Question

An amount of Rs 5000 is put into three investments at the rate of interest of 6%,7% and 8% per annum respectively. The total annual income is? 350. If the combined income from the first two investments is Rs 70 more than the income from the third .Find the amount of each investment ? ​​

Answer 1

Given:

An amount of rs 5000 is put into three investments at the rate of interests of 6.7%, 7.7% and 8% per annum respectively.

The total annual income is rs 350.

To find:

If the combined income from the first two investments is rs 70 more than the income from the third, find the amount of each investment by matrix method.

Solution:

Let the investments be x, y and z.

x = 6.7% × 5000 = 6.7/100 × 5000 = 335

y = 7.7% × 5000 = 7.7/100 × 5000 = 385

z = 8% × 5000 = 8/100 × 5000 = 400

6.7x + 7.7y + 8z = 100

x + y + z = 350

x + y - z = 70

AX = B

$\begin{gathered}\left[\begin{array}{ccc}6.7&7.7&8\\1&1&1\\1&1&-1\end{array}\right] \begin{bmatrix}x\\ y\\ z\end{bmatrix} = \begin{bmatrix}100\\ 350\\ 70\end{bmatrix}\end{gathered}$

⎣⎢⎡6.7117.71181−1⎦⎥⎤

⎣⎢⎡xyz⎦⎥⎤

=⎣⎢⎡10035070⎦⎥⎤

$\begin{gathered}\Delta =\begin{bmatrix}6.7&7.7&8\\ 1&1&1\\ 1&1&-1\end{bmatrix}\\=6.7\cdot \det \begin{pmatrix}1&1\\ 1&-1\end{pmatrix}-7.7\cdot \det \begin{pmatrix}1&1\\ 1&-1\end{pmatrix}+8\cdot \det \begin{pmatrix}1&1\\ 1&1\end{pmatrix}\\=6.7\left(-2\right)-7.7\left(-2\right)+8\cdot \:0\\\Delta = 2\end{gathered}$

Δ= ⎣⎢⎡6.7117.71181−1⎦⎥⎤

=6.7⋅det( 111−1 )−7.7⋅det( 111−1 )+8⋅det( 1111 )

=6.7(−2)−7.7(−2)+8⋅0

Δ=2

$\begin{gathered}\Delta_1 =\begin{bmatrix}100&7.7&8\\ 350&1&1\\ 70&1&-1\end{bmatrix}\\=100\cdot \det \begin{pmatrix}1&1\\ 1&-1\end{pmatrix}-7.7\cdot \det \begin{pmatrix}350&1\\ 70&-1\end{pmatrix}+8\cdot \det \begin{pmatrix}350&1\\ 70&1\end{pmatrix}\\=100\left(-2\right)-7.7\left(-420\right)+8\cdot \:280\ \\\Delta_1 = 5274\end{gathered}$

Δ 1 = ⎣⎢⎡100350707.71181−1⎦⎥⎤

=100⋅det( 11 1−1 )−7.7⋅det( 350701−1 )+8⋅det( 350711 )

=100(−2)−7.7(−420)+8⋅280

Δ 1 =5274

$\begin{gathered}\Delta_2 =\begin{bmatrix}6.7&100&8\\ \:\:1&350&1\\ \:\:1&70&-1\end{bmatrix}\\=6.7\cdot \det \begin{pmatrix}350&1\\ 70&-1\end{pmatrix}-100\cdot \det \begin{pmatrix}1&1\\ 1&-1\end{pmatrix}+8\cdot \det \begin{pmatrix}1&350\\ 1&70\end{pmatrix}\\=6.7\left(-420\right)-100\left(-2\right)+8\left(-280\right)\ \\\Delta_2 = -4854\end{gathered}$

Δ 2 = ⎣⎢⎡6.7111003507081−1⎦⎥⎤

=6.7⋅det( 350701−1 )−100⋅det( 111−1 )+8⋅det( 1135070 )

=6.7(−420)−100(−2)+8(−280)

