​Pls solve

Dear Student,
Please find below the solution to the asked query:

We have,a+xa-xa-xa-xa+xa-xa-xa-xa+x=0R1R1+R2+R33a-x3a-x3a-xa-xa+xa-xa-xa-xa+x=03a-x111a-xa+xa-xa-xa-xa+x=0C1C1-C2, C3C3-C23a-x010-2xa+x-2x0a-x2x=0Taking 2x common from C1 and C2, we get4x23a-x010-1a+x-10a-x1=0Expanding along R1, we get4x23a-x-1-1+0=04x23a-x=0x2=0 or 3a-x=0 x=0 or x=3a

Hope this information will clear your doubts about the topic.

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x=0,3a
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