A Matrix problem by hobart pao

Algebra Level 4

If A = ( 2 6 6 2 2 7 3 6 1 5 0 1 3 7 0 7 ) A=\begin{pmatrix} 2 & 6 & 6 & 2 \\ 2 & 7 & 3 & 6 \\ 1 & 5 & 0 & 1 \\ 3 & 7 & 0 & 7 \end{pmatrix} , what is the determinant of A − 1 A^{-1} ? Express the answer in the form − a b -\dfrac{a}{b} in lowest terms, and find the sum a + b a+b .


The answer is 169.

Hobart Pao by Hobart Pao , 23, USA

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1 solution

∣ A − 1 ∣ = 1 ∣ A ∣ ( p r o p e r t y o f d e t e r m i n a n t i f A i s n o n s i n g u l a r . ) ∣ A ∣ = ∣ 2 6 6 2 2 7 3 6 1 5 0 1 3 7 0 7 ∣ R 1 → R 1 − 2 R 2 ∣ A ∣ = ∣ − 2 − 8 0 − 10 2 7 3 6 1 5 0 1 3 7 0 7 ∣ E x p a n d i n g u s i n g 3 r d C o l u m n ∣ A ∣ = − 3. ∣ − 2 − 8 − 10 1 5 1 3 7 7 ∣ C 2 → C 2 − C 3 ∣ A ∣ = − 3. ∣ − 2 2 − 10 1 4 1 3 0 7 ∣ C 3 → C 3 − C 1 ∣ A ∣ = − 3. ∣ − 2 2 − 8 1 4 0 3 0 4 ∣ R 1 → R 1 − 2 R 3 ∣ A ∣ = − 3. ∣ 4 2 0 1 4 0 3 0 4 ∣ e x p a n d i n g u s i n g 3 r d c o l u m n ∣ A ∣ = − 3.4. ∣ 4 2 1 4 ∣ ∣ A ∣ = − 12. ( 16 − 2 ) = − 12.14 = − 168 ∣ A − 1 ∣ = 1 ∣ A ∣ = − 1 168 = − a b a + b = 169 |{ A }^{ -1 }|=\frac { 1 }{ |A| } \quad \quad \quad \qquad \qquad (property\quad of\quad determinant\quad if\quad A\quad is\quad non\quad singular.)\\ |A|=\left| \begin{matrix} 2 & 6 & 6 & 2 \\ 2 & 7 & 3 & 6 \\ 1 & 5 & 0 & 1 \\ 3 & 7 & 0 & 7 \end{matrix} \right| \\ { R }_{ 1 }\rightarrow { R }_{ 1 }-2{ R }_{ 2 }\\ |A|=\left| \begin{matrix} -2 & -8 & 0 & -10 \\ 2 & 7 & 3 & 6 \\ 1 & 5 & 0 & 1 \\ 3 & 7 & 0 & 7 \end{matrix} \right| \\ Expanding\quad using\quad 3rd\quad Column\\ \left| A \right| =-3.\left| \begin{matrix} -2 & -8 & -10 \\ 1 & 5 & 1 \\ 3 & 7 & 7 \end{matrix} \right| \\ { C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 3 }\\ \left| A \right| =-3.\left| \begin{matrix} -2 & 2 & -10 \\ 1 & 4 & 1 \\ 3 & 0 & 7 \end{matrix} \right| \\ { C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }\\ \left| A \right| =-3.\left| \begin{matrix} -2 & 2 & -8 \\ 1 & 4 & 0 \\ 3 & 0 & 4 \end{matrix} \right| \\ { R }_{ 1 }\rightarrow { R }_{ 1 }-2{ R }_{ 3 }\\ \left| A \right| =-3.\left| \begin{matrix} 4 & 2 & 0 \\ 1 & 4 & 0 \\ 3 & 0 & 4 \end{matrix} \right| \\ expanding\quad using\quad 3rd\quad column\\ |A|=-3.4.\left| \begin{matrix} 4 & 2 \\ 1 & 4 \end{matrix} \right| \\ |A|=-12.(16-2)=-12.14=-168\\ |{ A }^{ -1 }|=\frac { 1 }{ |A| } =-\frac { 1 }{ 168 } =-\frac { a }{ b } \\ a+b=\quad 169

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Moderator note:

Good approach simplifying the determinant calculation.

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