Given that M=[2012] and N =[20−12]\text{M} = \begin{bmatrix}[r] 2 & 0 \ 1 & 2 \end{bmatrix} \text{ and N }= \begin{bmatrix}[r] 2 & 0 \ -1 & 2 \end{bmatrix}M=[2102] and N =[2−102], find M + 2N.
41 Likes
M + 2N =[2012]+2[20−12]=[2012]+[40−24]=[2+40+01−22+4]=[60−16]\text{M + 2N }= \begin{bmatrix}[r] 2 & 0 \ 1 & 2 \end{bmatrix} + 2\begin{bmatrix}[r] 2 & 0 \ -1 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 & 0 \ 1 & 2 \end{bmatrix} + \begin{bmatrix}[r] 4 & 0 \ -2 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 + 4 & 0 + 0 \ 1 - 2 & 2 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 6 & 0 \ -1 & 6 \end{bmatrix} \\[1em]M + 2N =[2102]+2[2−102]=[2102]+[4−204]=[2+41−20+02+4]=[6−106]
Hence, the matrix M + 2N = [60−16]\begin{bmatrix}[r] 6 & 0 \ -1 & 6 \end{bmatrix}[6−106].
Answered By
28 Likes
Find the values of x, y, a and b if [x−2ya+2b3a−b]=[3151]\begin{bmatrix}[r] x - 2 & y \ a + 2b & 3a - b \end{bmatrix} = \begin{bmatrix}[r] 3 & 1 \ 5 & 1 \end{bmatrix}[x−2a+2by3a−b]=[3511].
Find the values of a, b, c and d if [a+b35+cab]=[6d−18].\begin{bmatrix}[r] a + b & 3 \ 5 + c & ab \end{bmatrix} = \begin{bmatrix}[r] 6 & d \ -1 & 8 \end{bmatrix} .[a+b5+c3ab]=[6−1d8].
If A =[20−31] and B =[01−23]\text{If A }= \begin{bmatrix}[r] 2 & 0 \ -3 & 1 \end{bmatrix} \text{ and B }= \begin{bmatrix}[r] 0 & 1 \ -2 & 3 \end{bmatrix}If A =[2−301] and B =[0−213], find 2A - 3B.
Simplify : sin A[sin A−cos Acos Asin A]+cos A[cos Asin A−sin Acos A]\text{sin A}\begin{bmatrix}[r] \text{sin A} & -\text{cos A} \ \text{cos A} & \text{sin A} \end{bmatrix} + \text{cos A}\begin{bmatrix}[r] \text{cos A} & \text{sin A} \ -\text{sin A} & \text{cos A} \end{bmatrix}sin A[sin Acos A−cos Asin A]+cos A[cos A−sin Asin Acos A].
