 & n^2 & n^2 + n + 2\end{vmatrix} and \sum^n_{k = 1} D_k = 48\], then n equals

 


If \[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix}\] , then x = 

 


The determinant  \[\begin{vmatrix}b^2 - ab & b - c & bc - ac \\ ab - a^2 & a - b & b^2 - ab \\ bc - ca & c - a & ab - a^2\end{vmatrix}\]


 


Solve the following system of equations by matrix method:

\[\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2; x, y, z \neq 0\]

 


Show that each one of the following systems of linear equation is inconsistent:
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3


If \[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\] , find A−1. Using A−1, solve the system of linear equations  x − 2y = 10, 2x + y + 3z = 8, −2y + z = 7.

Given \[A = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}, B = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}\] , find BA and use this to solve the system of equations  y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17


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


If \[A = \begin{bmatrix}2 & 4 \\ 4 & 3\end{bmatrix}, X = \binom{n}{1}, B = \binom{ 8}{11}\]  and AX = B, then find n.

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 + = 7.


Write the value of `|(a-b, b- c, c-a),(b-c, c-a, a-b),(c-a, a-b, b-c)|`


If A = `[(2, 0),(0, 1)]` and B = `[(1),(2)]`, then find the matrix X such that A−1X = B.


Using determinants, find the equation of the line joining the points (1, 2) and (3, 6).


The existence of unique solution of the system of linear equations x + y + z = a, 5x – y + bz = 10, 2x + 3y – z = 6 depends on 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×