Δ 2 =−4854

$\begin{gathered}\Delta_3 =\begin{bmatrix}6.7&7.7&100\\ 1&1&350\\ 1&1&70\end{bmatrix}\\=6.7\cdot \det \begin{pmatrix}1&350\\ 1&70\end{pmatrix}-7.7\cdot \det \begin{pmatrix}1&350\\ 1&70\end{pmatrix}+100\cdot \det \begin{pmatrix}1&1\\ 1&1\end{pmatrix}\\=6.7\left(-280\right)-7.7\left(-280\right)+100\cdot \:0\ \\\Delta_3 = 280\end{gathered}$

Δ 3 = ⎣⎢⎡6.7117.71110035070⎦⎢⎤

=6.7⋅det( 1135070−7.7⋅det( 1135070+100⋅det( 1111 )

=6.7(−280)−7.7(−280)+100⋅0

Δ 3 =280

Now let us consider,

Hence the investments are as follows:

x = Δ1/Δ = 5274/2 = 2637

y = Δ2/Δ = -4854/2 = -2427

z = Δ3/Δ = 280/2 = 140

Verification:

The combined income from the first two investments is rs 70 more than the income from the third.

x + y = z + 70

2637 + (-2427) = 140 + 70

210 = 210

Answer 2

Answer:

Given:

An amount of rs 5000 is put into three investments at the rate of interests of 6.7%, 7.7% and 8% per annum respectively.

The total annual income is rs 350.

To find:

If the combined income from the first two investments is rs 70 more than the income from the third, find the amount of each investment by matrix method.

Solution:

Let the investments be x, y and z.

x = 6.7% × 5000 = 6.7/100 × 5000 = 335

y = 7.7% × 5000 = 7.7/100 × 5000 = 385

z = 8% × 5000 = 8/100 × 5000 = 400

6.7x + 7.7y + 8z = 100

x + y + z = 350

x + y - z = 70

AX = B

\begin{gathered}\begin{gathered}\left[\begin{array}{ccc}6.7&7.7&8\\1&1&1\\1&1&-1\end{array}\right] \begin{bmatrix}x\\ y\\ z\end{bmatrix} = \begin{bmatrix}100\\ 350\\ 70\end{bmatrix}\end{gathered} \end{gathered}

⎣

⎡

6.7

1

1

7.7

1

1

8

1

−1

⎦

⎤

⎣

⎡

x

y

z

⎦

⎤

=

⎣

⎡

100

350

70

⎦

⎤

⎣⎢⎡6.7117.71181−1⎦⎥⎤

⎣⎢⎡xyz⎦⎥⎤

=⎣⎢⎡10035070⎦⎥⎤

\begin{gathered}\begin{gathered}\Delta =\begin{bmatrix}6.7&7.7&8\\ 1&1&1\\ 1&1&-1\end{bmatrix}\\=6.7\cdot \det \begin{pmatrix}1&1\\ 1&-1\end{pmatrix}-7.7\cdot \det \begin{pmatrix}1&1\\ 1&-1\end{pmatrix}+8\cdot \det \begin{pmatrix}1&1\\ 1&1\end{pmatrix}\\=6.7\left(-2\right)-7.7\left(-2\right)+8\cdot \:0\\\Delta = 2\end{gathered} \end{gathered}

Δ=

⎣

⎡

6.7

1

1

7.7

1

1

8

1

−1

⎦

⎤

=6.7⋅det(

1

1

1

−1

)−7.7⋅det(

1

1

1

−1

)+8⋅det(

1

1

1

1

)

=6.7(−2)−7.7(−2)+8⋅0

Δ=2

Δ= ⎣⎢⎡6.7117.71181−1⎦⎥⎤

=6.7⋅det( 111−1 )−7.7⋅det( 111−1 )+8⋅det( 1111 )

