Advertisements
Advertisements
प्रश्न
Solve the matrix equations:
`[x1][[1,0],[-2,-3]][[x],[5]]=0`
उत्तर
[x 1] `[[1,0],[-2,-3]]` `[[x],[5]]=0`
`⇒[x−2 0 -3 ] [(x), (5)] ` = 0
`⇒[x−2 -3 ] [[x],[5]]=0`
`⇒[x^2−2x -15 ]` =0
`⇒x^2−2x -15 `=0
`⇒x^2−5x +3x -15 `=0
`⇒x(x -5) +3(x -15 ) `=0
`⇒(x -5) (x+3)= 0 `
`⇒x -5 = 0 or x+3= 0 `
`⇒x = 5 or x = -3`
APPEARS IN
संबंधित प्रश्न
Compute the indicated product:
`[(a,b),(-b,a)][(a,-b),(b,a)]`
Compute the indicated product.
`[(2,1),(3,2),(-1,1)][(1,0,1),(-1,2,1)]`
Compute the indicated product:
`[(2,3,4),(3,4,5),(4,5,6)][(1,-3,5),(0,2,4), (3,0,5)]`
Show that AB ≠ BA in each of the following cases:
`A= [[5 -1],[6 7]]`And B =`[[2 1],[3 4]]`
Show that AB ≠ BA in each of the following cases
`A=[[-1 1 0],[0 -1 1],[2 3 4]]` and =B `[[1 2 3], [0 1 0],[1 1 0]]`
If A = `[[1 1],[0 1]]` show that A2 = `[[1 2],[0 1]]` and A3 = `[[1 3],[0 1]]`
For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A(BC):
`A =-[[1 2 0],[-1 0 1]]`,`B=[[1 0],[-1 2],[0 3]]` and C= `[[1],[-1]]`
For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC:
`A = [[1 -1],[0 2]] B= [[-1 0],[2 1]]`and `C= [[0 1],[1 -1]]`
If \[A = \begin{bmatrix}3 & - 5 \\ - 4 & 2\end{bmatrix}\] , find A2 − 5A − 14I.
If `[[2 3],[5 7]] [[1 -3],[-2 4]]-[[-4 6],[-9 x]]` find x.
\[A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\]show that A2 − 5A + 7I = O use this to find A4.
If
`A=[[1,2,2],[2,1,2],[2,2,1]]`, then prove that A2 − 4A − 5I = 0
Find the matrix A such that [2 1 3 ] `[[-1,0,-1],[-1,1,0],[0,1,1]] [[1],[0],[-1]]=A`
Find a 2 × 2 matrix A such that `A=[[1,-2],[1,4]]=6l_2`
If `P(x)=[[cos x,sin x],[-sin x,cos x]],` then show that `P(x),P(y)=P(x+y)=P(y)P(x).`
Give examples of matrices
A and B such that AB = O but A ≠ 0, B ≠ 0.
If A and B are square matrices of the same order, explain, why in general
(A + B)2 ≠ A2 + 2AB + B2
A trust fund has Rs 30000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30000 among the two types of bonds. If the trust fund must obtain an annual total interest of(ii) Rs 2000
Let `A =[[2,-3],[-7,5]]` And `B=[[1,0],[2,-4]]` verify that
(A − B)T = AT − BT
Let `A= [[1,-1,0],[2,1,3],[1,2,1]]` And `B=[[1,2,3],[2,1,3],[0,1,1]]` Find `A^T,B^T` and verify that (A + B)T = AT + BT
If \[A = \begin{bmatrix}\cos x & - \sin x \\ \sin x & \cos x\end{bmatrix}\] , find AAT
If \[\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}\begin{bmatrix}3 & 1 \\ 2 & 5\end{bmatrix} = \begin{bmatrix}7 & 11 \\ k & 23\end{bmatrix}\] ,then write the value of k.
Let A = \[\begin{bmatrix}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a\end{bmatrix}\], then An is equal to
If A = [aij] is a scalar matrix of order n × n such that aii = k, for all i, then trace of A is equal to
If \[A = \begin{bmatrix}2 & - 1 & 3 \\ - 4 & 5 & 1\end{bmatrix}\text{ and B }= \begin{bmatrix}2 & 3 \\ 4 & - 2 \\ 1 & 5\end{bmatrix}\] then
If A = `[[3,9,0] ,[1,8,-2], [7,5,4]]` and B =`[[4,0,2],[7,1,4],[2,2,6]]` , then find the matrix `B'A'` .
If A = `[(3, 5)]`, B = `[(7, 3)]`, then find a non-zero matrix C such that AC = BC.
Let A and B be square matrices of the order 3 × 3. Is (AB)2 = A2B2? Give reasons.
If A = `[(3, -5),(-4, 2)]`, then find A2 – 5A – 14I. Hence, obtain A3.
If A and B are square matrices of the same order, then (kA)′ = ______. (k is any scalar)
Three schools DPS, CVC, and KVS decided to organize a fair for collecting money for helping the flood victims. They sold handmade fans, mats, and plates from recycled material at a cost of Rs. 25, Rs.100, and Rs. 50 each respectively. The numbers of articles sold are given as
| School/Article | DPS | CVC | KVS |
| Handmade/fans | 40 | 25 | 35 |
| Mats | 50 | 40 | 50 |
| Plates | 20 | 30 | 40 |
Based on the information given above, answer the following questions:
- What is the total money (in Rupees) collected by the school DPS?
If A = `[(a, b),(b, a)]` and A2 = `[(α, β),(β, α)]`, then ______.
If A = `[(-3, -2, -4),(2, 1, 2),(2, 1, 3)]`, B = `[(1, 2, 0),(-2, -1, -2),(0, -1, 1)]` then find AB and use it to solve the following system of equations:
x – 2y = 3
2x – y – z = 2
–2y + z = 3
