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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Solve the equation for $x, y, z$ and $t$ if $ 2\begin{bmatrix}x & z \\ y & t \end{bmatrix} + 3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix} $

$\begin{array}{1 1} x=3 y=6 z=9 t=6 \\x=3 y=3 z=3 t=6 \\x=3 y=6 z=6 t=6 \\ x=3 y=6 z=6 t=9 \end{array} $

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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.
  • 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.
  • We can then match the corresponding elements and solve the resulting equations to find the values of x and y.
Given $2\begin{bmatrix}x & z \\ y & t \end{bmatrix} + 3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$:
$\Rightarrow \begin{bmatrix}2x & 2z \\ 2y & 2t \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix} $
$\Rightarrow \begin{bmatrix}2x+3 & 2z-3 \\ 2y+0 & 2t+6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix} $
Since these matrices are equal, we can an obtain the value of x, y, z and t by comparing the matrices' corresponding elements:
$2x+3 = 9 \rightarrow 2x = 9-3 = 6 \rightarrow x = \frac{6}{2} = 3$
$2y+0 = 12 \rightarrow 2y = 12-0 = 12 \rightarrow y = \frac{12}{2} = 6$
$2z-3 = 15 \rightarrow 2z = 15+3 = 18 \rightarrow z = \frac{18}{2} = 9$
$2t+6 = 18 \rightarrow 2t = 18-6 = 12 \rightarrow t = \frac{12}{2} = 6$
answered Feb 13, 2013 by sreemathi.v
edited Feb 27, 2013 by balaji.thirumalai
 

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