Advertisements
Advertisements
Question
x + y + z + w = 2
x − 2y + 2z + 2w = − 6
2x + y − 2z + 2w = − 5
3x − y + 3z − 3w = − 3
Advertisements
Solution
\[D = \begin{vmatrix}1 & 1 & 1 & 1 \\ 1 & - 2 & 2 & 2 \\ 2 & 1 & - 2 & 2 \\ 3 & - 1 & 3 & - 3\end{vmatrix}\]
\[1\begin{vmatrix}- 2 & 2 & 2 \\ 1 & - 2 & 2 \\ - 1 & 3 & - 3\end{vmatrix} - 1\begin{vmatrix}1 & 2 & 2 \\ 2 & - 2 & 2 \\ 3 & 3 & - 3\end{vmatrix} + 1\begin{vmatrix}1 & - 2 & 2 \\ 2 & 1 & 2 \\ 3 & - 1 & - 3\end{vmatrix} - 1\begin{vmatrix}1 & - 2 & 2 \\ 2 & 1 & - 2 \\ 3 & - 1 & 3\end{vmatrix}\]
\[ = 1\left[ - 2\left( 6 - 6 \right) - 2\left( - 3 + 2 \right) + 2\left( 3 - 2 \right) \right] - 1\left[ 1\left( 6 - 6 \right) - 2\left( - 6 - 6 \right) + 2\left( 6 + 6 \right) \right] + 1\left[ 1\left( - 3 + 2 \right) + 2\left( - 6 - 6 \right) + 2\left( - 2 - 3 \right) \right] - 1\left[ 1\left( 3 - 2 \right) + 2\left( 6 + 6 \right) + 2\left( - 2 - 3 \right) \right]\]
\[ = 4 - 48 - 35 - 15\]
\[ = - 94\]
\[ D_1 = \begin{vmatrix}2 & 1 & 1 & 1 \\ - 6 & - 2 & 2 & 2 \\ - 5 & 1 & - 2 & 2 \\ - 3 & - 1 & 3 & - 3\end{vmatrix}\]
\[2\begin{vmatrix}- 2 & 2 & 2 \\ 1 & - 2 & 2 \\ - 1 & 3 & - 3\end{vmatrix} - 1\begin{vmatrix}- 6 & 2 & 2 \\ - 5 & - 2 & 2 \\ - 3 & 3 & - 3\end{vmatrix} + 1\begin{vmatrix}- 6 & - 2 & 2 \\ - 5 & 1 & 2 \\ - 3 & - 1 & - 3\end{vmatrix} - 1\begin{vmatrix}- 6 & - 2 & 2 \\ - 5 & 1 & - 2 \\ - 3 & - 1 & 3\end{vmatrix}\]
\[ = 2\left[ - 2\left( 6 - 6 \right) - 2\left( - 3 + 2 \right) + 2\left( 3 - 2 \right) \right] - 1\left[ - 6\left( 6 - 6 \right) - 2\left( 15 + 6 \right) + 2\left( - 15 - 6 \right) \right] + 1\left[ - 6\left( - 3 + 2 \right) + 2\left( 15 + 6 \right) + 2\left( 5 + 3 \right) \right] - 1\left[ - 6\left( 3 - 2 \right) + 2\left( - 15 - 6 \right) + 2\left( 5 + 3 \right) \right]\]
\[ = 188\]
\[ D_2 = \begin{vmatrix}1 & 2 & 1 & 1 \\ 1 & - 6 & 2 & 2 \\ 2 & - 5 & - 2 & 2 \\ 3 & - 3 & 3 & - 3\end{vmatrix}\]
\[1\begin{vmatrix}- 6 & 2 & 2 \\ - 5 & - 2 & 2 \\ - 3 & 3 & - 3\end{vmatrix} - 2\begin{vmatrix}1 & 2 & 2 \\ 2 & - 2 & 2 \\ 3 & 3 & - 3\end{vmatrix} + 1\begin{vmatrix}1 & - 6 & 2 \\ 2 & - 5 & 2 \\ 3 & - 3 & - 3\end{vmatrix} - 1\begin{vmatrix}1 & - 6 & 2 \\ 2 & - 5 & - 2 \\ 3 & - 3 & 3\end{vmatrix}\]
\[1\left[ - 6\left( 6 - 6 \right) - 2\left( 15 + 6 \right) + 2\left( - 15 - 6 \right) \right] - 2\left[ 1\left( 6 - 6 \right) - 2\left( - 6 - 6 \right) + 2\left( 6 + 6 \right) \right] + 1\left[ 1\left( 15 + 6 \right) + 6\left( - 6 - 6 \right) + 2\left( - 6 + 15 \right) \right] - 1\left[ 1\left( - 15 - 6 \right) - 6\left( 6 + 6 \right) + 2\left( - 6 + 15 \right) \right]\]
