Let P = [[3, -1, -2], [2, 0, α], [3, -5, 0]], where α ∈ R. Suppose Q = [qij] is a matrix satisfying

Q: Wait, why is $q_{23} = \frac{k}{|P|} C_{32}$ and not $C_{23}$?

That is a fantastic question! Remember that the adjoint of a matrix, $\text{adj}(P)$, is defined as the transpose of the cofactor matrix. So, the element in the $i$-th row and $j$-th column of the adjoint matrix is actually the cofactor $C_{ji}$ (with indices swapped). Since $Q = \frac{k}{|P|} \text{adj}(P)$, the element $q_{ij}$ must be $\frac{k}{|P|} C_{ji}$. Thus, $q_{23}$ corresponds to $C_{32}$.

Q: Why can we just cancel $k$ from both sides of the equation?

We can safely cancel $k$ because the problem explicitly states that $k$ is a non-zero constant ($k \neq 0$). In algebra, dividing by a variable is only forbidden if that variable could be zero. Since we are guaranteed $k \neq 0$, it is perfectly valid to divide both sides by $k$ to simplify our equation.

Q: How did $|PQ| = |kI_3|$ become $|P||Q| = k^3$?

This relies on two key properties of determinants. First, the determinant of a product is the product of the determinants: $|PQ| = |P||Q|$. Second, for a $3 \times 3$ matrix, the determinant of a scalar multiple $kA$ is $k^3|A|$. Since the identity matrix $I_3$ has a determinant of 1, $|kI_3| = k^3 \cdot |I_3| = k^3 \cdot 1 = k^3$. Putting these together gives $|P||Q| = k^3$.

Question

Let $P = \begin{bmatrix} 3 & -1 & -2 \ 2 & 0 & \alpha \ 3 & -5 & 0 \end{bmatrix}$, where $\alpha \in R$. Suppose $Q = [q_{ij}]$ is a matrix satisfying $PQ = kI_3$ for some non-zero $k \in R$. If $q_{23} = -\frac{k}{8}$ and $|Q| = \frac{k^2}{2}$, then $\alpha^2 + k^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex system of linear transformations. You have a matrix $P$, a mysterious entity containing an unknown parameter $\alpha$, and another matrix $Q$ that acts as a scaled inverse.

The relationship $PQ = kI_3$ is not just an equation; it is a geometric statement about how these matrices interact. We know that $PQ = kI_3$ implies $Q = kP^{-1}$.

This is our master key. We know that the inverse of any non-singular matrix $P$ is given by $P^{-1} = \frac{1}{|P|} 	ext{adj}(P)$. Therefore, we have the following relationship:

$$Q = \frac{k}{|P|} 	ext{adj}(P)$$

This tells us that every element of $Q$ is essentially a scaled cofactor of $P$.

### The Engine of the Solution: The Determinant

Before we can unlock the secrets of $\alpha$, we must understand the structure of $P$. Let us calculate the determinant $|P|$ by expanding along the first row:

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

Simplifying this, we get $3(5\alpha) + 1(-3\alpha) + 20$, which boils down to a clean expression:

$$|P| = 12\alpha + 20$$

This expression is the heartbeat of our problem.

### The Cofactor Connection

Now, let us look at the specific element $q_{23} = -\frac{k}{8}$. As we discussed, $q_{23} = \frac{k}{|P|} C_{32}$.

To find $C_{32}$, we delete the third row and second column of $P$. The remaining $2 	imes 2$ determinant is:

$$\begin{vmatrix} 3 & -2 \ 2 & \alpha \end{vmatrix} = 3\alpha - (-4) = 3\alpha + 4$$

Applying the sign convention $(-1)^{3+2} = -1$, we find $C_{32} = -(3\alpha + 4)$. Now, we substitute everything into our equation:

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

Since $k 
eq 0$, we cancel $-k$ from both sides, leaving:

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

Cross-multiplying gives $8(3\alpha + 4) = 12\alpha + 20$, which simplifies to $24\alpha + 32 = 12\alpha + 20$. Solving for $\alpha$, we find $12\alpha = -12$, so **$\alpha = -1$**.

### The Final Synthesis

With $\alpha = -1$, we find $|P| = 12(-1) + 20 = 8$. We are given $|Q| = \frac{k^2}{2}$.

Using the property $|PQ| = |P||Q| = k^3$, we substitute our values:

$$8 \cdot \frac{k^2}{2} = k^3 \Rightarrow 4k^2 = k^3$$

Again, since $k 
eq 0$, we divide by $k^2$ to get **$k = 4$**. Finally, we calculate the target expression:

$$\alpha^2 + k^2 = (-1)^2 + 4^2 = 1 + 16 = 17$$

We have navigated the matrix landscape and arrived at the truth. The final result is **17**.

Enter Numerical Value:

Visualized Solution

Introduction to Matrices $P$ and $Q$

Relating $Q$ to the Adjoint of $P$

Calculating the Determinant $∣ P ∣$

Mapping Element $q_{23}$ to Cofactors

Finding the Cofactor $C_{32}$

Setting up the Equation for $α$

Solving for $α$

Using Determinant Properties to find $k$

Solving for $k$

Final Calculation and Conclusion

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Engine of the Solution

The Cofactor Connection

The Final Synthesis

Similar Questions

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Enter Numerical Value:

Visualized Solution

Introduction to Matrices .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q to the Adjoint of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

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

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

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

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

Using Determinant Properties to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

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