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Home  >>  CBSE XII  >>  Math  >>  Matrices
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If $ A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix} \text{ and } B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix} \text{ then compute } 3A - 5B $

$\begin{array}{1 1} 1 \\ I \\ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \\ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\end{array} $

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Toolbox:
  • The scalar multiplication $cA$ of a matrix $A$ and a number $c$ (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of $A$ by $c$.
  • The difference $A-B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A - B)_{i,j} = A_{i,j} - B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
$3A=3\begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$
$3A = \begin{bmatrix} 3\times \frac{2}{3} & 1\times 3 &3\times \frac{5}{3} \\ 3\times\frac{1}{3} &3\times \frac{2}{3} & 3\times\frac{4}{3} \\ 3\times\frac{7}{3} & 3\times 2 & 3\times\frac{2}{3} \end{bmatrix}$
$3A = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}$
$-5B = -5\begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$
$-5B = \begin{bmatrix} -5\times\frac{2}{5} &-5\times \frac{3}{5} & 1\times-5 \\ -5\times\frac{1}{5} & -5\times\frac{2}{5} & -5\times\frac{4}{5} \\ -5\times\frac{7}{5} & -5\times\frac{6}{5} & -5\times\frac{2}{5} \end{bmatrix}$
$-5B = \begin{bmatrix} -2 & -3 & -5 \\ -1 &- 2 & -4 \\ -7 & -6 & -2 \end{bmatrix}$
$3A-5B = 3A+(-5B)= \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}+ \begin{bmatrix} -2 & -3 & -5 \\ -1 & -2 & -4 \\ -7 & -6 & -2 \end{bmatrix}$
$3A-5B = \begin{bmatrix} 2+(-2) & 3+(-3) & 5+(-5) \\ 1+(-1) & 2+(-2) & 4+(-4) \\ 7+(-7) & 6+(-6) & 2+(-2) \end{bmatrix}$
$3A-5B = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
answered Feb 27, 2013 by balaji.thirumalai
 

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