$M \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, M \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} \text{ and } M \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}. \text{ If } M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}, \text{ then } x+y+z \text{ equals :}$
The problem involves a $3 \times 3$ matrix $M$ and its action on standard basis vectors. We need to find the sum $x+y+z$ where $M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}$.
Finding the Matrix M
The columns of matrix $M$ are determined by its action on the standard basis vectors:
- $M \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ gives the first column of $M$.
- $M \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}$ gives the second column of $M$.
- $M \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$ gives the third column of $M$.
Therefore, the matrix $M$ is:
$ M = \begin{pmatrix} 1 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{pmatrix} $Solving the Vector Equation
We are given the equation $M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}$. Substituting the matrix $M$ we found:
$ \begin{pmatrix} 1 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix} $This matrix equation corresponds to the following system of linear equations:
- $x - z = 1$
- $2x + y + z = 7$
- $3x + 2y + z = 11$
Calculating x, y, and z
We can solve this system. From equation (1), we get $x = 1 + z$. Substitute this into equations (2) and (3):
- Equation (2) becomes: $2(1+z) + y + z = 7 \implies 2 + 2z + y + z = 7 \implies y + 3z = 5$. (Equation 4)
- Equation (3) becomes: $3(1+z) + 2y + z = 11 \implies 3 + 3z + 2y + z = 11 \implies 2y + 4z = 8 \implies y + 2z = 4$. (Equation 5)
Now, solve the system formed by equations (4) and (5):
- $y + 3z = 5$
- $y + 2z = 4$
Subtracting equation (5) from equation (4) gives:
$ (y + 3z) - (y + 2z) = 5 - 4 $ $ z = 1 $Substitute $z=1$ back into equation (5):
$ y + 2(1) = 4 \implies y + 2 = 4 \implies y = 2 $Substitute $z=1$ back into the expression for $x$ ($x = 1+z$):
$ x = 1 + 1 \implies x = 2 $So, we have $x=2$, $y=2$, and $z=1$. Let's check this with the original vector equation:
$ M \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1(2)+0(2)-1(1) \\ 2(2)+1(2)+1(1) \\ 3(2)+2(2)+1(1) \end{pmatrix} = \begin{pmatrix} 2-1 \\ 4+2+1 \\ 6+4+1 \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix} $This matches the given condition.
Calculating the Sum x+y+z
Finally, calculate the sum $x+y+z$:
$ x+y+z = 2 + 2 + 1 = 5 $Similar Questions
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