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Sagot :
Let's address the problem step by step.
### Part 1: Show that [tex]\( 3(B - A) = 3B - 3A \)[/tex]
Given matrices:
[tex]\[ A = \begin{pmatrix} 3 & 1 & 2 \\ 0 & 5 & 7 \\ 9 & 1 & -4 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 7 & 1 & 9 \\ 3 & 0 & -1 \\ 4 & -6 & 2 \end{pmatrix} \][/tex]
First, we calculate [tex]\( B - A \)[/tex]:
[tex]\[ B - A = \begin{pmatrix} 7 & 1 & 9 \\ 3 & 0 & -1 \\ 4 & -6 & 2 \end{pmatrix} - \begin{pmatrix} 3 & 1 & 2 \\ 0 & 5 & 7 \\ 9 & 1 & -4 \end{pmatrix} \][/tex]
Subtracting the matrices element-wise:
[tex]\[ B - A = \begin{pmatrix} 7 - 3 & 1 - 1 & 9 - 2 \\ 3 - 0 & 0 - 5 & -1 - 7 \\ 4 - 9 & -6 - 1 & 2 - (-4) \end{pmatrix} = \begin{pmatrix} 4 & 0 & 7 \\ 3 & -5 & -8 \\ -5 & -7 & 6 \end{pmatrix} \][/tex]
Next, we multiply [tex]\( B - A \)[/tex] by 3:
[tex]\[ 3(B - A) = 3 \begin{pmatrix} 4 & 0 & 7 \\ 3 & -5 & -8 \\ -5 & -7 & 6 \end{pmatrix} \][/tex]
Multiplying each element by 3:
[tex]\[ 3(B - A) = \begin{pmatrix} 12 & 0 & 21 \\ 9 & -15 & -24 \\ -15 & -21 & 18 \end{pmatrix} \][/tex]
Now, we calculate [tex]\( 3B \)[/tex]:
[tex]\[ 3B = 3 \begin{pmatrix} 7 & 1 & 9 \\ 3 & 0 & -1 \\ 4 & -6 & 2 \end{pmatrix} = \begin{pmatrix} 21 & 3 & 27 \\ 9 & 0 & -3 \\ 12 & -18 & 6 \end{pmatrix} \][/tex]
And [tex]\( 3A \)[/tex]:
[tex]\[ 3A = 3 \begin{pmatrix} 3 & 1 & 2 \\ 0 & 5 & 7 \\ 9 & 1 & -4 \end{pmatrix} = \begin{pmatrix} 9 & 3 & 6 \\ 0 & 15 & 21 \\ 27 & 3 & -12 \end{pmatrix} \][/tex]
Next, we find [tex]\( 3B - 3A \)[/tex]:
[tex]\[ 3B - 3A = \begin{pmatrix} 21 & 3 & 27 \\ 9 & 0 & -3 \\ 12 & -18 & 6 \end{pmatrix} - \begin{pmatrix} 9 & 3 & 6 \\ 0 & 15 & 21 \\ 27 & 3 & -12 \end{pmatrix} \][/tex]
Subtracting the matrices element-wise:
[tex]\[ 3B - 3A = \begin{pmatrix} 21 - 9 & 3 - 3 & 27 - 6 \\ 9 - 0 & 0 - 15 & -3 - 21 \\ 12 - 27 & -18 - 3 & 6 - (-12) \end{pmatrix} = \begin{pmatrix} 12 & 0 & 21 \\ 9 & -15 & -24 \\ -15 & -21 & 18 \end{pmatrix} \][/tex]
We see that:
[tex]\[ 3(B - A) = 3B - 3A \][/tex]
### Part 2: Find the inverse of [tex]\( A = \begin{pmatrix} 2 & -3 & -5 \\ 0 & 5 & 7 \\ 9 & 1 & -4 \end{pmatrix} \)[/tex]
The inverse of matrix [tex]\( A \)[/tex] is calculated can be summarized step by step, here is the result for the inverse of the matrix [tex]\( A \)[/tex]:
[tex]\[ A^{-1} = \begin{pmatrix} 1.5 & 0.94444444 & -0.22222222 \\ -3.5 & -2.05555556 & 0.77777778 \\ 2.5 & 1.61111111 & -0.55555556 \end{pmatrix} \][/tex]
Thus, the inverse of [tex]\( A \)[/tex] is:
[tex]\[ \begin{pmatrix} 1.5 & 0.94444444 & -0.22222222 \\ -3.5 & -2.05555556 & 0.77777778 \\ 2.5 & 1.61111111 & -0.55555556 \end{pmatrix} \][/tex]
That completes the solution to the problem.
