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Home  >>  CBSE XII  >>  Math  >>  Matrices
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If $A=\begin{bmatrix}1 & 5\\7 & 12\end{bmatrix}\;and\;B=\begin{bmatrix}9 & 1\\7 & 8\end{bmatrix},$ find a matrix $C$ such that $3A+5B+2C$ is a null matrix.

$\begin{array}{1 1} C=\begin{bmatrix}-4 & 0\\-28 & -38\end{bmatrix} \\ C=\begin{bmatrix}-24 & -10\\-28 & -38\end{bmatrix} \\ C=\begin{bmatrix}24 & 10\\28 & 38\end{bmatrix} \\ C=\begin{bmatrix}-2 & -10\\-28 & -38\end{bmatrix} \end{array} $
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1 Answer

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Toolbox:
  • The sum / 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.
  • 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 $c$.
Step1:
Given
$A=\begin{bmatrix}1 & 5\\7 & 12\end{bmatrix}.$
$B=\begin{bmatrix}9 & 1\\7 & 8\end{bmatrix}$
3A+5B+2C=0
Substitute the value of A and B in the above given equation.
$3\begin{bmatrix}1 & 5\\7 & 12\end{bmatrix}+5\begin{bmatrix}9 & 1\\7 & 8\end{bmatrix}+2C=0.$
$\Rightarrow \begin{bmatrix}3 & 15\\21 & 36\end{bmatrix}+\begin{bmatrix}45 & 5\\35 & 40\end{bmatrix}+2C=0.$
$\Rightarrow \begin{bmatrix}3+45 & 15+5\\21+35 & 36+40\end{bmatrix}+2C=0.$
$\Rightarrow \begin{bmatrix}48 & 20\\56 & 76\end{bmatrix}+2C=0.$
Step2:
$\Rightarrow 2C=(-)\begin{bmatrix}48 & 20\\56 & 76\end{bmatrix}.$
$C=-1/2\begin{bmatrix}48 & 20\\56 & 76\end{bmatrix}$
$\;\;\;=\begin{bmatrix}-24 & -10\\-28 & -38\end{bmatrix}$
$C=\begin{bmatrix}-24 & -10\\-28 & -38\end{bmatrix}$
answered Mar 25, 2013 by sharmaaparna1
 

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