Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
Here,
\[ A = \begin{bmatrix}3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}\]
\[\left| A \right|=3 \left( 3 - 0 \right) + 4\left( 2 - 5 \right) + 2\left( 0 - 3 \right)\]
\[ = 9 - 12 - 6\]
\[ = - 9\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}3 & 5 \\ 0 & 1\end{vmatrix} = 3, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 5 \\ 1 & 1\end{vmatrix} = 3, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 3 \\ 1 & 0\end{vmatrix} = - 3\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 4 & 2 \\ 0 & 1\end{vmatrix} = 4, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix} = 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & - 4 \\ 1 & 0\end{vmatrix} = - 4\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 4 & 2 \\ 3 & 5\end{vmatrix} = - 26, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & 2 \\ 2 & 5\end{vmatrix} = - 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & - 4 \\ 2 & 3\end{vmatrix} = 17\]
\[adj A = \begin{bmatrix}3 & 3 & - 3 \\ 4 & 1 & - 4 \\ - 26 & - 11 & 17\end{bmatrix}^T \]
\[ = \begin{bmatrix}3 & 4 & - 26 \\ 3 & 1 & - 11 \\ - 3 & - 4 & 17\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 9}\begin{bmatrix}3 & 4 & - 26 \\ 3 & 1 & - 11 \\ - 3 & - 4 & 17\end{bmatrix}\]
\[AX = B\]
Here,
\[A = \begin{bmatrix}3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}- 1 \\ 7 \\ 2\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{- 9}\begin{bmatrix}3 & 4 & - 26 \\ 3 & 1 & - 11 \\ - 3 & - 4 & 17\end{bmatrix}\begin{bmatrix}- 1 \\ 7 \\ 2\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{- 9}\begin{bmatrix}- 3 + 28 - 52 \\ - 3 + 7 - 22 \\ 3 - 28 + 34\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}- 27 \\ - 18 \\ 9\end{bmatrix}\]
\[ \therefore x = 3, y = 2\text{ and }z = - 1\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following determinant:
\[\begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}\]
\[∆ = \begin{vmatrix}\cos \alpha \cos \beta & \cos \alpha \sin \beta & - \sin \alpha \\ - \sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{vmatrix}\]
Find the value of x, if
\[\begin{vmatrix}3 & x \\ x & 1\end{vmatrix} = \begin{vmatrix}3 & 2 \\ 4 & 1\end{vmatrix}\]
Find the integral value of x, if \[\begin{vmatrix}x^2 & x & 1 \\ 0 & 2 & 1 \\ 3 & 1 & 4\end{vmatrix} = 28 .\]
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:
\[\begin{vmatrix}8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3\end{vmatrix}\]
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:
\[\begin{vmatrix}\sin\alpha & \cos\alpha & \cos(\alpha + \delta) \\ \sin\beta & \cos\beta & \cos(\beta + \delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma + \delta)\end{vmatrix}\]
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:
Prove that :
Prove that :
2x − y = 17
3x + 5y = 6
x+ y = 5
y + z = 3
x + z = 4
5x − 7y + z = 11
6x − 8y − z = 15
3x + 2y − 6z = 7
x − y + 3z = 6
x + 3y − 3z = − 4
5x + 3y + 3z = 10
For what value of x, the following matrix is singular?
If \[\begin{vmatrix}2x + 5 & 3 \\ 5x + 2 & 9\end{vmatrix} = 0\]
For what value of x is the matrix \[\begin{bmatrix}6 - x & 4 \\ 3 - x & 1\end{bmatrix}\] singular?
If \[A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}\] , then for any natural number, find the value of Det(An).
If \[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix}\] , then x =
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:
\[\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10\]
\[\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10\]
\[\frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13\]
Solve the following system of equations by matrix method:
5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25
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:
8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5
Show that the following systems of linear equations is consistent and also find their solutions:
6x + 4y = 2
9x + 6y = 3
Show that the following systems of linear equations is consistent and also find their solutions:
2x + 2y − 2z = 1
4x + 4y − z = 2
6x + 6y + 2z = 3
Show that each one of the following systems of linear equation is inconsistent:
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
Two schools P and Q want to award their selected students on the values of Discipline, Politeness and Punctuality. The school P wants to award ₹x each, ₹y each and ₹z each the three respectively values to its 3, 2 and 1 students with a total award money of ₹1,000. School Q wants to spend ₹1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for three values as before). If the total amount of awards for one prize on each value is ₹600, using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.
2x − y + 2z = 0
5x + 3y − z = 0
x + 5y − 5z = 0
x + y + z = 0
x − y − 5z = 0
x + 2y + 4z = 0
x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0
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
If `|(2x, 5),(8, x)| = |(6, 5),(8, 3)|`, then find x
Prove that (A–1)′ = (A′)–1, where A is an invertible matrix.
`abs ((1, "a"^2 + "bc", "a"^3),(1, "b"^2 + "ca", "b"^3),(1, "c"^2 + "ab", "c"^3))`
Using the matrix method, solve the following system of linear equations:
`2/x + 3/y + 10/z` = 4, `4/x - 6/y + 5/z` = 1, `6/x + 9/y - 20/z` = 2.
