If $$S=0$$ $$1$$ $$11$$ $$0$$ $$11$$ $$1$$,$$A=12b+cc$$−ab−ac−$$bc+aa$$−bb−ca−$$ca+b$$, then SAS−$$1=$$

Question

If $$S=0$$    $$1$$    $$11$$    $$0$$    $$11$$    $$1$$,$$A=12b+cc$$−ab−ac−$$bc+aa$$−bb−ca−$$ca+b$$, then SAS−$$1=$$

Infinity Learn · Accepted Answer

$$S=011101110$$⇒det⁡$$S=0(0$$−$$1)$$−$$1(0$$−$$1)+1(1$$−$$0)$$ $$=0+1+1$$ $$=2The$$ cofactors of elements of A are $$A11=+(0$$−$$1)=$$−$$1$$,$$A12=$$−$$(0$$−$$1)=1$$,$$A13=+(1$$−$$0)=1A21=$$−$$1(0$$−$$1)=1$$,$$A22=+(0$$−$$1)=$$−$$1$$,$$A23=$$−$$(0$$−$$1)=1A31=+(1$$−$$0)=1$$,$$A32=$$−$$(0$$−$$1)=1$$,$$A33=+(0$$−$$1)=$$−$$1Adj$$⁡$$A=A11A21A31A12A22A32A13A23A33=$$−$$1111$$−$$1111$$−$$1$$∴A−$$1=AdjAdet$$⁡$$A=12$$−$$1111$$−$$1111$$−$$1SAS$$−$$1=01110111012b+cc$$−ab−ac−$$bc+aa$$−bb−ca−$$ca+b12$$−$$1111$$−$$1111$$−$$1=140+c$$−$$b+b$$−$$c0+c+a+a$$−$$c0+a$$−$$b+a+bb+c+0+b$$−ac−$$a+0+a$$−cb−$$a+0+a+bb+c+c$$−$$b+0c$$−$$a+c+a+0b$$−$$a+a$$−$$b+0$$−$$1111$$−$$1111$$−$$1=1402a2a2b02b2c2c0$$−$$1111$$−$$1111$$−$$1=140+2a+2a0$$−$$2a+2a0+2a$$−$$2a$$−$$2b+0+2b2b+0+2b2b+0$$−$$2b$$−$$2c+2c+02c$$−$$2c+02c+2c+0=144a0004b0004c=a$$    $$0$$    $$00$$    b    $$00$$    $$0$$    c

$If S = [\begin{array}{lll} 0 1 1 \\ 1 0 1 \\ 1 1 \end{array}], A = \frac{1}{2} [\begin{array}{ccc} b + c & c - a & b - a \\ c - b & c + a & a - b \\ b - c & a - c & a + b \end{array}], then S A S^{- 1} =$

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

If S=0 1 11 0 11 1,A=12b+cc−ab−ac−bc+aa−bb−ca−ca+b, then SAS−1=

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

$If S = [\begin{array}{lll} 0 1 1 \\ 1 0 1 \\ 1 1 \end{array}], A = \frac{1}{2} [\begin{array}{ccc} b + c & c - a & b - a \\ c - b & c + a & a - b \\ b - c & a - c & a + b \end{array}], then S A S^{- 1} =$