If `P=[(1,2,4),(3,1,0),(0,0,1)], Q=[(1,-2,-3),(-3,1,9),(0,0,-5)]`then `(PQ)^(-1)` equals to

Options

A. zero matrix
B. `I_(3)`
C. diag `[-5, -5, -5]`
D. `-(1)/(5)I_(3)`

Answer

To find \((PQ)^{-1}\) for the matrices \(P\) and \(Q\) given as:

\[ P = \begin{pmatrix} 1 & 2 & 4 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad Q = \begin{pmatrix} 1 & -2 & -3 \\ -3 & 1 & 9 \\ 0 & 0 & -5 \end{pmatrix} \]

we will follow these steps:

### Step 1: Calculate the product \(PQ\)

To find \(PQ\), we will multiply matrix \(P\) by matrix \(Q\).

\[ PQ = P \cdot Q = \begin{pmatrix} 1 & 2 & 4 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & -2 & -3 \\ -3 & 1 & 9 \\ 0 & 0 & -5 \end{pmatrix} \]

Calculating each element:

- First row:
- First column: \(1 \cdot 1 + 2 \cdot (-3) + 4 \cdot 0 = 1 - 6 + 0 = -5\)
- Second column: \(1 \cdot (-2) + 2 \cdot 1 + 4 \cdot 0 = -2 + 2 + 0 = 0\)
- Third column: \(1 \cdot (-3) + 2 \cdot 9 + 4 \cdot (-5) = -3 + 18 - 20 = -5\)

- Second row:
- First column: \(3 \cdot 1 + 1 \cdot (-3) + 0 \cdot 0 = 3 - 3 + 0 = 0\)
- Second column: \(3 \cdot (-2) + 1 \cdot 1 + 0 \cdot 0 = -6 + 1 + 0 = -5\)
- Third column: \(3 \cdot (-3) + 1 \cdot 9 + 0 \cdot (-5) = -9 + 9 + 0 = 0\)

- Third row:
- First column: \(0 \cdot 1 + 0 \cdot (-3) + 1 \cdot 0 = 0 + 0 + 0 = 0\)
- Second column: \(0 \cdot (-2) + 0 \cdot 1 + 1 \cdot 0 = 0 + 0 + 0 = 0\)
- Third column: \(0 \cdot (-3) + 0 \cdot 9 + 1 \cdot (-5) = 0 + 0 - 5 = -5\)

Thus, we have:

\[ PQ = \begin{pmatrix} -5 & 0 & -5 \\ 0 & -5 & 0 \\ 0 & 0 & -5 \end{pmatrix} \]

### Step 2: Calculate the determinant of \(PQ\)

The determinant of a \(3 \times 3\) matrix \(A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}\) is given by:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \(PQ\):

\[ \text{det}(PQ) = -5 \left((-5)(-5) - (0)(0)\right) - 0 + (-5)\left(0 - 0\right) \]
\[ = -5(25) = -125 \]

### Step 3: Calculate the adjoint of \(PQ\)

The adjoint of a matrix is the transpose of its cofactor matrix. For our \(PQ\) matrix:

\[ \text{Cofactor}(PQ) = \begin{pmatrix} (-5)(-5) & 0 & -(-5)(0) \\ 0 & -5(-5) & 0 \\ 0 & 0 & (-5)(-5) \end{pmatrix} = \begin{pmatrix} 25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25 \end{pmatrix} \]

Taking the transpose gives us:

\[ \text{adj}(PQ) = \begin{pmatrix} 25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25 \end{pmatrix} \]

### Step 4: Calculate \((PQ)^{-1}\)

Now we can find \((PQ)^{-1}\):

\[ (PQ)^{-1} = \frac{1}{\text{det}(PQ)} \cdot \text{adj}(PQ) = \frac{1}{-125} \cdot \begin{pmatrix} 25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25 \end{pmatrix} \]

This simplifies to:

\[ (PQ)^{-1} = \begin{pmatrix} \frac{25}{-125} & 0 & 0 \\ 0 & \frac{25}{-125} & 0 \\ 0 & 0 & \frac{25}{-125} \end{pmatrix} = \begin{pmatrix} -\frac{1}{5} & 0 & 0 \\ 0 & -\frac{1}{5} & 0 \\ 0 & 0 & -\frac{1}{5} \end{pmatrix} \]

### Final Answer:

\[ (PQ)^{-1} = -\frac{1}{5} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = -\frac{1}{5} I \]