Gaussian and Gauss-Jordan Elimination

Matrix Notation

Elimination methods for solving systems of equations are carried out more efficiently by means of matrices. A matrix is a rectangular array of numbers, arranged in rows and columns and enclosed in brackets, thus:

\[ \left[\begin{aligned}a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\end{aligned}\right] \]

The numbers are called elements of the matrix. The above matrix would be said to have three rows (first row: $ a_{11} $ $ a_{12} $ $ a_{13} $ $ a_{14} $ , and so on) and four columns, and would be called a matrix of dimension $ 3 $ . The elements are referred to by two subscripts; thus, the element in row 2, column 3 is element $ a_{23} $ . A matrix may have any number of rows and any number of columns; a general matrix is said to have dimension $ m n $ , thus, m rows and n columns.

Row-Equivalent Matrices

Two matrices are said to be row-equivalent if one can be transformed into the other by successive applications of the following row operations on matrices:

  1. Interchange two rows. (Symbol: R ↔︎ R)

  2. Replace a row by a nonzero multiple of itself. (Symbol: $ kR_{i} R_{i} $ )

  3. Replace a row by itself plus a multiple of another row. (Symbol: $ kR_{i} + R_{i} R_{i} $ )

Note the exact correspondence to the operations that result in equivalent systems of equations (Chapter 30).

EXAMPLE 31.1 Given the matrix \(\begin{bmatrix} 5 & -2 \\ 2 & 6 \end{bmatrix}\), show the result of applying, in turn, (a) \(R_{1} \leftrightarrow R_{2}\);

  1. $ R_{1}R_{1} $ , (c) $ -5R_{1}+R_{2}R_{2} $

  2. $ R_{1} R_{2} $ interchanges the two rows to yield $

    \[\begin{bmatrix} 2 & 6 \\ 5 & -2 \end{bmatrix}\]

    $ .

  3. $ R_{1}R_{1} $ replaces each element in the new first row 2 6 by one-half its value to yield $

    \[\begin{bmatrix}1&3\\ 5&-2\end{bmatrix}\]

    $ .

  4. $ -5R_{1} + R_{2} R_{2} $ replaces each element in the new second row by itself plus -5 times the corresponding element in the first row -5 -15 to yield $

    \[\begin{bmatrix} 1 & 3 \\ 0 & -17 \end{bmatrix}\]

    $ .

Matrices and Systems of Linear Equations

To every linear system of m equations in n variables in standard form there corresponds a matrix of dimension $ m n + 1 $ called the augmented matrix of the system. Thus to the system:

\[ \begin{aligned}3x+5y-2z&=4\\-2x-3y&=6\\2x+4y+z&=-3\end{aligned} \]

corresponds the augmented matrix

\[ \left[\begin{array}{r r r r|r}{{{3}}}&{{{5}}}&{{{-2}}}&{{{4}}} \\{{{-2}}}&{{{-3}}}&{{{0}}}&{{{6}}} \\{{{2}}}&{{{4}}}&{{{1}}}&{{{-3}}}\end{array}\right] \]

The vertical bar has no mathematical significance and serves only to separate the coefficients of the variables from the constant terms.

Row-Echelon Form of a Matrix

A matrix is in row-echelon form if it satisfies the following conditions:

  1. The first nonzero number in each row is a 1.

  2. The column containing the first nonzero number in each row is to the left of the column containing the first nonzero number in rows below it.

  3. Any row containing only zeros appears below any row having any nonzero numbers.

EXAMPLE 31.2 Perform row operations to find a matrix in row-echelon form that is row-equivalent to the matrix

\[ \begin{bmatrix}{{{1}}}&{{{-1}}}&{{{3}}} \\{{{3}}}&{{{2}}}&{{{-1}}}\end{bmatrix}. \]

\[ \begin{bmatrix}{{{1}}}&{{{-1}}}&{{{3}}} \\{{{3}}}&{{{2}}}&{{{-1}}}\end{bmatrix}-3R_{1}+R_{2}\rightarrow R_{2}\begin{bmatrix}{{{1}}}&{{{-1}}}&{{{3}}} \\{{{0}}}&{{{5}}}&{{{-10}}}\end{bmatrix}\frac{1}{5}R_{2}\rightarrow R_{2}\begin{bmatrix}{{{1}}}&{{{-1}}}&{{{3}}} \\{{{0}}}&{{{1}}}&{{{-2}}}\end{bmatrix} \]

Gaussian Elimination

Gaussian elimination (with back substitution) is the following process for solving systems of linear equations:

  1. Write the system in standard form.

  2. Write the augmented matrix of the system.

  3. Apply row operations to this augmented matrix to obtain a row-equivalent matrix in row-echelon form.

  4. Write the system of equations to which this matrix corresponds.

  5. Find the solution of this system; it can be solved readily by substituting values from each equation into the one above it, starting with the last nonzero equation.

EXAMPLE 31.3 Solve the system $ \[\begin{array}{r}x - y = 3 \\ 3x + 2y = -1\end{array}\]

$ by Gaussian elimination.

The system is in standard form. The augmented matrix of the system is $ \[\begin{bmatrix}1 & -1 & | & 3| \\ 3 & 2 & | & -1|\end{bmatrix}\]

$ , considered in the previous example.

Reducing this to row-echelon form yields $ \[\begin{bmatrix}1 & -1 & | & 3| \\ 0 & 1 & | & -2\end{bmatrix}\]

$ . This matrix corresponds to the system x - y = 3, y = -2.