### Part 1: Show that [tex]\( 3(B - A) = 3B - 3A \)[/tex]
Given matrices:
[tex]\[ A = \begin{pmatrix} 3 & 1 & 2 \\ 0 & 5 & 7 \\ 9 & 1 & -4 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 7 & 1 & 9 \\ 3 & 0 & -1 \\ 4 & -6 & 2 \end{pmatrix} \][/tex]
First, we calculate [tex]\( B - A \)[/tex]:
[tex]\[ B - A = \begin{pmatrix} 7 & 1 & 9 \\ 3 & 0 & -1 \\ 4 & -6 & 2 \end{pmatrix} - \begin{pmatrix} 3 & 1 & 2 \\ 0 & 5 & 7 \\ 9 & 1 & -4 \end{pmatrix} \][/tex]
Subtracting the matrices element-wise:
[tex]\[ B - A = \begin{pmatrix} 7 - 3 & 1 - 1 & 9 - 2 \\ 3 - 0 & 0 - 5 & -1 - 7 \\ 4 - 9 & -6 - 1 & 2 - (-4) \end{pmatrix} = \begin{pmatrix} 4 & 0 & 7 \\ 3 & -5 & -8 \\ -5 & -7 & 6 \end{pmatrix} \][/tex]
Next, we multiply [tex]\( B - A \)[/tex] by 3:
[tex]\[ 3(B - A) = 3 \begin{pmatrix} 4 & 0 & 7 \\ 3 & -5 & -8 \\ -5 & -7 & 6 \end{pmatrix} \][/tex]
Multiplying each element by 3:
[tex]\[ 3(B - A) = \begin{pmatrix} 12 & 0 & 21 \\ 9 & -15 & -24 \\ -15 & -21 & 18 \end{pmatrix} \][/tex]
Now, we calculate [tex]\( 3B \)[/tex]:
[tex]\[ 3B = 3 \begin{pmatrix} 7 & 1 & 9 \\ 3 & 0 & -1 \\ 4 & -6 & 2 \end{pmatrix} = \begin{pmatrix} 21 & 3 & 27 \\ 9 & 0 & -3 \\ 12 & -18 & 6 \end{pmatrix} \][/tex]
And [tex]\( 3A \)[/tex]:
[tex]\[ 3A = 3 \begin{pmatrix} 3 & 1 & 2 \\ 0 & 5 & 7 \\ 9 & 1 & -4 \end{pmatrix} = \begin{pmatrix} 9 & 3 & 6 \\ 0 & 15 & 21 \\ 27 & 3 & -12 \end{pmatrix} \][/tex]
Next, we find [tex]\( 3B - 3A \)[/tex]:
[tex]\[ 3B - 3A = \begin{pmatrix} 21 & 3 & 27 \\ 9 & 0 & -3 \\ 12 & -18 & 6 \end{pmatrix} - \begin{pmatrix} 9 & 3 & 6 \\ 0 & 15 & 21 \\ 27 & 3 & -12 \end{pmatrix} \][/tex]
Subtracting the matrices element-wise:
[tex]\[ 3B - 3A = \begin{pmatrix} 21 - 9 & 3 - 3 & 27 - 6 \\ 9 - 0 & 0 - 15 & -3 - 21 \\ 12 - 27 & -18 - 3 & 6 - (-12) \end{pmatrix} = \begin{pmatrix} 12 & 0 & 21 \\ 9 & -15 & -24 \\ -15 & -21 & 18 \end{pmatrix} \][/tex]
We see that:
[tex]\[ 3(B - A) = 3B - 3A \][/tex]
### Part 2: Find the inverse of [tex]\( A = \begin{pmatrix} 2 & -3 & -5 \\ 0 & 5 & 7 \\ 9 & 1 & -4 \end{pmatrix} \)[/tex]
The inverse of matrix [tex]\( A \)[/tex] is calculated can be summarized step by step, here is the result for the inverse of the matrix [tex]\( A \)[/tex]:
[tex]\[ A^{-1} = \begin{pmatrix} 1.5 & 0.94444444 & -0.22222222 \\ -3.5 & -2.05555556 & 0.77777778 \\ 2.5 & 1.61111111 & -0.55555556 \end{pmatrix} \][/tex]
Thus, the inverse of [tex]\( A \)[/tex] is:
[tex]\[ \begin{pmatrix} 1.5 & 0.94444444 & -0.22222222 \\ -3.5 & -2.05555556 & 0.77777778 \\ 2.5 & 1.61111111 & -0.55555556 \end{pmatrix} \][/tex]
That completes the solution to the problem.
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