IPMAT Indore Matrices & Determinants — PYPs

The Setup: This is a top-tier matrix determinant manipulation. When a determinant's column contains a sum or difference, it can be split into two separate determinants. The given $a=3$ is literal clickbait meant to throw you off; it has absolutely zero impact on the algebraic proof.
Step 1: Split the original determinant. We separate the third column $\left(a^3-1,\ b^3-1,\ c^3-1\right)$ into two determinants:

$$\Delta=\left|\begin{matrix}a&a^2&a^3\\b&b^2&b^3\\c&c^2&c^3\end{matrix}\right|-\left|\begin{matrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{matrix}\right|=0$$

Step 2: Factor the first determinant. In the first matrix, notice that row 1 shares an $a$, row 2 shares a $b$, and row 3 shares a $c$. Let's factor them out completely:

$$abc\left|\begin{matrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{matrix}\right|$$

Step 3: Align the second determinant. We want the second matrix to match the first, so we walk its last column to the front by two swaps.
Swap Column 3 with Column 2 (sign flips):

$$\left|\begin{matrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{matrix}\right|=-\left|\begin{matrix}a&1&a^2\\b&1&b^2\\c&1&c^2\end{matrix}\right|$$

Now swap Column 2 with Column 1 (sign flips back):

$$=+\left|\begin{matrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{matrix}\right|$$

Two swaps, two sign flips, so the second determinant is exactly the same object as the first one's cofactor - no leftover minus sign.
Step 4: Factor the entire equation. Let the aligned matrix be $V$ (a Vandermonde determinant). Our original equation $\Delta=0$ now looks like this:

$$abc(V)-V=0 \implies V(abc-1)=0$$

The Vandermonde determinant factors cleanly as $V=(b-a)(c-a)(c-b)$. Because the problem states $a$, $b$ and $c$ are distinct real numbers, every one of those three brackets is non-zero, so $V\neq 0$. Therefore the other multiplier must be zero:

$$abc-1=0\implies abc=1$$

Final Answer: 1

The Setup: A determinant-properties speedrun. Four rules do all the work: $\det(kM)=k^n\det M$ for an $n\times n$ matrix, $\det(XY)=\det X\det Y$, $\det(X^{-1})=\frac{1}{\det X}$, and $\det(X^T)=\det X$.
Step 1: Extract the scalar. The matrices are $4\times 4$, so $n=4$ and the scalar 2 comes out to the fourth power - not the first, which is the classic slip here:

$$\det\left(2AB^{-1}C^3B^T\right)=2^4\det\left(AB^{-1}C^3B^T\right)=16\det\left(AB^{-1}C^3B^T\right)$$

Step 2: Shatter the product. The determinant is multiplicative, so it distributes across the four factors:

$$16\times\det A\times\det\left(B^{-1}\right)\times\det\left(C^3\right)\times\det\left(B^T\right)$$

Step 3: Substitute the properties. $\det(B^{-1})=\frac{1}{\det B}$, $\det(C^3)=(\det C)^3$, and $\det(B^T)=\det B$:

$$16\times\det A\times\frac{1}{\det B}\times(\det C)^3\times\det B$$

Step 4: Cancel and calculate. The transpose contributes $\det B$ and the inverse contributes $\frac{1}{\det B}$, so they annihilate each other:

$$16\times\det A\times(\det C)^3=16\times 5\times\left(\frac{1}{2}\right)^3=80\times\frac{1}{8}=10$$

**Step 5: Notice what $\det B=-3$ was for.** Its *value* never reaches the answer - it cancels. But its being non-zero is exactly what guarantees $B$ is invertible, so $B^{-1}$ exists and the expression is meaningful at all. It is a load-bearing hypothesis disguised as a red herring.

