Let P = [3 -1 -2; 2 0 α; 3 -5 0], where α ∈ ℝ. Suppose Q = [qᵢⱼ] is a matrix such that PQ = kI, wher

Q: Why do we use the cofactor $C_{32}$ for the element $q_{23}$? Isn't it supposed to be $C_{23}$?

This is a classic trap! Remember that the adjoint matrix, $\text{adj}(P)$, is the *transpose* of the cofactor matrix. Therefore, the element at row $i$, column $j$ of the adjoint matrix is the cofactor $C_{ji}$. So, for $q_{23}$, we need the cofactor $C_{32}$. Always visualize the transpose operation!

Q: Why does the scalar $k$ become $k^3$ when we take the determinant of $Q = kP^{-1}$?

Great question! Remember the property $|kA| = k^n |A|$, where $n$ is the order of the matrix. Since $P$ is a $3 \times 3$ matrix, $n=3$. Thus, $|kP^{-1}| = k^3 |P^{-1}|$. Always keep the order of the matrix in mind when pulling scalars out of determinants!

Q: Is it possible to solve for $k$ without finding $\alpha$ first?

Technically, no. The determinant of $P$ depends on $\alpha$. Since the expression for $|Q|$ involves $|P|$, you need the numerical value of $|P|$ to isolate $k$. Solving for $\alpha$ is the necessary first step to unlock the rest of the problem.

Question

Let $P = \begin{bmatrix} 3 & -1 & -2 \ 2 & 0 & \alpha \ 3 & -5 & 0 \end{bmatrix}$, where $\alpha \in \mathbb{R}$. Suppose $Q = [q_{ij}]$ is a matrix such that $PQ = kI$, where $k \in \mathbb{R}, k 
eq 0$ and $I$ is the identity matrix of order $3$. If $q_{23} = -k/8$ and $\det(Q) = k^2/2$, then

Options:

(A) $\alpha = 0, k = 8$

(B) $4\alpha - k + 8 = 0$

(C) $\det(P 	ext{ adj } (Q)) = 2^9$

(D) $\det(Q 	ext{ adj } (P)) = 2^{13}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a matrix $P = \begin{bmatrix} 3 & -1 & -2 \ 2 & 0 & \alpha \ 3 & -5 & 0 \end{bmatrix}$ and the relationship $PQ = kI$. This equation implies that $Q$ is a scaled version of the inverse of $P$.

Specifically, by pre-multiplying both sides by $P^{-1}$, we obtain the relation:
$$Q = kP^{-1}$$

### The Determinant: The Heartbeat of the Matrix

To proceed, we must calculate the determinant of $P$, denoted as $|P|$. Expanding along the first row:

$$|P| = 3(0 - (-5\alpha)) - (-1)(0 - 3\alpha) + (-2)(-10 - 0)$$

Simplifying this expression, we arrive at:
$$|P| = 15\alpha + 3\alpha + 20 = 18\alpha + 20$$

*(Correction: Re-evaluating the expansion: $3(0 - (-5\alpha)) = 15\alpha$; $(-1)(-1)(0 - 3\alpha) = -3\alpha$; $(-2)(-10) = 20$. Thus, $|P| = 12\alpha + 20$.)*

### The Cofactor Hunt

We are given the specific element $q_{23} = -k/8$. Since $Q = \frac{k}{|P|} 	ext{adj}(P)$, the element $q_{23}$ corresponds to the entry in the second row and third column of the adjoint matrix.

The adjoint matrix is the transpose of the cofactor matrix, so the element at $(2,3)$ in the adjoint is the cofactor $C_{32}$ of the original matrix $P$. We calculate $C_{32}$ as follows:

$$C_{32} = (-1)^{3+2} \det \begin{bmatrix} 3 & -2 \ 2 & \alpha \end{bmatrix} = -(3\alpha - (-4)) = -(3\alpha + 4)$$

### The Algebraic Dance

Substituting our expressions into the relation $q_{23} = \frac{k}{|P|} C_{32}$, we get:

$$\frac{k(-(3\alpha + 4))}{12\alpha + 20} = -\frac{k}{8}$$

Since $k 
eq 0$, we cancel $k$ and the negative signs from both sides:

$$\frac{3\alpha + 4}{12\alpha + 20} = \frac{1}{8}$$

Cross-multiplying yields $24\alpha + 32 = 12\alpha + 20$, which simplifies to $12\alpha = -12$. Therefore, we find:
$$\alpha = -1$$

Substituting $\alpha = -1$ back into our determinant expression, we get $|P| = 12(-1) + 20 = 8$.

### The Final Victory

We are given $|Q| = k^2/2$. We also know that $|Q| = |kP^{-1}| = k^3 |P^{-1}| = \frac{k^3}{|P|}$. Equating these two expressions:

$$\frac{k^3}{8} = \frac{k^2}{2}$$

Since $k 
eq 0$, we divide by $k^2$ to obtain $\frac{k}{8} = \frac{1}{2}$, which results in:
$$k = 4$$

We can now verify the properties of the system. For instance, evaluating $4\alpha - k + 8$ with our findings:
$$4(-1) - 4 + 8 = -4 - 4 + 8 = 0$$

The logic holds, confirming that **$\alpha = -1$** and **$k = 4$**.

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Visualized Solution

Matrix $P$ and Condition

Expressing $Q$ in terms of $P$

Determinant of $P$

Finding Element $q_{23}$

Calculating Cofactor $C_{32}$

Solving for $α$

Updating $∣ P ∣$

Determinant of $Q$

Solving for $k$

Checking Option (b)

Checking Option (c)

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Determinant

The Cofactor Hunt

The Algebraic Dance

The Final Victory

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Visualized Solution

Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and Condition

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Determinant of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Finding Element .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q23​

Calculating Cofactor .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C32​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Updating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣P∣

Determinant of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k