Example 7 Find 2a B If A 1 2 3 2 3 1 B 3 1 3

Example 7 Find 2a B If A 1 2 3 2 3 1 B 3 1 3 Example 7 if a = [ 8 (1&2&3@2&3&1)] and b = [ 8 (3&−1&3@−1&0&2)] then find 2a – b. finding 2a 2a = 2 [ 8 (1&2&3@2&3&1)] = [ 8 (2 × 1&2 × 2&2 × 3@2 × 2&2 × 3&2 × 1)] = [ 8 (𝟐&𝟒&𝟔@𝟒&𝟔&𝟐)] finding 2a – b 2a – b = [ 8 (𝟐&𝟒&𝟔@𝟒&𝟔&𝟐)] − [ 8 (𝟑&−𝟏&𝟑@−𝟏&𝟎&𝟐)] = [ 8 (2−3&4− (−1)&6−3@4− (−1)&6−0&2−2)] = [ 8 (−1&4 1&3@4 1&6&0)] = [ 8 (−𝟏&𝟓&𝟑@𝟓&𝟔&𝟎)]. Quickmath will automatically answer the most common problems in algebra, equations and calculus faced by high school and college students. the algebra section allows you to expand, factor or simplify virtually any expression you choose.

Example 7 Find 2a B If A 1 2 3 2 3 1 B 3 1 3 Example 14 if a = [ 8(2&3@1&−4)] and b = [ 8(1&−2@−1&3)] , then verify that (ab) 1 = b 1 a 1 solving l.h.s (ab) –1 first calculating ab ab = [ 8(2&3@1&−4)] [ 8(1&−2@−1&3)] = [ 8(2(1) 3(−1)&2(−2) 3(3)@1(1) ( −4)(−1)&1(−2) (⤶7−4)3)] = [ 8(2−3&−4 9@1 4&−2−12)] = [ 8(−𝟏&𝟓@𝟓&−𝟏𝟒)] now, (ab) 1. If a = `[(1,2,3), (2,k,2), (5,7,3)]` is a singular matrix then find the value of 'k'. if `"a" = [(1,2, 3),(5,4,0)] , "b" = [(1,4,3),( 2,5,0)]`, then find 2a 3b. `[(2, 1),(5, 1)]` construct a 2 x 2 matrix whose elements a ij are given by a ij = 2i – j. construct a matrix a = [a ij] 3 × 2 whose element a ij is given by. a ij = `(("i" "j. Ex 3.3, 4 if a’ = [ 8(−2&3@1&2)] and b = [ 8(−1&0@1&2)] , then find (a 2b)’ we need to find (a 2b)’ finding a a’ = [ 8(−2&3@1&2)] a = (a’)’ = [ 8(−𝟐&𝟏@𝟑&𝟐)] also, b = [ 8(−1&0@1&2)] 2b = 2 [ 8(−1&0@1&2)] = [ 8(2(−1)&2(0)@2(1)&2(2))] = [ 8(−𝟐&𝟎@𝟐&𝟒)] now a 2b = [ 8(−2&1@3&2)] [ 8(−2. 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 (2a) t = 2a t. if \[a = \begin{bmatrix}4 & 3 \\ 1 & 2\end{bmatrix} and b = \binom{ 4}{ 3}\] write ab .

Solved 11 If A 1 2 3 And B 1 2 3 4 Find A A B B A A C B Ex 3.3, 4 if a’ = [ 8(−2&3@1&2)] and b = [ 8(−1&0@1&2)] , then find (a 2b)’ we need to find (a 2b)’ finding a a’ = [ 8(−2&3@1&2)] a = (a’)’ = [ 8(−𝟐&𝟏@𝟑&𝟐)] also, b = [ 8(−1&0@1&2)] 2b = 2 [ 8(−1&0@1&2)] = [ 8(2(−1)&2(0)@2(1)&2(2))] = [ 8(−𝟐&𝟎@𝟐&𝟒)] now a 2b = [ 8(−2&1@3&2)] [ 8(−2. 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 (2a) t = 2a t. if \[a = \begin{bmatrix}4 & 3 \\ 1 & 2\end{bmatrix} and b = \binom{ 4}{ 3}\] write ab . If `[(a, 3),(4, 1)] [(2, b),(1, 2)] [(1, 1),( 2, c)] = [(5, 0),(7, 3)]`, find the values of a, b and c. if `a = [(2, 5),(1, 3)]`, `b = [(4, 2),( 1, 3)]` and i is identity matrix of same order and `a^t` is the transpose of matrix a find `a^t.b bi`. If (a 2b) = \( \begin{pmatrix} 1 & 2 \\ 3 & 0 \\ \end{pmatrix} \) and (2a 3b) = \( \begin{pmatrix} 2 & 2 \\ 3 & 3 \\ \end{pmatrix} \) then b = ? a. \( \begin{pmatrix} 6 & 4 \\ 3 & 3 \\ \end{pmatrix} \). There are 3 steps to solve this one. let a = [1 2 − 3] and b = [− 4 2 6]. find, if possible, a b, a b, 2a, 2a b, and b za. (if not possible, enter impossible in any single cell.) 1 2 3 b = [ 4 2 6] a) a b (b) a b it (c) 2a (d) 24 b (2) b 14 need help? read it. not the question you’re looking for?. Example 17 find |𝑎 ⃗ − 𝑏 ⃗| , if two vectors a and b are such that |𝑎 ⃗| = 2, |𝑏 ⃗| = 3 and 𝑎 ⃗ ⋅ 𝑏 ⃗ = 4. given, |𝒂 ⃗ | = 2 , |𝒃 ⃗ | = 2 & 𝒂 ⃗. 𝒃 ⃗ = 4 we need to find |𝑎 ⃗−𝑏 ⃗ | taking square, |𝒂 ⃗−𝒃 ⃗ |2 = (𝒂 ⃗−𝒃 ⃗).