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I am first stating the question:

Let $A=\{a_{ij}\}$ be a $3\times 3$ matrix, where $$a_{ij}=\begin{cases} (-1)^{j-i}&\text{if $i<j$,}\\ 2&\text{if $i=j$,}\\ (-1)^{i-j}&\text{if $i>j$,} \end{cases}$$ then $\det(3\,\text{adj}(2A^{-1}))$ is equal to __________

I solved this in the following manner: $$ A=\left[\begin{array}{lcc} 2 & (-1)^{2-1} & (-1)^{3-1} \\ (-1)^{2+1} & 2 & (-1)^{3-2} \\ (-1)^{3+1} & (-1)^{3 + 2} & 2 \end{array}\right]=\left[\begin{array}{ccc} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right] $$ $$\begin{aligned}|A| &=2(4-1)+(-2+1)+(1-2) \\ &=6-1-1=4 \end{aligned}$$

$$ \begin{aligned} & \operatorname{det}\left(3 \operatorname{adj}\left(2 A^{-1}\right)\right) \\ =& 3^{3}\left|\operatorname{adj}\left(2 A^{-1}\right)\right| \\ =& 3^{3}\left|2^{3} \operatorname{adj}\left(A^{-1}\right)\right| \\ =&(3 \times 2)^{3} \times\left(\left|A^{-1}\right|\right)^{2}\\=&6^3\times\Big(\frac14\Big)\\=&13.5 \end{aligned} $$ Original image

Is my solution correct?

Note: The problem came in the JEE Main Exam of India, on the 20th of July. The answer given for this question in the Answer Key is 108.

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    $\begingroup$ @Wolgwang Excellent editing. Well done! $\endgroup$
    – Robert Z
    Jul 31 at 9:11
  • $\begingroup$ Since you are new here, let me tell you that the person who asked can mark one answer as "accepted". Please see math.stackexchange.com/tour $\endgroup$
    – Robert Z
    Jul 31 at 11:38
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No, your solution is not correct, but you are almost done.

Looking through the properties of adjugate matrix, we note that if $A$ is a $n\times n$ matrix then $\text{adj}(cA)=c^{n-1}\text{adj}(A)$ (not $c^n$ as you did) and $\text{adj}(A^{-1})=\det(A^{-1})A$. Therefore $$\begin{align}\det(3\,\text{adj}(2A^{-1}))&=\det(3\cdot 2^{n-1}\det(A^{-1})A)\\ &=(3\cdot 2^{n-1}\det(A^{-1}))^{n}\det(A) = \frac{3^n2^{n(n-1)}}{\det(A)^{n-1}}. \end{align}$$ Hence when $n=3$ and $\det(A)=4$, we find $$\det(3\,\text{adj}(2A^{-1})=\frac{3^32^{6}}{4^{2}}=27\cdot 4=108$$ which is precisely the given answer.

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Alternatively, use the properties: $$\begin{align}\text{adj}(cA)&=c^{n-1}\text{adj}(A) \quad (1)\\ \det(cA)&=c^n\det(A) \quad (2)\\ \det(\text{adj}(A))&=(\det(A))^{n-1} \quad (3)\\ \det(A^{-1})&=(\det(A))^{-1} \quad (4) \end{align}$$ to get: $$\det(3\,\text{adj}(2A^{-1}))\stackrel{(1)}{=}\\ \det(3\cdot 2^2\,\text{adj}(A^{-1}))\stackrel{(2)}{=}\\ 12^3\det(\text{adj}(A^{-1}))\stackrel{(3)}{=}\\ 12^3(\det(A^{-1}))^2\stackrel{(4)}{=}\\ 12^3((\det(A))^{-1})^2=\\ 12^3\cdot 4^{-2}=\\ 108.$$

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