A=[} 2&3&-4 -&0&6 -2&1&5 ] Find 2A-3B Solution First calculate 2.A and 3B 20.2.12:2]-(xu si 2x)=[2=cm 3.8=[} 5&1&2 6&-1&4 5&-4 ] 2A-3A=[} 4&6&-8 2&6&12 -4&2&16 ] 14) Find B.A B=[} 5&1&2 6&-1&4 5&3&-4 ] A=[} 2&3&-4 1&0&6 -2&1&5 ] (5)(3)+(1)(0)+(2)(c)5) (-4)-10(6)+(27)(5) B. 4= ) (5)(a)+c)(c)+(a.2(-2) (5)(s)+(1)(0)+(2)(1) (k)(a)+(-1)(c)+4(-2))(5)(-2)+(-1)(-1)(-1)(-3)(c)+( (b) (-4)+(-1)(6)+1(4) (5)(3)+(3)(0)+(-4)(i)(5) (-4)+(3)(6)+(-4)

Let's solve the problem step by step. We need to find \(2A - 3B\) given matrices \(A\) and \(B\).<br /><br />### Step 1: Calculate \(2A\)<br /><br />Given:<br />\[ A = \begin{bmatrix} 2 & 3 & -4 \\ 0 & 0 & 6 \\ -2 & 1 & 5 \end{bmatrix} \]<br /><br />To find \(2A\), multiply each element of \(A\) by 2:<br /><br />\[ 2A = \begin{bmatrix} 2 \times 2 & 2 \times 3 & 2 \times (-4) \\ 2 \times 0 & 2 \times 0 & 2 \times 6 \\ 2 \times (-2) & 2 \times 1 & 2 \times 5 \end{bmatrix} = \begin{bmatrix} 4 & 6 & -8 \\ 0 & 0 & 12 \\ -4 & 2 & 10 \end{bmatrix} \]<br /><br />### Step 2: Calculate \(3B\)<br /><br />Given:<br />\[ B = \begin{bmatrix} 5 & 1 & 2 \\ 6 & -1 & 4 \\ -5 & 3 & -4 \end{bmatrix} \]<br /><br />To find \(3B\), multiply each element of \(B\) by 3:<br /><br />\[ 3B = \begin{bmatrix} 3 \times 5 & 3 \times 1 & 3 \times 2 \\ 3 \times 6 & 3 \times (-1) & 3 \times 4 \\ 3 \times (-5) & 3 \times 3 & 3 \times (-4) \end{bmatrix} = \begin{bmatrix} 15 & 3 & 6 \\ 18 & -3 & 12 \\ -15 & 9 & -12 \end{bmatrix} \]<br /><br />### Step 3: Calculate \(2A - 3B\)<br /><br />Now, subtract \(3B\) from \(2A\):<br /><br />\[ 2A - 3B = \begin{bmatrix} 4 & 6 & -8 \\ 0 & 0 & 12 \\ -4 & 2 & 10 \end{bmatrix} - \begin{bmatrix} 15 & 3 & 6 \\ 18 & -3 & 12 \\ -15 & 9 & -12 \end{bmatrix} \]<br /><br />Perform the subtraction element-wise:<br /><br />\[ 2A - 3B = \begin{bmatrix} 4 - 15 & 6 - 3 & -8 - 6 \\ 0 - 18 & 0 - (-3) & 12 - 12 \\ -4 - (-15) & 2 - 9 & 10 - (-12) \end{bmatrix} \]<br /><br />\[ 2A - 3B = \begin{bmatrix} -11 & 3 & -14 \\ -18 & 3 & 0 \\ 11 & -7 & 22 \end{bmatrix} \]<br /><br />Thus, the result of \(2A - 3B\) is:<br /><br />\[ \begin{bmatrix} -11 & 3 & -14 \\ -18 & 3 & 0 \\ 11 & -7 & 22 \end{bmatrix} \]