\[ = 1\]
\[ D_3 = \begin{vmatrix}1 & 1 & 2 & 1 \\ 1 & - 2 & - 6 & 2 \\ 2 & 1 & - 5 & 2 \\ 3 & - 1 & - 3 & - 3\end{vmatrix}\]
\[1\begin{vmatrix}- 2 & - 6 & 2 \\ 1 & - 5 & 2 \\ - 1 & - 3 & - 3\end{vmatrix} - 1\begin{vmatrix}1 & - 6 & 2 \\ 2 & - 5 & 2 \\ 3 & - 3 & - 3\end{vmatrix} + 2\begin{vmatrix}1 & - 2 & 2 \\ 2 & 1 & 2 \\ 3 & - 1 & - 3\end{vmatrix} - 1\begin{vmatrix}1 & - 2 & - 6 \\ 2 & 1 & - 5 \\ 3 & - 1 & - 3\end{vmatrix}\]
\[ = 1\left[ - 2\left( 15 + 6 \right) + 6\left( - 3 + 2 \right) + 2\left( - 3 - 5 \right) \right] - 1\left[ 1\left( 15 + 6 \right) + 6\left( - 6 - 6 \right) + 2\left( - 6 + 15 \right) \right] + 2\left[ 1\left( - 3 + 2 \right) + 2\left( - 6 - 6 \right) + 2\left( - 2 - 3 \right) \right] - 1\left[ 1\left( - 3 - 5 \right) + 2\left( - 6 + 15 \right) - 6\left( - 2 - 3 \right) \right]\]
\[ = - 141\]
\[ D_4 = \begin{vmatrix}1 & 1 & 1 & 2 \\ 1 & - 2 & 2 & - 6 \\ 2 & 1 & - 2 & - 5 \\ 3 & - 1 & 3 & - 3\end{vmatrix}\]
\[1\begin{vmatrix}- 2 & 2 & - 6 \\ 1 & - 2 & - 5 \\ - 1 & 3 & - 3\end{vmatrix} - 1\begin{vmatrix}1 & 2 & - 6 \\ 2 & - 2 & - 5 \\ 3 & 3 & - 3\end{vmatrix} + 1\begin{vmatrix}1 & - 2 & - 6 \\ 2 & 1 & - 5 \\ 3 & - 1 & - 3\end{vmatrix} - 2\begin{vmatrix}1 & - 2 & 2 \\ 2 & 1 & - 2 \\ 3 & - 1 & - 3\end{vmatrix}\]
\[1\left[ - 2\left( 6 + 15 \right) - 2\left( - 3 - 5 \right) - 6\left( 3 - 2 \right) \right] - 1\left[ 1\left( 6 + 15 \right) - 2\left( - 6 + 15 \right) - 6\left( 6 + 6 \right) \right] + 1\left[ 1\left( - 3 - 5 \right) + 2\left( - 6 + 15 \right) - 6\left( - 2 - 3 \right) \right] - 2\left[ 1\left( - 3 - 2 \right) + 2\left( - 6 + 6 \right) + 2\left( - 2 - 3 \right) \right]\]
\[ = 47\]
Thus,
\[x = \frac{D_1}{D} = \frac{188}{- 94} = - 2\]
\[y = \frac{D_2}{D} = \frac{- 282}{- 94} = 3\]
\[z = \frac{D_3}{D} = \frac{- 141}{- 94} = 1 . 5\]
\[w = \frac{D_4}{D} = \frac{47}{- 94} = - 0 . 5\]
APPEARS IN
RELATED QUESTIONS
The sum of three numbers is 6. When second number is subtracted from thrice the sum of first and third number, we get number 10. Four times the sum of third number is subtracted from five times the sum of first and second number, the result is 3. Using above information, find these three numbers by matrix method.
Find the inverse of the matrix, `A=[[1,3,3],[1,4,3],[1,3,4]]`by using column transformations.
Solve the following equations by the method of reduction :
2x-y + z=1, x + 2y +3z = 8, 3x + y-4z=1.
