Advertisements
Advertisements
प्रश्न
Give examples of matrices
A, B and C such that AB = AC but B ≠ C, A ≠ 0.
उत्तर
\[\left( iv \right) Let A = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} , B = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} \text{and C} = \begin{bmatrix}0 & 0 \\ 0 & 2\end{bmatrix}\]
\[ \therefore AB = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\]
\[ \Rightarrow AB = \begin{bmatrix}0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\]
\[and AC = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\begin{bmatrix}0 & 0 \\ 0 & 2\end{bmatrix}\]
\[ \Rightarrow AC = \begin{bmatrix}0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0\end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\]
\[\]
Thus,
AB = AC
But B ≠ C and A ≠ 0.
APPEARS IN
संबंधित प्रश्न
Let `A = [(2,4),(3,2)] , B = [(1,3),(-2,5)], C = [(-2,5),(3,4)]`
Find AB
Compute the indicated product.
`[(1),(2),(3)] [2,3,4]`
Compute the indicated products:
`[[a b],[-b a]][[a -b],[b a]]`
Compute the indicated products:
`[[1 -2],[2 3]][[1 2 3],[-3 2 -1]]`
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]]`
Evaluate the following:
`([[1 3],[-1 -4]]+[[3 -2],[-1 1]])[[1 3 5],[2 4 6]]`
If A = `[[1 0],[0 1]]`,B`[[1 0],[0 -1]]`
and C= `[[0 1],[1 0]]`
, then show that A2 = B2 = C2 = I2.
If A = `[[ cos 2θ sin 2θ],[ -sin 2θ cos 2θ]]`, find A2.
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=[[2 -1],[1 1],[-1 2]]` `B=[[0 1],[1 1]]` C=`[[1 -1],[0 1]]`
Compute the elements a43 and a22 of the matrix:`A=[[0 1 0],[2 0 2],[0 3 2],[4 0 4]]` `[[2 -1],[-3 2],[4 3]] [[0 1 -1 2 -2],[3 -3 4 -4 0]]`
If [x 4 1] `[[2 1 2],[1 0 2],[0 2 -4]]` `[[x],[4],[-1]]` = 0, find x.
\[A = \begin{bmatrix}3 & - 2 \\ 4 & - 2\end{bmatrix} and \text{ I }= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\], then prove that A2 − A + 2I = O.
\[A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix} and \text{ I} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
If `[[2 3],[5 7]] [[1 -3],[-2 4]]-[[-4 6],[-9 x]]` find x.
If A=, find k such that A2 = kA − 2I2
If f (x) = x2 − 2x, find f (A), where A=
`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`
Let A and B be square matrices of the order 3 × 3. Is (AB)2 = A2 B2? Give reasons.
If A and B are square matrices of the same order, explain, why in general
(A − B)2 ≠ A2 − 2AB + B2
If A and B are square matrices of the same order such that AB = BA, then show that (A + B)2 = A2 + 2AB + B2.
Three shopkeepers A, B and C go to a store to buy stationary. A purchases 12 dozen notebooks, 5 dozen pens and 6 dozen pencils. B purchases 10 dozen notebooks, 6 dozen pens and 7 dozen pencils. C purchases 11 dozen notebooks, 13 dozen pens and 8 dozen pencils. A notebook costs 40 paise, a pen costs Rs. 1.25 and a pencil costs 35 paise. Use matrix multiplication to calculate each individual's bill.
In a legislative assembly election, a political group hired a public relations firm to promote its candidates in three ways: telephone, house calls and letters. The cost per contact (in paise) is given matrix A as
Cost per contact
`A=[[40],[100],[50]]` `[["Teliphone"] ,["House call "],[" letter"]]`
The number of contacts of each type made in two cities X and Y is given in matrix B as
Telephone House call Letter
`B= [[ 1000, 500, 5000],[3000,1000, 10000 ]]`
Find the total amount spent by the group in the two cities X and Y.
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
(AB)T = BT AT
If `A= [[3],[5],[2]]` And B=[1 0 4] , Verify that `(AB)^T=B^TA^T`
If \[A = \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}\] write AAT.
If A = [aij] is a 2 × 2 matrix such that aij = i + 2j, write A.
If \[A = \begin{bmatrix}1 & a \\ 0 & 1\end{bmatrix}\]then An (where n ∈ N) equals
If A is a square matrix such that A2 = A, then (I + A)3 − 7A 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 is a matrix of order m × n and B is a matrix such that ABT and BTA are both defined, then the order of matrix B is
Disclaimer: option (a) and (d) both are the same.
If X = `[(3, 1, -1),(5, -2, -3)]` and Y = `[(2, 1, -1),(7, 2, 4)]`, find 2X – 3Y
Give an example of matrices A, B and C such that AB = AC, where A is nonzero matrix, but B ≠ C.
Prove by Mathematical Induction that (A′)n = (An)′, where n ∈ N for any square matrix A.
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:
- How many articles (in total) are sold by three schools?