Final Answer: 10

The Setup: A binomial expansion on matrices. The condition $A^2=0$ (a nilpotent matrix) is a kill switch: every power of $A$ from the second onward vanishes, truncating a 51-term expansion into two terms.
Step 0: Check the binomial theorem is even allowed. For matrices, $(X+Y)^n$ expands binomially **only if $X$ and $Y$ commute** - otherwise the cross terms will not collect. Here $X=I$, and the identity commutes with every matrix, so $IA=AI=A$ and the expansion is valid. Skipping this check is how the same move goes wrong on a harder problem.
Step 1: Expand.

$$(I+A)^{50}=I^{50}+\binom{50}{1}I^{49}A+\binom{50}{2}I^{48}A^2+\cdots+A^{50}$$

Step 2: Trigger the kill switch. From $A^2=0$ it follows that $A^3=A\cdot A^2=A\cdot 0=0$, and inductively every higher power is zero too. So every term from the $A^2$ one onward dies, and $I^m=I$:

$$(I+A)^{50}=I+50A$$

Step 3: Simplify the expression.

$$(I+A)^{50}-50A=(I+50A)-50A=I$$

The $50A$ subtraction is built into the question precisely to cancel the surviving term, leaving the identity.
Step 4: Take the determinant. The determinant of the identity matrix is 1 in any dimension:

$$\det(I)=1$$

Note the answer never depended on *which* nilpotent $A$ was chosen, only on $A^2=0$ - the 'non-zero' in the stem just stops the problem being trivial.

Final Answer: 1

The Setup: We are literally entering the Matrix for this one. This is a linear algebra scaling problem where we need to find the cube of a $2 \times 2$ matrix, equate it to a scaled target matrix, and extract the hidden variables. I have fully audited the matrix multiplication below to ensure zero dropped frames.

Step 1: Squaring the Matrix (Level 1)
First, we need to find $A^2$ by multiplying matrix $A$ by itself. We use the standard row-by-column dot product method.
$$A^2 = \begin{bmatrix} 2 & n \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 2 & n \\ 4 & 1 \end{bmatrix}$$
$$A^2 = \begin{bmatrix} (2)(2)+(n)(4) & (2)(n)+(n)(1) \\ (4)(2)+(1)(4) & (4)(n)+(1)(1) \end{bmatrix}$$
$$A^2 = \begin{bmatrix} 4+4n & 3n \\ 12 & 4n+1 \end{bmatrix}$$

Step 2: Cubing the Matrix (Boss Phase)
Now we multiply our $A^2$ result by $A$ to reach the final form, $A^3$.
$$A^3 = \begin{bmatrix} 4+4n & 3n \\ 12 & 4n+1 \end{bmatrix} \begin{bmatrix} 2 & n \\ 4 & 1 \end{bmatrix}$$
Let's calculate each element one by one to avoid throwing:
* Top-Left: $(4+4n)(2) + (3n)(4) = 8 + 8n + 12n = 8+20n$
* Top-Right: $(4+4n)(n) + (3n)(1) = 4n^2 + 4n + 3n = 4n^2+7n$
* Bottom-Left: $(12)(2) + (4n+1)(4) = 24 + 16n + 4 = 16n+28$
* Bottom-Right: $(12)(n) + (4n+1)(1) = 12n + 4n + 1 = 16n+1$

So, our fully loaded $A^3$ matrix is:
$$A^3 = \begin{bmatrix} 8+20n & 4n^2+7n \\ 16n+28 & 16n+1 \end{bmatrix}$$

Step 3: Decoding the Target Matrix
The problem gives us the target state for $A^3$. Let's distribute that scalar multiplier (27) into the matrix so we can match it element for element.
$$A^3 = 27\begin{bmatrix} 4 & q \\ p & r \end{bmatrix} = \begin{bmatrix} 108 & 27q \\ 27p & 27r \end{bmatrix}$$

Step 4: Extracting the Lore (Solving for Variables)
We now set the elements of our calculated $A^3$ equal to the target matrix. The top-left element is the key to unlocking the entire board because it only relies on $n$.
$$8+20n = 108$$
$$20n = 100$$
$$n = 5$$

Now substitute $n=5$ into the other elements to find $p$, $q$, and $r$:
* **Top-Right ($q$):** $4(5)^2 + 7(5) = 100 + 35 = 135$
$$27q = 135 \rightarrow q = 5$$
* **Bottom-Left ($p$):** $16(5) + 28 = 80 + 28 = 108$
$$27p = 108 \rightarrow p = 4$$
* **Bottom-Right ($r$):** $16(5) + 1 = 80 + 1 = 81$
$$27r = 81 \rightarrow r = 3$$

Step 5: The Final Calculation
The question asks for the sum of $p+q+r$. Just add up our freshly extracted stats.
$$4 + 5 + 3 = 12$$

Step 6: The Audit (Double Check Protocol)
Let's run a determinant check to confirm the matrix integrity.
If $n=5$, matrix $A = \begin{bmatrix} 2 & 5 \\ 4 & 1 \end{bmatrix}$.
The determinant of $A$ is $(2)(1) - (5)(4) = -18$.
Using determinant properties, $\det(A^3) = (\det(A))^3 = (-18)^3 = -5832$.
Now let's check the determinant of our target matrix $27\begin{bmatrix} 4 & 5 \\ 4 & 3 \end{bmatrix}$:
$$\det = 27^2 \times ((4)(3) - (5)(4)) = 729 \times (12 - 20) = 729 \times (-8) = -5832$$.
The determinants match perfectly. The math is completely flawless.

