Advertisements
Advertisements
Question
An amount of Rs 10,000 is put into three investments at the rate of 10, 12 and 15% per annum. The combined income is Rs 1310 and the combined income of first and second investment is Rs 190 short of the income from the third. Find the investment in each using matrix method.
Advertisements
Solution
Let x , y and z be the investments at the rates of interest of 10%, 12% and 15% per annum respectively.
Total investment = Rs 10,000 \[\Rightarrow x + y + z = 10, 000\]
\[\text{ Income from the first investment of Rs }x = Rs\frac{10x}{100} = Rs 0 . 1x\]
\[\text{ Income from the second investment of Rs }x = Rs\frac{12y}{100} = Rs 0 . 12y\]
\[\text{ Income from the third investment of Rs }x = Rs\frac{15z}{100} = Rs 0 . 15z\]
\[ \therefore\text{ Total annual income }= Rs \left( 0 . 1x + 0 . 12y + 0 . 15z \right)\]
\[ \Rightarrow 0 . 1x + 0 . 12y + 0 . 15z = 1310 \left( \because \text{ Total annual income }= Rs 1310 \right)\]
It is given that the combined income from the first two incomes is Rs 190 short of the income from the third .
\[ \therefore 0 . 1x + 0 . 12y = 0 . 15z - 190\]
\[ \Rightarrow - 0 . 1x - 0 . 12y + 0 . 15z = 190\]
Thus, we obtain the following system of simultaneous linear equations:
\[x + y + z = 10000\]
\[0 . 1x + 0 . 12y + 0 . 15z = 1310\]
\[ - 0 . 1x - 0 . 12y + 0 . 15z = 190\]
The given system of equation can be written in matrix form as follows:
\[ \begin{bmatrix}1 & 1 & 1 \\ 0 . 1 & 0 . 12 & 0 . 15 \\ - 0 . 1 & - 0 . 12 & 0 . 15\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}10000 \\ 1310 \\ 190\end{bmatrix}\]
\[AX = B\]
Here,
\[A = \begin{bmatrix}1 & 1 & 1 \\ 0 . 1 & 0 . 12 & 0 . 15 \\ - 0 . 1 & - 0 . 12 & 0 . 15\end{bmatrix}, X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}\text{ and }B = \begin{bmatrix}10000 \\ 1310 \\ 190\end{bmatrix}\]
\[\left| A \right|=1 \left( 0 . 15 \times 0 . 12 + 0 . 15 \times 0 . 12 \right) - 1\left( 0 . 15 \times 0 . 1 + 0 . 15 \times 0 . 1 \right) + 1\left( - 0 . 1 \times 0 . 12 + 0 . 12 \times 0 . 1 \right)\]
\[ = 0 . 036 - 0 . 03 + 0\]
\[ = 0 . 006\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}0 . 12 & 0 . 15 \\ - 0 . 12 & 0 . 15\end{vmatrix} = 0 . 036, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}0 . 1 & 0 . 15 \\ - 0 . 1 & 0 . 15\end{vmatrix} = - 0 . 03, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}0 . 1 & 0 . 12 \\ - 0 . 1 & - 0 . 12\end{vmatrix} = 0\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ - 0 . 12 & 0 . 15\end{vmatrix} = - 0 . 27, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ - 0 . 1 & 0 . 15\end{vmatrix} = 0 . 25, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ - 0 . 1 & - 0 . 12\end{vmatrix} = 0 . 02\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 0 . 12 & 0 . 15\end{vmatrix} = 0 . 03, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 0 . 1 & 0 . 15\end{vmatrix} = - 0 . 05, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 0 . 1 & 0 . 12\end{vmatrix} = 0 . 02\]
\[adj A = \begin{bmatrix}0 . 036 & - 0 . 03 & 0 \\ - 0 . 27 & 0 . 25 & 0 . 02 \\ 0 . 03 & - 0 . 05 & 0 . 02\end{bmatrix}^T \]
\[ = \begin{bmatrix}0 . 036 & - 0 . 27 & 0 . 03 \\ - 0 . 03 & 0 . 25 & - 0 . 05 \\ 0 & 0 . 02 & 0 . 02\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{0 . 006}\begin{bmatrix}0 . 036 & - 0 . 27 & 0 . 03 \\ - 0 . 03 & 0 . 25 & - 0 . 05 \\ 0 & 0 . 02 & 0 . 02\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{0 . 006}\begin{bmatrix}0 . 036 & - 0 . 27 & 0 . 03 \\ - 0 . 03 & 0 . 25 & - 0 . 05 \\ 0 & 0 . 02 & 0 . 02\end{bmatrix}\begin{bmatrix}10000 \\ 1310 \\ 190\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{0 . 006}\begin{bmatrix}360 - 353 . 7 + 5 . 7 \\ - 300 + 327 . 5 - 9 . 5 \\ 0 + 26 . 2 + 3 . 8\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1000}{6}\begin{bmatrix}12 \\ 18 \\ 30\end{bmatrix}\]
\[ \therefore x = 2000, y = 3000\text{ and }z = 5000\]
Thus, the three investments are of Rs 2000, Rs 3000 and Rs 5000, respectively .
