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For a system of equations in matrix form $AX=B$,

If $|A|\neq0$, there exists a unique solution $X=A^{-1}B$.

That is fine i understand this.

If $|A|=0$,

Case 1: $(\operatorname{adj} A)\cdot B\neq O$,

then solution does not exist and the system of equations is called inconsistent.

Case 2: $(\operatorname{adj} A)\cdot B=O$,

then system may be either consistent or inconsistent according as the system have either infinitely many solutions or no solution.

How does $(\operatorname{adj} A)\cdot B$ comes into the problem and how does the case 1 and 2 statements emerge ?

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You have $(\operatorname{adj}\,A)\,A=|A|$. When you multiply $Ax=b$ by $\operatorname{adj}\,A$, you get $$ |A|=(\operatorname{adj}\,A)\,b. $$ If $|A|=0$, you need the right-hand-side to be zero.

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  • $\begingroup$ thanx. So if $(\operatorname{adj}\,A)b\neq0$ then there is no solutions as LHS is equal to zero right ?. But why is there a case where if $(\operatorname{adj}\,A)b=0$. then system may or may not be consistent according as the system have either infinitely many solutions or no solution?. I thought in that case $x$ can take any values. $\endgroup$ – ss1729 Feb 16 '18 at 22:08
  • $\begingroup$ No, the second case is the case where you still have $|A|=0$, but the system may be inconsistent or not. You can even have $A=0$ and the system be incosistent; even in dimension one: $0x=1$ is inconsistent. $\endgroup$ – Martin Argerami Feb 16 '18 at 22:39
  • $\begingroup$ actually my doubt is: If $|A|=0$ and $(\text{adj }A)b\neq{0}$ then the system is inconsistent clearly. If $|A|=0$ and $(\text{adj }A)b=0$ how can i reach the conclusion that "system may or may not be consistent according as the system have either infinitely many solutions or no solution" from $|A|x=(\text{adj }A)b$ ? $\endgroup$ – ss1729 Feb 16 '18 at 22:51
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    $\begingroup$ Because there are examples of both: $0x=1$ is inconsistent, $0x=0$ is consistent. $\endgroup$ – Martin Argerami Feb 16 '18 at 23:12

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