Answered

Given two matrices

[tex]\[P=\left[\begin{array}{cc}-3 & 4 \\ 14 & -1\end{array}\right]\][/tex]

and

[tex]\[Q=\left[\begin{array}{cc}4 & -2 \\ -7 & 3\end{array}\right]\][/tex]

a. Determine if [tex]\(PQ = QP\)[/tex]

b. Find [tex]\(m\)[/tex] and [tex]\(n\)[/tex] such that [tex]\(P = mQ + nI_{2}\)[/tex]



Answer :

GinnyAnswer
Certainly! Let's analyze the given problem step by step.

### Part (a): Determine if [tex]\( P Q = Q P \)[/tex]

Given matrices:
[tex]\[ P = \begin{bmatrix} -3 & 4 \\ 14 & -1 \end{bmatrix} \][/tex]
[tex]\[ Q = \begin{bmatrix} 4 & -2 \\ -7 & 3 \end{bmatrix} \][/tex]

First, we need to compute [tex]\( P Q \)[/tex]:
[tex]\[ PQ = \begin{bmatrix} -3 & 4 \\ 14 & -1 \end{bmatrix} \begin{bmatrix} 4 & -2 \\ -7 & 3 \end{bmatrix} \][/tex]

Performing the matrix multiplication:

Element at (1,1):
[tex]\[ (-3 \times 4) + (4 \times (-7)) = -12 - 28 = -40 \][/tex]

Element at (1,2):
[tex]\[ (-3 \times (-2)) + (4 \times 3) = 6 + 12 = 18 \][/tex]

Element at (2,1):
[tex]\[ (14 \times 4) + (-1 \times (-7)) = 56 + 7 = 63 \][/tex]

Element at (2,2):
[tex]\[ (14 \times (-2)) + (-1 \times 3) = -28 - 3 = -31 \][/tex]

Thus:
[tex]\[ PQ = \begin{bmatrix} -40 & 18 \\ 63 & -31 \end{bmatrix} \][/tex]

Next, we need to compute [tex]\( Q P \)[/tex]:
[tex]\[ QP = \begin{bmatrix} 4 & -2 \\ -7 & 3 \end{bmatrix} \begin{bmatrix} -3 & 4 \\ 14 & -1 \end{bmatrix} \][/tex]

Performing the matrix multiplication:

Element at (1,1):
[tex]\[ (4 \times (-3)) + (-2 \times 14) = -12 - 28 = -40 \][/tex]

Element at (1,2):
[tex]\[ (4 \times 4) + (-2 \times (-1)) = 16 + 2 = 18 \][/tex]

Element at (2,1):
[tex]\[ (-7 \times (-3)) + (3 \times 14) = 21 + 42 = 63 \][/tex]

Element at (2,2):
[tex]\[ (-7 \times 4) + (3 \times (-1)) = -28 - 3 = -31 \][/tex]

Thus:
[tex]\[ QP = \begin{bmatrix} -40 & 18 \\ 63 & -31 \end{bmatrix} \][/tex]

Since:
[tex]\[ PQ = QP = \begin{bmatrix} -40 & 18 \\ 63 & -31 \end{bmatrix} \][/tex]

We conclude that [tex]\( PQ = QP \)[/tex]. Therefore, the result is:

[tex]\[ \text{Yes, } PQ = QP \][/tex]

### Part (b): Find [tex]\( m \)[/tex] and [tex]\( n \)[/tex] such that [tex]\( P = mQ + nI_2 \)[/tex]

We need to express matrix [tex]\( P \)[/tex] as [tex]\( P = mQ + nI_2 \)[/tex] where [tex]\( I_2 \)[/tex] is the [tex]\(2 \times 2\)[/tex] identity matrix.
[tex]\[ P = \begin{bmatrix} -3 & 4 \\ 14 & -1 \end{bmatrix} \][/tex]
[tex]\[ Q = \begin{bmatrix} 4 & -2 \\ -7 & 3 \end{bmatrix} \][/tex]
[tex]\[ I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \][/tex]

Writing [tex]\( P = mQ + nI_2 \)[/tex]:
[tex]\[ \begin{bmatrix} -3 & 4 \\ 14 & -1 \end{bmatrix} = m \begin{bmatrix} 4 & -2 \\ -7 & 3 \end{bmatrix} + n \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \][/tex]

Expanding this, we get:
[tex]\[ \begin{bmatrix} -3 & 4 \\ 14 & -1 \end{bmatrix} = \begin{bmatrix} 4m + n & -2m \\ -7m & 3m + n \end{bmatrix} \][/tex]

From here, comparing the elements of both matrices, we have the following system of equations:
1. [tex]\( 4m + n = -3 \)[/tex]
2. [tex]\( -2m = 4 \)[/tex]
3. [tex]\( -7m = 14 \)[/tex]
4. [tex]\( 3m + n = -1 \)[/tex]

From equation (2),
[tex]\[ -2m = 4 \][/tex]
[tex]\[ m = -2 \][/tex]

From equation (3),
[tex]\[ -7(-2) = 14 \][/tex]
[tex]\[ 14 = 14 \][/tex]
This equation is consistent with [tex]\( m = -2 \)[/tex].

Now, substitute [tex]\( m = -2 \)[/tex] into equations (1) and (4):

1. [tex]\( 4(-2) + n = -3 \)[/tex]
[tex]\[ -8 + n = -3 \][/tex]
[tex]\[ n = 5 \][/tex]

4. [tex]\( 3(-2) + n = -1 \)[/tex]
[tex]\[ -6 + n = -1 \][/tex]
[tex]\[ n = 5 \][/tex]

Both equations confirm that [tex]\( n = 5 \)[/tex].

Thus, the values are:
[tex]\[ m \approx -2 \][/tex]
[tex]\[ n \approx 5 \][/tex]

So, the solution to the equation [tex]\( P = mQ + nI_2 \)[/tex] is:
[tex]\[ m \approx -2 \][/tex]
[tex]\[ n \approx 5 \][/tex]

Other Questions