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SECTION B

QUESTION 4

(a) Find [tex]\( x, y, z, t \)[/tex] where

[tex]\[ 3\begin{pmatrix} x & y \\ z & t \end{pmatrix} = \begin{pmatrix} x & 6 \\ -1 & 2t \end{pmatrix} + \begin{pmatrix} 4 & x+y \\ z+t & 3 \end{pmatrix} \][/tex]

(b) If [tex]\( A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix} \)[/tex] and [tex]\( B = \begin{pmatrix} -3 & -2 \\ 1 & -5 \\ 4 & 3 \end{pmatrix} \)[/tex], find [tex]\( D = \begin{pmatrix} p & q \\ r & s \\ t & z \end{pmatrix} \)[/tex] such that [tex]\( A + B - D = 0 \)[/tex]

Sagot :

GinnyAnsweridn
Let's solve this step by step:

### Part (a)
We need to find the values of [tex]\( x, y, z, \)[/tex] and [tex]\( t \)[/tex] in the given matrix equation:

[tex]\[ 3 \begin{pmatrix} x & y \\ z & t \end{pmatrix} = \begin{pmatrix} x & 6 \\ -1 & 2t \end{pmatrix} + \begin{pmatrix} 4 & x + y \\ z + t & 3 \end{pmatrix} \][/tex]

First, let's write the equation for each element in the matrices:

1. For the (1,1) element:
[tex]\[ 3x = x + 4 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 3x - x = 4 \][/tex]
[tex]\[ 2x = 4 \][/tex]
[tex]\[ x = 2 \][/tex]

2. For the (1,2) element:
[tex]\[ 3y = 6 + (x + y) \][/tex]
Substituting [tex]\( x = 2 \)[/tex]:
[tex]\[ 3y = 6 + (2 + y) \][/tex]
[tex]\[ 3y = 8 + y \][/tex]
[tex]\[ 3y - y = 8 \][/tex]
[tex]\[ 2y = 8 \][/tex]
[tex]\[ y = 4 \][/tex]

3. For the (2,1) element:
[tex]\[ 3z = -1 + (z + t) \][/tex]
[tex]\[ 3z = -1 + z + t \][/tex]
[tex]\[ 3z - z = -1 + t \][/tex]
[tex]\[ 2z = -1 + t \][/tex]
Solving for [tex]\( z \)[/tex]:
[tex]\[ z = \frac{-1 + t}{2} \][/tex]

4. For the (2,2) element:
[tex]\[ 3t = 2t + 3 \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ 3t - 2t = 3 \][/tex]
[tex]\[ t = 3 \][/tex]

Now, substituting [tex]\( t = 3 \)[/tex] into the equation for [tex]\( z \)[/tex]:
[tex]\[ z = \frac{-1 + 3}{2} \][/tex]
[tex]\[ z = \frac{2}{2} \][/tex]
[tex]\[ z = 1 \][/tex]

So, we have:
[tex]\[ x = 2, \, y = 4, \, z = 1, \, t = 3 \][/tex]

### Part (b)
We are given matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex], and need to find matrix [tex]\( D \)[/tex] such that:

[tex]\[ A + B - D = 0 \][/tex]
This implies:
[tex]\[ D = A + B \][/tex]

Let's calculate [tex]\( D \)[/tex]:

Given:
[tex]\[ A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} -3 & -2 \\ 1 & -5 \\ 4 & 3 \end{pmatrix} \][/tex]

Adding [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:

[tex]\[ D = \begin{pmatrix} 1 + (-3) & 2 + (-2) \\ 3 + 1 & 4 + (-5) \\ 5 + 4 & 6 + 3 \end{pmatrix} \][/tex]
[tex]\[ D = \begin{pmatrix} -2 & 0 \\ 4 & -1 \\ 9 & 9 \end{pmatrix} \][/tex]

So, matrix [tex]\( D \)[/tex] is:
[tex]\[ D = \begin{pmatrix} p & q \\ r & s \\ t & z \end{pmatrix} = \begin{pmatrix} -2 & 0 \\ 4 & -1 \\ 9 & 9 \end{pmatrix} \][/tex]

Thus, the values are [tex]\( p = -2 \)[/tex], [tex]\( q = 0 \)[/tex], [tex]\( r = 4 \)[/tex], [tex]\( s = -1 \)[/tex], [tex]\( t = 9 \)[/tex], and [tex]\( z = 9 \)[/tex].

### Final Answers:
- Part (a): [tex]\( x = 2 \)[/tex], [tex]\( y = 4 \)[/tex], [tex]\( z = 1 \)[/tex], [tex]\( t = 3 \)[/tex]
- Part (b): [tex]\( p = -2 \)[/tex], [tex]\( q = 0 \)[/tex], [tex]\( r = 4 \)[/tex], [tex]\( s = -1 \)[/tex], [tex]\( t = 9 \)[/tex], [tex]\( z = 9 \)[/tex]
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