APPEARS IN
RELATED QUESTIONS
The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?
Find the adjoint of the matrices.
`[(1,-1,2),(2,3,5),(-2,0,1)]`
Verify A(adj A) = (adj A)A = |A|I.
`[(2,3),(-4,-6)]`
If A = `[(3,1),(-1,2)]` show that A2 – 5A + 7I = 0. Hence, find A–1.
If A is an invertible matrix of order 2, then det (A−1) is equal to ______.
If A−1 = `[(3,-1,1),(-15,6,-5),(5,-2,2)]` and B = `[(1,2,-2),(-1,3,0),(0,-2,1)]`, find (AB)−1.
Find the adjoint of the following matrix:
\[\begin{bmatrix}a & b \\ c & d\end{bmatrix}\]
If \[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\] , show that adj A = 3AT.
Find the inverse of the following matrix:
Find the inverse of the following matrix.
\[\begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{bmatrix}\]
Find the inverse of the following matrix.
For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]
\[A = \begin{bmatrix}2 & 1 \\ 5 & 3\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 5 \\ 3 & 4\end{bmatrix}\]
If \[A = \begin{bmatrix}4 & 5 \\ 2 & 1\end{bmatrix}\] , then show that \[A - 3I = 2 \left( I + 3 A^{- 1} \right) .\]
Show that
Show that the matrix, \[A = \begin{bmatrix}1 & 0 & - 2 \\ - 2 & - 1 & 2 \\ 3 & 4 & 1\end{bmatrix}\] satisfies the equation, \[A^3 - A^2 - 3A - I_3 = O\] . Hence, find A−1.
prove that \[A^{- 1} = A^3\]
If \[A = \begin{bmatrix}- 1 & 2 & 0 \\ - 1 & 1 & 1 \\ 0 & 1 & 0\end{bmatrix}\] , show that \[A^2 = A^{- 1} .\]
If \[A = \begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}\] , find \[A^{- 1}\] and prove that \[A^2 - 4A - 5I = O\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & 5 \\ 1 & 3\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{bmatrix}\]
If adj \[A = \begin{bmatrix}2 & 3 \\ 4 & - 1\end{bmatrix}\text{ and adj }B = \begin{bmatrix}1 & - 2 \\ - 3 & 1\end{bmatrix}\]
If A is a square matrix, then write the matrix adj (AT) − (adj A)T.
If A is an invertible matrix of order 3, then which of the following is not true ?
If A is a singular matrix, then adj A is ______.
If \[A = \begin{bmatrix}2 & 3 \\ 5 & - 2\end{bmatrix}\] be such that \[A^{- 1} = kA\], then k equals ___________ .
If \[A = \begin{bmatrix}1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2\end{bmatrix},\text{ then aI + bA + 2 }A^2\] equals ____________ .
If a matrix A is such that \[3A^3 + 2 A^2 + 5 A + I = 0,\text{ then }A^{- 1}\] equal to _______________ .
If \[A = \begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix}\], find A−1 and hence solve the system of linear equations 2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y + 2z = −3
Using matrix method, solve the following system of equations:
x – 2y = 10, 2x + y + 3z = 8 and -2y + z = 7
If A = `[(0, 1, 3),(1, 2, x),(2, 3, 1)]`, A–1 = `[(1/2, -4, 5/2),(-1/2, 3, -3/2),(1/2, y, 1/2)]` then x = 1, y = –1.
If A and B are invertible matrices, then which of the following is not correct?
A square matrix A is invertible if det A is equal to ____________.
Find the value of x for which the matrix A `= [(3 - "x", 2, 2),(2,4 - "x", 1),(-2,- 4,-1 - "x")]` is singular.
A and B are invertible matrices of the same order such that |(AB)-1| = 8, If |A| = 2, then |B| is ____________.
If A is a square matrix of order 3 and |A| = 5, then |adj A| = ______.
