हिंदी

If a = [Aij] is a 3 × 3 Scalar Matrix Such that A11 = 2, Then Write the Value of |A|.

Advertisements
Advertisements

प्रश्न

If A = [aij] is a 3 × 3 scalar matrix such that a11 = 2, then write the value of |A|.

 
Advertisements

उत्तर

A scalar matrix is a diagonal matrix, in which all the diagonal elements are equal to a given scalar number.
\[\text{ Given: }A = \left[ a_{i j} \right]\text{ is 3} \times\text{ 3 matrix, where }a_{11} = 2 \] 
\[ \Rightarrow A = \begin{bmatrix} 2 & 0 & 0\\0 & 2 & 0\\0 & 0 & 2 \end{bmatrix} \] 
\[ \Rightarrow \left| A \right| = \begin{vmatrix} 2 & 0 & 0\\0 & 2 & 0\\ 0 & 0 & 2 \end{vmatrix}\] 
\[ = 2 \times \begin{vmatrix} 2 & 0\\ 0 & 2 \end{vmatrix} \left[\text{ Expanding along }C_1 \right]\] 
\[ = 2 \times 2 \times 2 = 8\] 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Determinants - Exercise 6.6 [पृष्ठ ९०]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 5 Determinants
Exercise 6.6 | Q 14 | पृष्ठ ९०

संबंधित प्रश्न

Solve the system of linear equations using the matrix method.

5x + 2y = 3

3x + 2y = 5


Evaluate the following determinant:

\[\begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}\]


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

\[\begin{vmatrix}a & b & c \\ a + 2x & b + 2y & c + 2z \\ x & y & z\end{vmatrix}\]


Evaluate :

\[\begin{vmatrix}x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda\end{vmatrix}\]


Evaluate the following:

\[\begin{vmatrix}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{vmatrix}\]


Evaluate the following:

\[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix}\]


Prove the following identities:

\[\begin{vmatrix}y + z & z & y \\ z & z + x & x \\ y & x & x + y\end{vmatrix} = 4xyz\]


Prove the following identity:

\[\begin{vmatrix}2y & y - z - x & 2y \\ 2z & 2z & z - x - y \\ x - y - z & 2x & 2x\end{vmatrix} = \left( x + y + z \right)^3\]


Prove the following identity:

`|(a^3,2,a),(b^3,2,b),(c^3,2,c)| = 2(a-b) (b-c) (c-a) (a+b+c)`

 


​Solve the following determinant equation:

\[\begin{vmatrix}1 & 1 & x \\ p + 1 & p + 1 & p + x \\ 3 & x + 1 & x + 2\end{vmatrix} = 0\]

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:

(3, −2), (8, 8) and (5, 2)


Prove that :

\[\begin{vmatrix}a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b\end{vmatrix} = \left( a + b + c \right)^3\]

 


Prove that :

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

 


Prove that :

\[\begin{vmatrix}a^2 & bc & ac + c^2 \\ a^2 + ab & b^2 & ac \\ ab & b^2 + bc & c^2\end{vmatrix} = 4 a^2 b^2 c^2\]

Prove that :

\[\begin{vmatrix}x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4\end{vmatrix} = 16 \left( 3x + 4 \right)\]

x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1


Find the value of the determinant 
\[\begin{bmatrix}101 & 102 & 103 \\ 104 & 105 & 106 \\ 107 & 108 & 109\end{bmatrix}\]

 


Write the value of 

\[\begin{vmatrix}\sin 20^\circ & - \cos 20^\circ\\ \sin 70^\circ& \cos 70^\circ\end{vmatrix}\]

Find the value of the determinant \[\begin{vmatrix}243 & 156 & 300 \\ 81 & 52 & 100 \\ - 3 & 0 & 4\end{vmatrix} .\]


The value of the determinant

\[\begin{vmatrix}a^2 & a & 1 \\ \cos nx & \cos \left( n + 1 \right) x & \cos \left( n + 2 \right) x \\ \sin nx & \sin \left( n + 1 \right) x & \sin \left( n + 2 \right) x\end{vmatrix}\text{ is independent of}\]

 


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

 


The value of the determinant  

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




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


Solve the following system of equations by matrix method:

3x + 4y + 7z = 14

2x − y + 3z = 4

x + 2y − 3z = 0


Solve the following system of equations by matrix method:
 x − y + z = 2
2x − y = 0
2y − z = 1


Solve the following system of equations by matrix method:
 x + y + z = 6
x + 2z = 7
3x + y + z = 12


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 school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.


x + y − 6z = 0
x − y + 2z = 0
−3x + y + 2z = 0


x + y + z = 0
x − y − 5z = 0
x + 2y + 4z = 0


Let a, b, c be positive real numbers. The following system of equations in x, y and z 

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1, \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1, - \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \text { has }\]
(a) no solution
(b) unique solution
(c) infinitely many solutions
(d) finitely many solutions

x + y = 1
x + z = − 6
x − y − 2z = 3


Solve the following equations by using inversion method.

x + y + z = −1, x − y + z = 2 and x + y − z = 3


`abs ((("b" + "c"^2), "a"^2, "bc"),(("c" + "a"^2), "b"^2, "ca"),(("a" + "b"^2), "c"^2, "ab")) =` ____________.


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 


The value (s) of m does the system of equations 3x + my = m and 2x – 5y = 20 has a solution satisfying the conditions x > 0, y > 0.


The system of simultaneous linear equations kx + 2y – z = 1,  (k – 1)y – 2z = 2 and (k + 2)z = 3 have a unique solution if k equals:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×