Show that if `lambda_(1), lambda_(2), ...., lamnda_(n)` are `n` eigenvalues of a square matrix a of order n, then the eigenvalues of the matric `A^(2)` are `lambda_(1)^(2), lambda_(2)^(2),..., lambda_(n)^(2)`.
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`AX=lambdaX`
or `A(AX)=A(lambdaX)`
or `A^(2)X=lambda(AX)=lambda(lambdaX)=lambda^(2)X`
i.e., `A^(2)X=lambda^(2)X`
Hence, eigenvalue of `A^(2)` is `lambda^(2)`. Thus, if `lambda_(1), lambda_(2), ..., lambda_(n)` are eigenvalue of A, then `lambda_(1)^(2), lambda_(2)^(2), ..., lambda_(n)^(2)` are eigenvalues of `A^(2)`.
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The wavelength of the short radio waves, microwaves, ultraviolet waves are `lambda_(1),lambda_(2)` and `lambda_(3)`, respectively, Arrange them in dec
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If `A` and `B` are matrices of the same order, then `A B^T-B^T A` is a (a) skew-symmetric matrix (b) null matrix (c) unit matrix (d) symmetric matrix
Let `P=(AB^(T)-BA^(T))` `:. P^(T)=(AB^(T)-BA^(T))^(T)=(AB^(T))^(T)-(BA^(T))^(T)` `=(B^(T))^(T) (A)^(T)-(A^(T))^(T)B^(T)=BA^(T)-AB^(T)` `=-(AB^(T)-BA^(T))=-P` Hence, `(AB^(T)-BA^(T))` is a skew-symmetric matric.
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If A is a square matric of order 5 and `2A^(-1)=A^(T)`, then the remainder when `|"adj. (adj. (adj. A))"|` is divided by 7 is
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Correct Answer - (a) `5 eV` , (b) `n = 6, Z = 3`
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Let `A` be a square matrix of order `3` so that sum of elements of each row is `1`. Then the sum elements of matrix `A^(2)` is
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A be a square matrix of order `2` with `|A| ne 0` such that `|A+|A|adj(A)|=0`, where `adj(A)` is a adjoint of matrix `A`, then the value of `|A-|A|adj
Correct Answer - D `(d)` Let `A=[{:(m,n),(p,q):}]`,`adj(a)=[{:(q,-n),(-p,m):}]` Let `|A|=d=mq-np` `|A+dadjA|=|{:(m+qd,n(1-d)),(p(1-d),q+md):}|=0` `impliesmq+m^(2)d+q^(2)d+mqd^(2)-np+2npd-npd^(2)=0` `implies(mq-np)+(mq0np)d^(2)+m^(2)d+q^(2)d+2mqd-2d^(2)=0` `implies(d+d^(3)-2d^(2))+d(m^(2)+q^(2)+2mq)=0` Now, `|A- d adj|A|=-(m+q)^(2)+4(mq-np)=4d=4`
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