1. If [tex]A=\left[\begin{array}{ccc}3 & 1 & 2 \\ 0 & 5 & 7 \\ 9 & 1 & -4\end{array}\right][/tex] and [tex]B=\left[\begin{array}{ccc}7 & 1 & 9 \\ 3 & 0 & -1 \\ 4 & -6 & 2\end{array}\right][/tex], show that [tex]3(B-A)=3B-3A[/tex].

2. If [tex]A=\left[\begin{array}{ccc}2 & -3 & -5 \\ 0 & 5 & 7 \\ 9 & 1 & -4\end{array}\right][/tex], then find the inverse of [tex]A[/tex].



Answer :

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.