Question #245553

You are given that A and B are two square matrices of the same order, such that


det(A1B)=20anddet(B)=5.\mathrm{det}\left(A^{-1}B\right)=20\quad \mathrm{and} \quad \mathrm{det} \left(B\right)=5.


Which of the following is true?

  1. det(A)=15andadj(A)=115\mathrm{det}(A)=-15\quad \text{and} \quad \mathrm{adj}(A)=-\frac{1}{15}
  2. det(A)=4andadj(A)=4A1\mathrm{det}(A)=4\quad \text{and} \quad\mathrm{adj}(A)=4A^{-1}
  3. det(A)=14andadj(A)=14A1\mathrm{det}(A)=\frac{1}{4}\quad \text{and} \quad \mathrm{adj}(A)=\frac{1}{4}A^{-1}
  4. det(A)=25andadj(A)=25\mathrm{det}(A)=25\quad \text{and} \quad \mathrm{adj}(A)=25
  5. det(A)=15andadj(A)=15A1\mathrm{det}(A)=15\quad \text{and} \quad \mathrm{adj}(A)=15 A^{-1}

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1
Expert's answer
2021-10-06T16:20:55-0400

det(A1B)=detA1detB=detBdetA20=5detAdetA=14Inverse formula A1=adj(A)det(A)adj(A)=(detA)A1=14A1\det(A^{-1}B)=\det{A^{-1}}\det{B}\\=\dfrac{\det{B}}{\det{A}}\\ 20=\dfrac{5}{\det{A}}\\ \det{A}=\dfrac{1}{4}\\ \text{Inverse formula }A^{-1}=\dfrac{\text{adj(A)}}{\det(A)}\\ \text{adj(A)}=(\det{A})A^{-1}=\dfrac{1}{4}A^{-1}

True statement is 3


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