The sum of three numbers is 9. If we multiply third number by 3 and add to the second number, we get 16. By adding the first and the third number and then subtracting twice the second number from this sum, we get 6. Use this information and find the system of linear equations. Hence, find the three numbers using matrices.
Express the following equations in the matrix form and solve them by method of reduction :
2x- y + z = 1, x + 2y + 3z = 8, 3x + y - 4z =1
Use elementary column operation C2 → C2 + 2C1 in the following matrix equation :
`[[2,1],[2,0]] = [[3,1],[2,0]] [[1,0],[-1,1]]`
If `A=|[2,0,-1],[5,1,0],[0,1,3]|` , then find A-1 using elementary row operations
Using elementary transformations, find the inverse of the matrix A = `((8,4,3),(2,1,1),(1,2,2))`and use it to solve the following system of linear equations :
8x + 4y + 3z = 19
2x + y + z = 5
x + 2y + 2z = 7
Using properties of determinants, prove that :
`|[1+a,1,1],[1,1+b,1],[1,1,1+c]|=abc + bc + ca + ab`
The cost of 2 books, 6 notebooks and 3 pens is Rs 40. The cost of 3 books, 4 notebooks and 2 pens is Rs 35, while the cost of 5 books, 7 notebooks and 4 pens is Rs 61. Using this information and matrix method, find the cost of 1 book, 1 notebook and 1 pen separately.
Using elementary row transformations, find the inverse of the matrix A = `[(1,2,3),(2,5,7),(-2,-4,-5)]`
Prove that :
2x − 3z + w = 1
x − y + 2w = 1
− 3y + z + w = 1
x + y + z = 1
2x − y = 5
4x − 2y = 7
In the following matrix equation use elementary operation R2 → R2 + R1 and the equation thus obtained:
Use elementary column operation C2 → C2 + 2C1 in the following matrix equation : \[\begin{bmatrix} 2 & 1 \\ 2 & 0\end{bmatrix} = \begin{bmatrix}3 & 1 \\ 2 & 0\end{bmatrix}\begin{bmatrix}1 & 0 \\ - 1 & 1\end{bmatrix}\]
Use elementary column operations \[C_2 \to C_2 - 2 C_1\] in the matrix equation \[\begin{pmatrix}4 & 2 \\ 3 & 3\end{pmatrix} = \begin{pmatrix}1 & 2 \\ 0 & 3\end{pmatrix}\begin{pmatrix}2 & 0 \\ 1 & 1\end{pmatrix}\] .
If three numbers are added, their sum is 2. If two times the second number is subtracted from the sum of the first and third numbers, we get 8, and if three times the first number is added to the sum of the second and third numbers, we get 4. Find the numbers using matrices.
Apply the given elementary transformation on each of the following matrices `[(3, -4),(2, 2)]`, R1 ↔ R2.
Apply the given elementary transformation on each of the following matrices `[(2, 4),(1, -5)]`, C1 ↔ C2.
Apply the given elementary transformation on each of the following matrices `[(3, 1, -1),(1, 3, 1),(-1, 1, 3)]`, 3R2 and C2 ↔ C2 – 4C1.
Transform `[(1, -1, 2),(2, 1, 3),(3, 2, 4)]` into an upper traingular matrix by suitable row transformations.
Find the cofactor matrix, of the following matrices : `[(1, 2),(5, -8)]`
Find the cofactor matrix, of the following matrices: `[(5, 8, 7),(-1, -2, 1),(-2, 1, 1)]`
Find the adjoint of the following matrices : `[(2, -3),(3, 5)]`
Choose the correct alternative.
If A = `[(2, 5),(1, 3)]`, then A–1 = _______
Fill in the blank :
Order of matrix `[(2, 1, 1),(5, 1, 8)]` is _______
State whether the following is True or False :
Single element matrix is row as well as column matrix.
Solve the following :
If A = `[(1, 0, 0),(2, 1, 0),(3, 3, 1)]`, the reduce it to unit matrix by using row transformations.
Choose the correct alternative:
If A = `[(1, 2),(2, -1)]`, then adj (A) = ______
The suitable elementary row transformation which will reduce the matrix `[(1, 0),(2, 1)]` into identity matrix is ______
Find the inverse of matrix A = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)]` by using elementary row transformations
If A is a 3 × 3 matrix and |A| = 2, then the matrix represented by A (adj A) is equal to.
