Advertisements
Advertisements
Question
Advertisements
Solution
Here,
\[ A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\]
\[\left| A \right|=1 \left( 1 + 6 \right) + 2\left( 2 - 0 \right) + 0\left( - 4 - 0 \right)\]
\[ = 7 + 4 + 0\]
\[ = 11\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 3 \\ - 2 & 1\end{vmatrix} = 7, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 3 \\ 0 & 1\end{vmatrix} = - 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 1 \\ 0 & - 2\end{vmatrix} = - 4\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 2 & 0 \\ - 2 & 1\end{vmatrix} = 2, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix} = 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 2 \\ 0 & - 2\end{vmatrix} = 2\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 2 & 0 \\ 1 & 3\end{vmatrix} = - 6, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 0 \\ 2 & 3\end{vmatrix} = - 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 2 \\ 2 & 1\end{vmatrix} = 5\]
\[ \therefore adj A = \begin{bmatrix}7 & - 2 & - 4 \\ 2 & 1 & 2 \\ - 6 & - 3 & 5\end{bmatrix}^T \]
\[ = \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{11}\begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\]
\[\text{ or,} AX = B\]
\[\text{ where, } A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} \text{ and }B = \begin{bmatrix}10 \\ 8 \\ 7\end{bmatrix}\]
Now,
\[ \therefore X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{11}\begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\begin{bmatrix}10 \\ 8 \\ 7\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{11}\begin{bmatrix}70 + 16 - 42 \\ - 20 + 8 - 21 \\ - 40 + 16 + 35\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{11}\begin{bmatrix}44 \\ - 33 \\ 11\end{bmatrix}\]
\[ \therefore x = 4, y = - 3\text{ and }z = 1\]
APPEARS IN
RELATED QUESTIONS
If `|[2x,5],[8,x]|=|[6,-2],[7,3]|`, write the value of x.
Examine the consistency of the system of equations.
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
Solve the system of linear equations using the matrix method.
4x – 3y = 3
3x – 5y = 7
Solve the system of linear equations using the matrix method.
5x + 2y = 3
3x + 2y = 5
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 the cost of each item per kg by matrix method.
Evaluate
\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}^2 .\]
\[∆ = \begin{vmatrix}\cos \alpha \cos \beta & \cos \alpha \sin \beta & - \sin \alpha \\ - \sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda\end{vmatrix}\]
Evaluate the following:
\[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix}\]
\[If ∆ = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix}, ∆_1 = \begin{vmatrix}1 & 1 & 1 \\ yz & zx & xy \\ x & y & z\end{vmatrix},\text{ then prove that }∆ + ∆_1 = 0 .\]
Solve the following determinant equation:
Solve the following determinant equation:
Find the area of the triangle with vertice at the point:
(2, 7), (1, 1) and (10, 8)
3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11
x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1
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.
For what value of x, the following matrix is singular?
If \[A = \begin{bmatrix}0 & i \\ i & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\] , find the value of |A| + |B|.
Evaluate \[\begin{vmatrix}4785 & 4787 \\ 4789 & 4791\end{vmatrix}\]
If \[A = \left[ a_{ij} \right]\] is a 3 × 3 diagonal matrix such that a11 = 1, a22 = 2 a33 = 3, then find |A|.
Evaluate: \[\begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}\]
The value of \[\begin{vmatrix}5^2 & 5^3 & 5^4 \\ 5^3 & 5^4 & 5^5 \\ 5^4 & 5^5 & 5^6\end{vmatrix}\]
The number of distinct real roots of \[\begin{vmatrix}cosec x & \sec x & \sec x \\ \sec x & cosec x & \sec x \\ \sec x & \sec x & cosec x\end{vmatrix} = 0\] lies in the interval
\[- \frac{\pi}{4} \leq x \leq \frac{\pi}{4}\]
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
Solve the following system of equations by matrix method:
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
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 shopkeeper has 3 varieties of pens 'A', 'B' and 'C'. Meenu purchased 1 pen of each variety for a total of Rs 21. Jeevan purchased 4 pens of 'A' variety 3 pens of 'B' variety and 2 pens of 'C' variety for Rs 60. While Shikha purchased 6 pens of 'A' variety, 2 pens of 'B' variety and 3 pens of 'C' variety for Rs 70. Using matrix method, find cost of each variety of pen.
3x + y − 2z = 0
x + y + z = 0
x − 2y + z = 0
The number of solutions of the system of equations:
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
x + y = 1
x + z = − 6
x − y − 2z = 3
If A = `[(2, 0),(0, 1)]` and B = `[(1),(2)]`, then find the matrix X such that A−1X = B.
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 of λ, such that the following system of equations has no solution, is
`2x - y - 2z = - 5`
`x - 2y + z = 2`
`x + y + lambdaz = 3`
A set of linear equations is represented by the matrix equation Ax = b. The necessary condition for the existence of a solution for this system is
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 ______.
