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
Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
Solve the system of linear equations using the matrix method.
x − y + z = 4
2x + y − 3z = 0
x + y + z = 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|.
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:
\[\begin{vmatrix}1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin^2 23^\circ & \sin^2 67^\circ & \cos180^\circ \\ - \sin^2 67^\circ & - \sin^2 23^\circ & \cos^2 180^\circ \\ \cos180^\circ & \sin^2 23^\circ & \sin^2 67^\circ\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}a & b & c \\ c & a & b \\ b & c & a\end{vmatrix}\]
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:
Solve the following determinant equation:
Using determinants show that the following points are collinear:
(1, −1), (2, 1) and (4, 5)
Using determinants, find the equation of the line joining the points
(1, 2) and (3, 6)
Prove that :
Prove that :
Prove that :
Prove that :
A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission
| Month | Sale of units | Total commission drawn (in Rs) |
||
| A | B | C | ||
| Jan | 90 | 100 | 20 | 800 |
| Feb | 130 | 50 | 40 | 900 |
| March | 60 | 100 | 30 | 850 |
Find out the rates of commission on items A, B and C by using determinant method.
Write the cofactor of a12 in the following matrix \[\begin{bmatrix}2 & - 3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & - 7\end{bmatrix} .\]
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\[f\left( x \right) = \begin{vmatrix}0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0\end{vmatrix}\]
Solve the following system of equations by matrix method:
5x + 7y + 2 = 0
4x + 6y + 3 = 0
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.
x + y + z = 0
x − y − 5z = 0
x + 2y + 4z = 0
2x + 3y − z = 0
x − y − 2z = 0
3x + y + 3z = 0
The system of linear equations:
x + y + z = 2
2x + y − z = 3
3x + 2y + kz = 4 has a unique solution if
The existence of the unique solution of the system of equations:
x + y + z = λ
5x − y + µz = 10
2x + 3y − z = 6
depends on
System of equations x + y = 2, 2x + 2y = 3 has ______
Solve the following equations by using inversion method.
x + y + z = −1, x − y + z = 2 and x + y − z = 3
If `|(2x, 5),(8, x)| = |(6, -2),(7, 3)|`, then value of x is ______.
Solve the following system of equations x − y + z = 4, x − 2y + 2z = 9 and 2x + y + 3z = 1.
If the system of equations x + ky - z = 0, 3x - ky - z = 0 & x - 3y + z = 0 has non-zero solution, then k is equal to ____________.
If A = `[(1,-1,0),(2,3,4),(0,1,2)]` and B = `[(2,2,-4),(-4,2,-4),(2,-1,5)]`, then:
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:
For what value of p, is the system of equations:
p3x + (p + 1)3y = (p + 2)3
px + (p + 1)y = p + 2
x + y = 1
consistent?
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 ab + bc + ca equals ______.
