IPMAT Indore 2024 · Quantitative Ability · Question 12

This is a 3\times 3 magic square with three entries given: {a}_{13}=7, {a}_{32}=8, {a}_{33}=3. Call the common sum s.

Step 1 — the anti-diagonal pins the centre. Row 3 gives {z}_{1}+8+3=s, so {z}_{1}=s-11. The anti-diagonal {a}_{13}+{a}_{22}+{a}_{31} gives

7+{y}_{2}+\left( s-11 \right)=s \Rightarrow \boxed{{y}_{2}=4}

(The centre of a magic square never depends on s once two entries on a through-line are known.)

Step 2 — work out the rest in terms of s.

Column 3: 7+{y}_{3}+3=s\Rightarrow{y}_{3}=s-10

Main diagonal: {x}_{1}+4+3=s\Rightarrow{x}_{1}=s-7

Column 1: {x}_{1}+{y}_{1}+{z}_{1}=s\Rightarrow \left( s-7 \right)+{y}_{1}+\left( s-11 \right)=s\Rightarrow{y}_{1}=18-s

Step 3 — row 2 fixes s.

{y}_{1}+{y}_{2}+{y}_{3}=\left( 18-s \right)+4+\left( s-10 \right)=12

The s terms cancel, leaving 12=s. The magic sum is forced to be 12.

Step 4 — fill in. {x}_{1}=5, {y}_{1}=6, {y}_{2}=4, {y}_{3}=2, {z}_{1}=1, and row 1 gives {x}_{2}=12-5-7=0:

A=\left[ \begin{matrix}5 & 0 & 7 \\ 6 & 4 & 2 \\ 1 & 8 & 3\end{matrix}\right]

Check column 2: 0+4+8=12 ✔

Step 5 — expand the determinant.

\left| A \right|=5\left( 4\cdot 3-2\cdot 8 \right)-0\left( 6\cdot 3-2\cdot 1 \right)+7\left( 6\cdot 8-4\cdot 1 \right)

=5\left( -4 \right)+7\left( 44 \right)=-20+308=288

Answer: 288

Shortcut. Two magic-square facts end this fast.

The centre is always s/3. Here the anti-diagonal 7+{y}_{2}+{z}_{1}=s together with row 3 ({z}_{1}=s-11) gives {y}_{2}=4, so s=12 immediately.

Once s=12 the grid fills itself by subtraction — no simultaneous equations.

Then \det =288. Note also that 288=2\times{12}^{2}, an instance of the general identity \det =2{s}^{2}\cdot \frac{\text{(something)}}{} for magic squares built this way — but filling the grid is faster than recalling it.