If matrix A = [1 2 3], write AA’.

1.	If matrix A = [1 2 3], write AA’.
Answer» Given A = [1 2 3] We will find A’ to calculate AA’, A' = \(\begin{bmatrix}1 \\[0.3em]2 \\[0.3em]3\end{bmatrix}\) Now AA' = [1 2 3]\(\begin{bmatrix}1 \\[0.3em]2 \\[0.3em]3\end{bmatrix}\) \(\Rightarrow\) [1 + 4 + 9] \(\Rightarrow\) [14]

Answer»

Given A = [1 2 3]

We will find A’ to calculate AA’,

A' = \(\begin{bmatrix}1 \\[0.3em]2 \\[0.3em]3\end{bmatrix}\)

Now

AA' = [1 2 3]\(\begin{bmatrix}1 \\[0.3em]2 \\[0.3em]3\end{bmatrix}\)

\(\Rightarrow\) [1 + 4 + 9]

\(\Rightarrow\) [14]

Discussion

No Comment Found

Related InterviewSolutions

Let `A=" diag "[3,-5,7]" and "B=" diag "[-1,2,4].` ltbr. Find (i) (A+B) (ii) (A-B) (iii) -5A (iv) (2A+3B).`
A2=A THEN FIND VALUE OF (I+A)2 -3A
If `A=[(1,2),(2,1)]` and `f(x)=(1+x)/(1-x)`, then f(A) isA. `[(1,1),(1,1)]`B. `[(2,2),(2,2)]`C. `[(-1,-1),(-1,-1)]`D. none of these
Id `[1//25 0x1//25]=[5 0-a5]^(-2)`, then the value of `x`is`a//125`b. `2a//125`c. `2a//25`d. none of theseA. `a//125`B. `2a//125`C. `2a//25`D. none of these
If `A=[(a+ib,c+id),(-c+id,a-ib)]` and `a^(2)+b^(2)+c^(2)+d^(2)=1`, then `A^(-1)` is equal toA. `[(a-ib,-c-id),(c-id,a+ib)]`B. `[(a+ib,-c+id),(-c+id,a-ib)]`C. `[(a-ib,-c-id),(-c-id,a+ib)]`D. none of these
If the matrix A is such that `({:(1,3),(0,1):})A=({:(1,1),(0,-1):})`, then what is A equal to ?A. `{:[(1,4),(-1,0)]:}`B. `{:[(1,-4),(1,0)]:}`C. `{:[(1,-4),(0,-1)]:}`D. none of these
If `A=[[cos alpha, -sin alpha] , [sin alpha, cos alpha]], B=[[cos2beta, sin 2beta] , [sin 2 beta, -cos2beta]]` where `0 lt beta lt pi/2` then prove that `BAB=A^(-1)` Also find the least positive value of `alpha` for which `BA^4B= A^(-1)`A.B.C.D.
If `A=[[cos alpha, -sin alpha] , [sin alpha, cos alpha]], B=[[cos2beta, sin 2beta] , [sin 2 beta, -cos2beta]]` where `0 lt beta lt pi/2` then prove that `BAB=A^(-1)` Also find the least positive value of `alpha` for which `BA^4B= A^(-1)`
If `A(alpha, beta)=[("cos" alpha,sin alpha,0),(-sin alpha,cos alpha,0),(0,0,e^(beta))]`, then `A(alpha, beta)^(-1)` is equal toA. `A(-alpha, -beta)`B. `A(-alpha, beta)`C. `A(alpha, -beta)`D. `A(alpha, beta)`
`{:A=[(cos^2 alpha,cosalphasinalpha),(cosalphasinalpha,sin^2alpha)]:}` `{:B=[(cos^2 beta,cosbetasinbeta),(cosbetasinbeta,sin^2beta)]:}` are two matrices such that the product AB is the null matrix, then `(alpha-beta)` is

If matrix A = [1 2 3], write AA’.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment