Mathematics - Matrices Question with Solution | TestHub

Given $A\cdot \text{adj}(\text{adj}(A))=32I$. For a $3\times 3$ matrix, the property is $\text{adj}(\text{adj}(A))=(\det A{)}^{3-2}A=(\det A)A$. Substituting this into the given equation: $A\cdot (\det A)A=32I\Rightarrow (\det A)A^2=32I$. Let $\det A=d$. Then $d{A}^2=32I$. Taking the determinant of both sides:$\det (d{A}^2)=\det (32I)\Rightarrow d^3(\det A{)}^2=32^3(\det I)\Rightarrow d^3{d}^2=32^3\Rightarrow d^5=(2^5{)}^3=2^{15}$. Thus,$d=(2^{15}{)}^{1/5}=2^3=8$. So, $\det A=8$. P. $\det (A)=8$. (Matches 2) Q. $\det (\text{adj}(A))=(\det A{)}^{3-1}=(\det A{)}^2=8^2=64$. (Matches 3) R. $\det (A^{-1})=\frac{1}{\det A}=\frac{1}{8}$. (Matches 1)

S. From $d{A}^2=32I$, and $d=8$, we have $8A^2=32I\Rightarrow A^2=4I$. Since $\det A=8\ne 0$, $A$ is invertible. Multiplying $A^2=4I$ by $A^{-1}$, we get $A=4A^{-1}$, or $A^{-1}=\frac{1}{4}A$. Then$A^{-1}+A=\frac{1}{4}A+A=\frac{5}{4}A$.$\det (A^{-1}+A)=\det (\frac{5}{4}A)=(\frac{5}{4}{)}^3\det A=\frac{125}{64}\times 8=\frac{125}{8}$. (Matches 4)