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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Find the matrix \( X \) so that \( X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \: = \: \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix} \)

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Toolbox:
  • If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
  • If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij}=B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
Step 1: Let x be order of $m\times n$ and the given matrix $\begin{bmatrix}1 & 2& 3\\4 & 5& 6\end{bmatrix}$ is of order $2\times 3\Rightarrow n=2$
Given
$\begin{bmatrix}-7 & -8 & -9\\2 & 4 & 6\end{bmatrix}$ is of order $2\times 3$ m=2.
Hence X is of the over $2\times 2$
$Let X=\begin{bmatrix}a & b\\c & d\end{bmatrix}$
Given
$x\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9\\2 & 4 & 6\end{bmatrix}$
Replace X as $\begin{bmatrix}a & b\\c& d\end{bmatrix}$ in the above equation
$\begin{bmatrix}a & b\\c& d\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}a+4b &2a+5b & 3a+6b\\c+4d & 2c+5d & 3c+6d\end{bmatrix}$
Step 2:
Given
$\begin{bmatrix}a+4b &2a+5b & 3a+6b\\c+4d & 2c+5d & 3c+6d\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9\\2 & 4 & -6\end{bmatrix}$
The above given two matrices are equal hence there corresponding elements should be equal
Equate the 1st row with the corresponding elements
a+4b=-7-----(1)
2a+5b=-8-----(2)
3a+6b=-9------(3)
Consider eq(1) multiply by 2
2a+8b=-14------(4)
subtract (2) from (4)
2a+8b=-14
2a+5b=-8
________________
3b=-6
b=-6/3=-2
b=-2.
Substitute the value of b in (1)
a+4b=-7
a+4(-2)=-7
a-8=-7
a=-7+8
a=1.
Step 3: Equate the 2nd row with there corresponding element
c+4d=2-----(5)
2c+5d=4------(6)
3c+6d=-6-----(7)
Consider equation (5) and multiply by 2
2c+8d=4-------(8)
subtracting (6 from (8)
-2c+8d=4
2c+5d=4
______________
3d=0
d=0.
substitute the value of d in equation (5)
c+4d=2
c+4(0)=2
c=2.
Step 4: Thus a=1
b=-2
c=2
d=0
Hence $X=\begin{bmatrix}1 & -2\\2 & 0\end{bmatrix}$

 

answered Mar 4, 2013 by sharmaaparna1
edited Mar 20, 2013 by sharmaaparna1
 

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