=6.7(−2)−7.7(−2)+8⋅0

Δ=2

\begin{gathered}\begin{gathered}\Delta_1 =\begin{bmatrix}100&7.7&8\\ 350&1&1\\ 70&1&-1\end{bmatrix}\\=100\cdot \det \begin{pmatrix}1&1\\ 1&-1\end{pmatrix}-7.7\cdot \det \begin{pmatrix}350&1\\ 70&-1\end{pmatrix}+8\cdot \det \begin{pmatrix}350&1\\ 70&1\end{pmatrix}\\=100\left(-2\right)-7.7\left(-420\right)+8\cdot \:280\ \\\Delta_1 = 5274\end{gathered} \end{gathered}

Δ

1

=

⎣

⎡

100

350

70

7.7

1

1

8

1

−1

⎦

⎤

=100⋅det(

1

1

1

−1

)−7.7⋅det(

350

70

1

−1

)+8⋅det(

350

70

1

1

)

=100(−2)−7.7(−420)+8⋅280

Δ

1

=5274

Δ 1 = ⎣⎢⎡100350707.71181−1⎦⎥⎤

=100⋅det( 11 1−1 )−7.7⋅det( 350701−1 )+8⋅det( 350711 )

=100(−2)−7.7(−420)+8⋅280

Δ 1 =5274

\begin{gathered}\begin{gathered}\Delta_2 =\begin{bmatrix}6.7&100&8\\ \:\:1&350&1\\ \:\:1&70&-1\end{bmatrix}\\=6.7\cdot \det \begin{pmatrix}350&1\\ 70&-1\end{pmatrix}-100\cdot \det \begin{pmatrix}1&1\\ 1&-1\end{pmatrix}+8\cdot \det \begin{pmatrix}1&350\\ 1&70\end{pmatrix}\\=6.7\left(-420\right)-100\left(-2\right)+8\left(-280\right)\ \\\Delta_2 = -4854\end{gathered} \end{gathered}

Δ

2

=

⎣

⎡

6.7

1

1

100

350

70

8

1

−1

⎦

⎤

=6.7⋅det(

350

70

1

−1

)−100⋅det(

1

1

1

−1

)+8⋅det(

1

1

350

70

)

=6.7(−420)−100(−2)+8(−280)

Δ

2

=−4854

Δ 2 = ⎣⎢⎡6.7111003507081−1⎦⎥⎤

=6.7⋅det( 350701−1 )−100⋅det( 111−1 )+8⋅det( 1135070 )

=6.7(−420)−100(−2)+8(−280)

Δ 2 =−4854

\begin{gathered}\begin{gathered}\Delta_3 =\begin{bmatrix}6.7&7.7&100\\ 1&1&350\\ 1&1&70\end{bmatrix}\\=6.7\cdot \det \begin{pmatrix}1&350\\ 1&70\end{pmatrix}-7.7\cdot \det \begin{pmatrix}1&350\\ 1&70\end{pmatrix}+100\cdot \det \begin{pmatrix}1&1\\ 1&1\end{pmatrix}\\=6.7\left(-280\right)-7.7\left(-280\right)+100\cdot \:0\ \\\Delta_3 = 280\end{gathered} \end{gathered}

Δ

3

=

⎣

⎡

6.7

1

1

7.7

1

1

100

350

70

⎦

⎤

=6.7⋅det(

1

1

350

70

)−7.7⋅det(

1

1

350

70

)+100⋅det(

1

1

1

1

)

=6.7(−280)−7.7(−280)+100⋅0

Δ

3

=280

Δ 3 = ⎣⎢⎡6.7117.71110035070⎦⎢⎤

=6.7⋅det( 1135070−7.7⋅det( 1135070+100⋅det( 1111 )

=6.7(−280)−7.7(−280)+100⋅0

Δ 3 =280

Now let us consider,

Hence the investments are as follows:

x = Δ1/Δ = 5274/2 = 2637

y = Δ2/Δ = -4854/2 = -2427

z = Δ3/Δ = 280/2 = 140

Verification:

The combined income from the first two investments is rs 70 more than the income from the third.

x + y = z + 70

2637 + (-2427) = 140 + 70

210 = 210