Math, asked by akarshakd1110, 7 hours ago

If A,B,C are non zero square matrices of same order,then which one of the foolowing is not true
a)AB=BA
b)A(B+C)=AB+AC

Answers

Answered by singhpunam1985
0

Answer:

Answer

Let us assume that A is non-singular i.e. |A|

= 0 and hence A

−1

exists such that AA

−1

=I.

∴ AB=0

⟹ A

−1

(AB)=(A

−1

A)B=IB=B=0

Above shows that B is a null matrix which is a contradiction.

Similarly, if B is non-singular then as above we will have A=0 which is again a contradiction. hence, both A and B must be singular.

Step-by-step explanation:

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Answered by pulakmath007
2

SOLUTION

TO DETERMINE

If A , B , C are non zero square matrices of same order , then which one of the following is not true

a) AB = BA

b) A(B+C) = AB + AC

CONCEPT TO BE IMPLEMENTED

Matrix :

We know that a system of mn numbers arranged in a rectangular formation along M rows and n columns and bounded by the brackets is called m by n matrix which is written as m × n matrix.

A Matrix having n rows and n columns is called a square matrix of order n

EVALUATION

a) Matrix multiplication is not commutative

So AB = BA does not hold

For example let us assume that

\displaystyle \sf{}A=  \displaystyle\begin{pmatrix} 1 & 2\\ 3 & 1 \end{pmatrix}

\displaystyle \sf{B= \displaystyle\begin{pmatrix} 0 & 5\\ 1& 4 \end{pmatrix} }

Then

\displaystyle \sf{AB=  \displaystyle\begin{pmatrix} 1 & 2\\ 3 & 1 \end{pmatrix} \displaystyle\begin{pmatrix} 0 & 5\\ 1& 4 \end{pmatrix}  = \displaystyle\begin{pmatrix} 2 & 13\\ 1& 17 \end{pmatrix}}

\displaystyle \sf{BA= \displaystyle\begin{pmatrix} 0 & 5\\ 1& 4 \end{pmatrix} \displaystyle\begin{pmatrix} 1 & 2\\ 3 & 1 \end{pmatrix}   = \displaystyle\begin{pmatrix} 15 & 5\\ 13&6 \end{pmatrix}}

Thus AB ≠ BA

b) Matrix follows the distributive property

FINAL ANSWER

The untrue property on matrix is AB = BA

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