Final Answer: 12

The Setup: A $3 \times 3$ matrix is given with elements $[x_1, x_2, 7]$, $[y_1, y_2, y_3]$, and $[z_1, 8, 3]$ such that the sums across any row, column, or diagonal are identical. We must find its determinant.

Step 1: Determine the magic sum $S$ and the center element.
Let $S$ equal the constant sum of any row, column, or diagonal.
Analyze the bottom row:
$$z_1 + 8 + 3 = S \Rightarrow z_1 = S - 11$$
Analyze the right-to-left diagonal:
$$7 + y_2 + z_1 = S$$
Substitute $z_1$ into the diagonal equation:
$$7 + y_2 + (S - 11) = S \Rightarrow y_2 - 4 = 0 \Rightarrow y_2 = 4$$
In a $3 \times 3$ magic square, the central element is always exactly $\frac{1}{3}$ of the magic sum.
$$S = 3(y_2) = 12$$

Step 2: Populate the remaining elements of matrix $A$ using $S = 12$.
From Step 1, $z_1 = 12 - 11 = 1$.
Left-to-Right Diagonal:
$$x_1 + y_2 + 3 = 12 \Rightarrow x_1 + 4 + 3 = 12 \Rightarrow x_1 = 5$$
Top Row:
$$5 + x_2 + 7 = 12 \Rightarrow x_2 = 0$$
Left Column:
$$5 + y_1 + 1 = 12 \Rightarrow y_1 = 6$$
Right Column:
$$7 + y_3 + 3 = 12 \Rightarrow y_3 = 2$$

Step 3: Construct the populated matrix $A$.
$$A = \begin{bmatrix} 5 & 0 & 7 \\ 6 & 4 & 2 \\ 1 & 8 & 3 \end{bmatrix}$$

Step 4: Calculate the determinant $|A|$ by expanding along the top row.
$$|A| = 5((4)(3) - (8)(2)) - 0 + 7((6)(8) - (1)(4))$$
$$|A| = 5(12 - 16) + 7(48 - 4)$$
$$|A| = 5(-4) + 7(44) = -20 + 308 = 288$$

Final Answer: 288

The Setup: A specific permutation matrix $A$ is given. We must establish the cyclical property of its powers to evaluate the absolute determinant of a matrix polynomial $(A^9+A^6+A^3+A)$.

Step 1: Establish the power cycle of Matrix $A$.

$$A = \begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}$$

Calculate $A^2$:

$$A^2 = \begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix} \begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} = I$$

Since $A^2 = I$, the powers of $A$ alternate: odd powers yield $A$, and even powers yield the Identity matrix $I$.

Step 2: Simplify the matrix polynomial.
The expression is $A^9 + A^6 + A^3 + A$.
Apply the parity rules:
* $A^9 = A$ (odd)
* $A^6 = I$ (even)
* $A^3 = A$ (odd)
* $A = A$ (odd)

$$\text{Sum Matrix} = A + I + A + A = 3A + I$$

Step 3: Construct the elements of the final matrix.

$$3A = \begin{bmatrix}3&0&0\\0&0&3\\0&3&0\end{bmatrix}$$

Add the Identity matrix $I$:

$$3A + I = \begin{bmatrix}4&0&0\\0&1&3\\0&3&1\end{bmatrix}$$

Step 4: Calculate the determinant.
Expand along the top row:

$$\det(3A+I) = 4(1 \times 1 - 3 \times 3) = 4(1 - 9) = 4(-8) = -32$$

The absolute value of the determinant is $|-32| = 32$.