If `overlinea = hati + hatj + hatk, overlinea . overlineb = 1` and `overlinea xx overlineb = hatj - hatk,` then `overlineb` = ______
If A = `[(1, 1, -1), (1, -2, 1), (2, -1, -3)]`, then (adj A)A = ______
Let F(α) = `[(cosalpha, -sinalpha, 0), (sinalpha, cosalpha, 0), (0, 0, 1)]` where α ∈ R. Then [F(α)]-1 is equal to ______
If `[(1, 0, -1),(0, 2, 1),(1, -2, 0)] [(x),(y),(z)] = [(1),(2),(3)]`, then the values of x, y, z respectively are ______.
If `[(2, 3), (3, 1)][(x), (y)] = [(-5), (3)]`, then the values of x and y respectively are ______
If A = `[(1, 2, 1), (3, 2, 3), (2, 1, 2)]`, then `a_11A_11 + a_21A_21 + a_31A_31` = ______
In the matrix A = `[("a", 1, x),(2, sqrt(3), x^2 - y),(0, 5, (-2)/5)]`, write: The number of elements
Construct a 3 × 2 matrix whose elements are given by aij = ei.x sinjx.
Find the values of a and b if A = B, where A = `[("a" + 4, 3"b"),(8, -6)]`, B = `[(2"a" + 2, "b"^2 + 2),(8, "b"^2 - 5"b")]`
Find the matrix A satisfying the matrix equation:
`[(2, 1),(3, 2)] "A" [(-3, 2),(5, -3)] = [(1, 0),(0, 1)]`
Solve for x and y: `x[(2),(1)] + y[(3),(5)] + [(-8),(-11)]` = O
If A = `[(0, -1, 2),(4, 3, -4)]` and B = `[(4, 0),(1, 3),(2, 6)]`, then verify that: (A′)′ = A
If A = `[(0, -1, 2),(4, 3, -4)]` and B = `[(4, 0),(1, 3),(2, 6)]`, then verify that: (A′)′ = (AB)' = B'A'
Find the matrix A such that `[(2, -1),(1, 0),(-3, 4)] "A" = [(-1, -8, -10),(1, -2, -5),(9, 22, 15)]`
If possible, using elementary row transformations, find the inverse of the following matrices
`[(2, 3, -3),(-1, 2, 2),(1, 1, -1)]`
If possible, using elementary row transformations, find the inverse of the following matrices
`[(2, 0, -1),(5, 1, 0),(0, 1, 3)]`
If `[(2x + y, 4x),(5x - 7, 4x)] = [(7, 7y - 13),(y, x + 6)]`, then the value of x + y is ______.
If A = `1/pi [(sin^-1(xpi), tan^-1(x/pi)),(sin^-1(x/pi), cot^-1(pix))]`, B = `1/pi [(-cos^-1(x/pi), tan^-1 (x/pi)),(sin^-1(x/pi),-tan^-1(pix))]`, then A – B is equal to ______.
On using elementary column operations C2 → C2 – 2C1 in the following matrix equation `[(1, -3),(2, 4)] = [(1, -1),(0, 1)] [(3, 1),(2, 4)]`, we have: ______.
On using elementary row operation R1 → R1 – 3R2 in the following matrix equation: `[(4, 2),(3, 3)] = [(1, 2),(0, 3)] [(2, 0),(1, 1)]`, we have: ______.
If (AB)′ = B′ A′, where A and B are not square matrices, then number of rows in A is equal to number of columns in B and number of columns in A is equal to number of rows in B.
If A = `[(2, 3, -1),(1, 4, 2)]` and B = `[(2, 3),(4, 5),(2, 1)]`, then AB and BA are defined and equal.
If A = `[(0,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)],` then ____________.
If `[(2, 0, 7),(0, 1, 0),(1, -2, 1)] [(-x, 14x, 7x),(0, 1, 0),(x, -4x, -2x)] = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`then find the value of x
If f(x) = `|(1 + sin^2x, cos^2x, 4 sin 2x),(sin^2x, 1 + cos^2x, 4 sin 2x),(sin^2 x, cos^2 x, 1 + 4 sin 2x)|`
What is the maximum value of f(x)?
if `A = [(2,5),(1,3)] "then" A^-1` = ______
If `[(3,0),(0,2)][(x),(y)] = [(3),(2)], "then" x = 1 "and" y = -1`
