हिंदी

For What Value of X is the Matrix [ 6 − X 4 3 − X 1 ] Singular? - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\begin{vmatrix}6 - x & 4 \\ 3 - x & 1\end{vmatrix}\text{ is singular when its determinant is 0 .} \] 
\[ \Rightarrow \begin{vmatrix}6 - x & 4 \\ 3 - x & 1\end{vmatrix} = 0\] 
\[ \Rightarrow \left( 6 - x \right) - 4\left( 3 - x \right) = 0\] 
\[ \Rightarrow 6 - x - 12 + 4x = 0\] 
\[ \Rightarrow 3x - 6 = 0\] 
\[ \Rightarrow 3x = 6\] 
\[ \Rightarrow x = \frac{6}{3} = 2\] 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Determinants - Exercise 6.6 [पृष्ठ ९१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 6 Determinants
Exercise 6.6 | Q 41 | पृष्ठ ९१

संबंधित प्रश्न

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


Solve system of linear equations, using matrix method.

2x – y = –2

3x + 4y = 3


Solve the system of linear equations using the matrix method.

x − y + z = 4

2x + y − 3z = 0

x + y + z = 2


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 cost of each item per kg by matrix method.


Evaluate the following determinant:

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


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|.

 

For what value of x the matrix A is singular? 

\[A = \begin{bmatrix}x - 1 & 1 & 1 \\ 1 & x - 1 & 1 \\ 1 & 1 & x - 1\end{bmatrix}\]


Evaluate the following determinant:

\[\begin{vmatrix}1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38\end{vmatrix}\]


Evaluate :

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


Prove the following identity:

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

 


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

 

Prove that :

\[\begin{vmatrix}b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c\end{vmatrix} = 3abc - a^3 - b - c^3\]

 


Prove that :

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

 


2x + 3y = 10
x + 6y = 4


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


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

 

Find the value of x from the following : \[\begin{vmatrix}x & 4 \\ 2 & 2x\end{vmatrix} = 0\]


If ω is a non-real cube root of unity and n is not a multiple of 3, then  \[∆ = \begin{vmatrix}1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^n \\ \omega^n & \omega^{2n} & 1\end{vmatrix}\] 


Show that the following systems of linear equations is consistent and also find their solutions:
2x + 3y = 5
6x + 9y = 15


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


\[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\], find AB. Hence, solve the system of equations: x − 2y = 10, 2x + y + 3z = 8 and −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.


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.

 

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 − 2y + z = 0
3x + 6y − 5z = 0


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


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


Solve the following for x and y: \[\begin{bmatrix}3 & - 4 \\ 9 & 2\end{bmatrix}\binom{x}{y} = \binom{10}{ 2}\]


Consider the system of equations:
a1x + b1y + c1z = 0
a2x + b2y + c2z = 0
a3x + b3y + c3z = 0,
if \[\begin{vmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix}\]= 0, then the system has


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

 

If A = `[[1,1,1],[0,1,3],[1,-2,1]]` , find A-1Hence, solve the system of equations: 

x +y + z = 6

y + 3z = 11

and x -2y +z = 0


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


System of equations x + y = 2, 2x + 2y = 3 has ______


If `|(2x, 5),(8, x)| = |(6, 5),(8, 3)|`, then find x


`abs ((1, "a"^2 + "bc", "a"^3),(1, "b"^2 + "ca", "b"^3),(1, "c"^2 + "ab", "c"^3))`


If the system of equations 2x + 3y + 5 = 0, x + ky + 5 = 0, kx - 12y - 14 = 0 has non-trivial solution, then the value of k is ____________.


What is the nature of the given system of equations

`{:(x + 2y = 2),(2x + 3y = 3):}`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×