Final Answer: 32

The Setup: A matrix built under severe constraints. Three entries drawn from $\{+1,-1\}$ can only sum to 1 as $(+1)+(+1)+(-1)$ - no other combination works, since three $+1$s give 3 and anything with two $-1$s gives $-1$. So **every row and every column holds exactly two $+1$s and one $-1$. Step 1: See the structure.** Exactly one $-1$ per row and exactly one per column means the positions of the $-1$s form a permutation of the three columns across the three rows. There are therefore exactly $3!=6$ matrices satisfying the constraints - the problem is far more rigid than it first looks.
Step 2: Compute the prototype. Take the permutation that puts the $-1$s on the main diagonal:

$$M=\begin{pmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{pmatrix}$$

Row sums: $-1+1+1=1$. Column sums: likewise 1. A valid build. Expanding along the first row:

$$|M|=-1(1-1)-1(-1-1)+1(1-(-1))=-1(0)-1(-2)+1(2)=0+2+2=4$$

Step 3: Show the value is forced, not lucky. Every other valid matrix comes from this one by permuting rows (equivalently, by relabelling which column carries each row's $-1$). A row swap flips the sign of a determinant and nothing else, so across the whole family the determinant is $\pm 4$ and its absolute value is invariant at 4. Since $3!=6$ is a small family, this is checkable exhaustively rather than by assertion: all six matrices give $|\det|=4$, three with $\det=+4$ and three with $\det=-4$, matching the even and odd permutations.
That is exactly why the question asks for the *absolute* value - the determinant itself is not determined by the constraints, but its magnitude is.

Final Answer: 4

The Setup: Matrix algebra usually blocks you from treating matrix multiplication like standard variables because $AB \neq BA$. However, the prompt gives us the $AB=BA$ buff. This means $A$ and $B$ commute. Even better, since $A(A+B)=A^2+AB=A^2+BA=(A+B)A$, matrix $A$ also commutes with $(A+B)$ and its inverse $(A+B)^{-1}$.
Step 1: Group the complicated inverse terms together to see what we're working with. Because $A$ commutes with $(A+B)^{-1}$, we can rewrite $A(A+B)^{-1}A$ as $A^2(A+B)^{-1}$. The identical argument runs for $B$ (it commutes with $A+B$ for exactly the same reason), so $B(A+B)^{-1}B$ becomes $B^2(A+B)^{-1}$.

$$-A^2(A+B)^{-1}+B^2(A+B)^{-1}=(B^2-A^2)(A+B)^{-1}$$

Step 2: Factor the difference of squares. Because $AB=BA$, the cross terms cancel out perfectly, allowing standard factorization.

$$B^2-A^2=(B-A)(B+A)$$

Step 3: Substitute the factored form back in. Notice that matrix addition is always commutative, so $(B+A)=(A+B)$.

$$(B-A)(A+B)(A+B)^{-1}$$

Since $(A+B)$ multiplied by its inverse is the Identity matrix ($I$), this entire chunk simplifies to just $(B-A)$.
Step 4: Plug this beautifully simplified chunk back into the main equation.

$$2A-B+(B-A)$$

$$2A-B+B-A=A$$

Final Answer: A

The Setup: We are given a $2 \times 2$ matrix $A$ with a variable $a$. We must find a valid value for $a$ given that $\det(A^3-3A^2-5A)=0$.

Step 1: Factor the matrix polynomial.

$$A^3 - 3A^2 - 5A = A(A^2 - 3A - 5I)$$

Using the determinant product rule $\det(XY) = \det(X)\det(Y)$, the condition becomes:

$$\det(A) \cdot \det(A^2 - 3A - 5I) = 0$$

This means that either $\det(A) = 0$ or $\det(A^2 - 3A - 5I) = 0$.

Step 2: Test the first determinant condition.
Calculate the determinant of matrix $A$:

$$A = \begin{bmatrix} 1 & 2 \\ 3 & a \end{bmatrix}$$

$$\det(A) = (1)(a) - (2)(3) = a - 6$$

If $\det(A) = 0$, then $a - 6 = 0 \implies a = 6$.
Checking the multiple-choice options provided in the prompt, $6$ is a listed option.

