मराठी

If ⎡ ⎢ ⎣ 1 0 0 0 1 0 0 0 1 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ X Y Z ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ 1 − 1 0 ⎤ ⎥ ⎦ , Find X, Y and Z. - Mathematics

Advertisements
Advertisements

प्रश्न

If \[\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ - 1 \\ 0\end{bmatrix}\], find x, y and z.

उत्तर

Here,
\[\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ - 1 \\ 0\end{bmatrix} \]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ - 1 \\ 0\end{bmatrix}\]
\[ \therefore x = 1, y = - 1\text{ and }z = 0\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Solution of Simultaneous Linear Equations - Exercise 8.3 [पृष्ठ २१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 8 Solution of Simultaneous Linear Equations
Exercise 8.3 | Q 1 | पृष्ठ २१

संबंधित प्रश्‍न

If `|[x+1,x-1],[x-3,x+2]|=|[4,-1],[1,3]|`, then write the value of x.


Solve the system of linear equations using the matrix method.

x − y + 2z = 7

3x + 4y − 5z = −5

2x − y + 3z = 12


Evaluate

\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}^2 .\]


Find the value of x, if
\[\begin{vmatrix}2 & 4 \\ 5 & 1\end{vmatrix} = \begin{vmatrix}2x & 4 \\ 6 & x\end{vmatrix}\]


Evaluate the following determinant:

\[\begin{vmatrix}6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2\end{vmatrix}\]


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

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


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


Using properties of determinants prove that

\[\begin{vmatrix}x + 4 & 2x & 2x \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4\end{vmatrix} = \left( 5x + 4 \right) \left( 4 - x \right)^2\]


If \[\begin{vmatrix}a & b - y & c - z \\ a - x & b & c - z \\ a - x & b - y & c\end{vmatrix} =\] 0, then using properties of determinants, find the value of  \[\frac{a}{x} + \frac{b}{y} + \frac{c}{z}\]  , where \[x, y, z \neq\] 0


Find the value of \[\lambda\]  so that the points (1, −5), (−4, 5) and \[\lambda\]  are collinear.


Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, −6) and (5, 4).


Using determinants, find the area of the triangle with vertices (−3, 5), (3, −6), (7, 2).


Using determinants, find the equation of the line joining the points

(1, 2) and (3, 6)


Using determinants, find the equation of the line joining the points

(3, 1) and (9, 3)


Prove that :

\[\begin{vmatrix}1 & a^2 + bc & a^3 \\ 1 & b^2 + ca & b^3 \\ 1 & c^2 + ab & c^3\end{vmatrix} = - \left( a - b \right) \left( b - c \right) \left( c - a \right) \left( a^2 + b^2 + c^2 \right)\]

 


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

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


2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11


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


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.


If w is an imaginary cube root of unity, find the value of \[\begin{vmatrix}1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w\end{vmatrix}\]


Write the value of the determinant \[\begin{vmatrix}2 & - 3 & 5 \\ 4 & - 6 & 10 \\ 6 & - 9 & 15\end{vmatrix} .\]


Write the value of the determinant \[\begin{vmatrix}2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x\end{vmatrix}\]


If \[\begin{vmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{vmatrix} = 8\] , write the value of x.


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

 


Let \[A = \begin{bmatrix}1 & \sin \theta & 1 \\ - \sin \theta & 1 & \sin \theta \\ - 1 & - \sin \theta & 1\end{bmatrix},\text{ where 0 }\leq \theta \leq 2\pi . \text{ Then,}\]




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:
 x + y − z = 3
2x + 3y + z = 10
3x − y − 7z = 1


Solve the following system of equations by matrix method:
 8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5


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.

A company produces three products every day. Their production on a certain day is 45 tons. It is found that the production of third product exceeds the production of first product by 8 tons while the total production of first and third product is twice the production of second product. Determine the production level of each product using matrix method.


The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.

 

3x − y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z = 0


For the system of equations:
x + 2y + 3z = 1
2x + y + 3z = 2
5x + 5y + 9z = 4


If A = `[(1, -1, 2),(3, 0, -2),(1, 0, 3)]`, verify that A(adj A) = (adj A)A


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 `|(x + a, beta, y),(a, x + beta, y),(a, beta, x + y)|` = 0, then 'x' is equal to


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 ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×