Solution of Simultaneous Linear Equations

An amount of ₹10,000 is put into three investments at the rate of 10, 12 and 15% per annum. The combined incomes are ₹1310 and the combined income of first and second investment is ₹ 190 short of the income from the third. Find the investment in each using matrix method.

Solution: Suppose the numbers as $x$, $y$, and $z$ $x+y+z=10,000 \ldots \ldots \text { (i) }$ Also, $0.1 \mathrm{x}+0.12 \mathrm{y}+0.15 z=1310 \ldots \ldots \text { (ii) }$ Again, $0.1 x+0.12...

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The sum of three numbers is 2. If twice the second number is added to the sum of first and third, the sum is 1. By adding second and third number to five times the first number, we get 6. Find the three numbers by using matrices.

Solution: Suppose the numbers as $x$, $y$, $z$ $\begin{array}{l} x+y+z=2 \\ \ldots \cdots(i) \end{array}$ Also, $2 y+(x+z)+1$ $x+2 y+z=1 \ldots \ldots \text { (ii) }$ Again, $\begin{array}{l}...

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Show that each one of the following systems of linear equations is consistent and also find their solutions:
(i) 5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
(ii) x + y + z = 6
x + 2y + 3z = 14
x + 4y + 7z = 30

Solution: (i) Given that$5 x+3 y+7 z=4$ $\begin{array}{l} 3 x+26 y+2 z=9 \\ 7 x+2 y+10 z=5 \end{array}$ We can write this as: $\begin{array}{l} {\left[\begin{array}{ccc} 5 & 3 & 7 \\ 3 &...

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Solve the following system of equations by matrix method:
(i) (2/x) + (3/y) + (10/z) = 4,
(4/x) – (6/y) + (5/z) = 1,
(6/x) + (9/y) – (20/z) = 2, x, y, z ≠ 0
(ii) x – y + 2z = 7
3x + 4y – 5z = -5
2x – y + 3z = 12

Solution: (i) Given that $(2 / x)+(3 / y)+(10 / z)=4$ $\begin{array}{l} (4 / x)-(6 / y)+(5 / z)=1 \\ (6 / x)+(9 / y)-(20 / z)=2, x, y, z \neq 0 \end{array}$ We can write the given system in matrix...

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