Final Answer: 6

The Setup: A $2\times 2$ matrix $M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is singular when its determinant vanishes, i.e. $ad-bc=0$, so we need the main-diagonal product to equal the anti-diagonal product: $ad=bc$. The whole question reduces to finding which four numbers from $\{1,2,3,4,5,6\}$ can be split into two disjoint pairs of equal product.
Step 1: Calculate the total sample space. We pick 4 distinct numbers from 6 and place them in 4 labelled positions, so order matters - this is a permutation:

$$\text{Total outcomes}=P(6,4)=6\times 5\times 4\times 3=360$$

Step 2: Find every pair of pairs with equal products. Rather than assert it, list all 15 pairs and their products outright:

Exactly two product values repeat, and in both cases the two pairs are disjoint (they share no element, which is required since all four entries must be distinct):
* Product $6$: $\{1,6\}$ and $\{2,3\}$
* Product $12$: $\{2,6\}$ and $\{3,4\}$

Every other product in the table is unique, so no further singular matrices exist.
Step 3: Count matrix arrangements. Take product 6. Put $\{1,6\}$ on the main diagonal $(a,d)$ and $\{2,3\}$ on the anti-diagonal $(b,c)$:
* Arrangements of $\{1,6\}$ across $a,d$: $2!=2$
* Arrangements of $\{2,3\}$ across $b,c$: $2!=2$

That gives $2\times 2=4$ matrices. But the two pairs can also switch diagonals - $\{2,3\}$ on the main, $\{1,6\}$ on the anti - for another 4. So product 6 contributes $8$ matrices, and by identical logic so does product 12.
Step 4: Final probability.

$$\text{Singular matrices}=8+8=16 \implies P=\frac{16}{360}=\frac{2}{45}$$

Final Answer: $\frac{2}{45}$

The Setup: A specific circulant $3 \times 3$ matrix $A$ is populated with integers $a > b > c > 0$. We need to evaluate the algebraic sign of its determinant.

Step 1: Expand the determinant of Matrix $A$.

$$A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$$

Calculate the determinant using the standard top-row expansion:

$$\det(A) = a(bc - a^2) - b(b^2 - ac) + c(ab - c^2)$$

$$\det(A) = abc - a^3 - b^3 + abc + abc - c^3$$

$$\det(A) = 3abc - a^3 - b^3 - c^3$$

Step 2: Factor the resulting polynomial.
The expression $a^3 + b^3 + c^3 - 3abc$ possesses a highly standardized factorization identity:

$$a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2]$$

Notice that our determinant is the exact algebraic negative of this expression:

$$\det(A) = - \left( \frac{1}{2}(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2] \right)$$

Step 3: Evaluate the bounds of the factored terms.
We are strictly given $a > b > c > 0$.
Because all terms are strictly positive, their sum $(a+b+c)$ is strictly positive.
Because the terms are strictly unequal, the squared differences $(a-b)^2, (b-c)^2, (c-a)^2$ are strictly positive non-zeros.
Thus, the entire bracketed term evaluates to a strictly positive number. Because a negative sign sits in front of it, the total determinant is strictly negative.

Final Answer: is negative

The Setup: We are given $A = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 0 \end{bmatrix}$ and must identify $A^{2022}$ among the listed options. The decisive detail is the bottom-right entry, which is $0$.

Step 1: Compute $A^2$.

$$A^2 = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 0 \end{bmatrix} = \begin{bmatrix} 1(1) + 0\left(\frac{1}{2}\right) & 1(0) + 0(0) \\ \frac{1}{2}(1) + 0\left(\frac{1}{2}\right) & \frac{1}{2}(0) + 0(0) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 0 \end{bmatrix} = A$$

Because the second column of $A$ is entirely zero, multiplying by $A$ reproduces $A$ exactly.

Step 2: Extend the result to every power.
The matrix is therefore idempotent ($A^2 = A$). By induction, every higher power collapses back to $A$:

$$A^3 = A^2 A = A A = A, \qquad A^4 = A, \qquad \dots, \qquad A^n = A \text{ for all } n \ge 1$$

The bottom-left entry never grows.

$$A^{2022} = A = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & 0 \end{bmatrix}$$

Step 3: Compare against the given options.
The three explicit choices offer bottom-left entries of $\frac{1}{2^{2022}}$, $1011$ and $2022$. Our result has bottom-left entry $\frac{1}{2}$, which matches none of them.

*A note on the trap:* had the bottom-right entry been $1$ rather than $0$, the matrix would satisfy $A^n = \begin{bmatrix} 1 & 0 \\ \frac{n}{2} & 1 \end{bmatrix}$, giving a bottom-left entry of $\frac{2022}{2} = 1011$. That is precisely the decoy offered in the second option. With the printed $0$ in the bottom-right, that reasoning does not apply and the power never accumulates.