Thus y = -2. Substituting this into the first equation yields x - (-2) = 3 or x = 1. Thus the solution of the system is (1, -2).

Reduced Row-Echelon Form

A matrix is in reduced row-echelon form (often called just reduced form) if it satisfies the conditions for row-echelon form and, in addition, the entries above the first 1 in each row are all 0.

EXAMPLE 31.4 Find a matrix in reduced form that is row-equivalent to the matrix $ \[\begin{bmatrix}1 & -1 & 3 \\ 0 & 1 & -2\end{bmatrix}\]

$ from Example 31.2.

\[ \begin{bmatrix}{{{1}}}&{{{-1}}}&{{{3}}} \\{{{0}}}&{{{1}}}&{{{-2}}}\end{bmatrix}\boldsymbol{R}_{2}+\boldsymbol{R}_{1}\rightarrow\boldsymbol{R}_{1}\begin{bmatrix}{{{1}}}&{{{0}}}&{{{1}}} \\{{{0}}}&{{{1}}}&{{{-2}}}\end{bmatrix} \]

Note that the solution of the system can be read off immediately after writing the system that corresponds to this matrix $ (x = 1, y = -2) $ .

Gauss-Jordan Elimination

Gauss-Jordan elimination is the following process for solving systems of linear equations:

  1. Write the system in standard form.

  2. Write the augmented matrix of the system.

  3. Apply row operations to this augmented matrix to obtain a row-equivalent matrix in reduced row-echelon form.

  4. Write the system of equations to which this matrix corresponds.

  5. Find the solution of this system. If there is a unique solution, it can be read off immediately. If there are infinite solutions, the system will be such that after assigning arbitrary real values to undetermined variables, the other variables are immediately expressed in terms of these.

The process of finding a matrix in row-echelon form or reduced row-echelon form that is row-equivalent to a given matrix thus plays a key role in solving systems of linear equations. This process is usually abbreviated as “Transform to row-echelon (or reduced row-echelon) form.”

SOLVED PROBLEMS

31.1. Show the result of applying (a) $ R_{1} R_{3} $ ; (b) $ -R_{2} R_{2} $ ; (c) $ 2R_{2} + R_{1} R_{1} $ to the matrix

\[ \begin{bmatrix}5&3&-2&\begin{array}{r}3\\ -3&6&12&-3\\ 1&0&-4&5\end{array}\end{bmatrix}. \]

  1. $ R_{1} R_{3} $ interchanges rows 1 and 3, yielding $

    \[\begin{bmatrix} 1 & 0 & -4 \\ -3 & 6 & 12 \\ 5 & 3 & -2 \end{bmatrix} \begin{bmatrix} 5 \\ -3 \\ 3 \end{bmatrix}\]

    $ .

  2. $ -R_{2} R_{2} $ replaces row 2 by $ - $ times itself, yielding $

    \[\begin{bmatrix}5 & 3 & -2 & | & 3 \\ 1 & -2 & -4 & | & 1 \\ 1 & 0 & -4 & | & 5\end{bmatrix}\]

    $ .

  3. $ 2R_{2} + R_{1} R_{1} $ adds row -6 12 24 | -6 to the existing row 1, yielding $

    \[\begin{bmatrix}-1 & 15 & 22 & | & -3 \\ -3 & 6 & 12 & | & -3 \\ 1 & 0 & -4 & | & 5\end{bmatrix}\]

    $ .

31.2. Transform the matrix $ \[\begin{bmatrix}1&2&-2&3\\ 2&5&0&-7\\ 3&7&-2&-4\end{bmatrix}\]

$ to row-echelon form.

The first element in row 1 is a 1. Use this to produce zeros in the first position in the lower rows:

\[ \begin{bmatrix}{{{1}}}&{{{2}}}&{{{-2}}}&{{{3}}} \\{{{2}}}&{{{5}}}&{{{0}}}&{{{-7}}} \\{{{3}}}&{{{7}}}&{{{-2}}}&{{{-4}}}\end{bmatrix}\begin{aligned}R_{2}+(-2)R_{1}&\rightarrow R_{2}\\ R_{3}+(-3)R_{1}&\rightarrow R_{3}\end{aligned}\begin{bmatrix}{{{1}}}&{{{2}}}&{{{-2}}}&{{{3}}} \\{{{0}}}&{{{1}}}&{{{4}}}&{{{-13}}} \\{{{0}}}&{{{1}}}&{{{4}}}&{{{-13}}}\end{bmatrix} \]

The first nonzero element in row 2 is now a 1. Use this to produce a zero in the corresponding position in the last row.

\[ \begin{bmatrix}{{{1}}}&{{{2}}}&{{{-2}}}&{{{3}}} \\{{{0}}}&{{{1}}}&{{{4}}}&{{{-13}}} \\{{{0}}}&{{{1}}}&{{{4}}}&{{{-13}}}\end{bmatrix}\boldsymbol{R}_{3}+(-1)\boldsymbol{R}_{2}\rightarrow\boldsymbol{R}_{3}\begin{bmatrix}{{{1}}}&{{{2}}}&{{{-2}}}&{{{3}}} \\{{{0}}}&{{{1}}}&{{{4}}}&{{{-13}}} \\{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}\end{bmatrix} \]

This matrix is in row-echelon form, and is row-equivalent to the original matrix.

31.3. Generalize the procedure of the previous problem to a general strategy for transforming the matrix of an arbitrary system to row-echelon form.

  1. By interchanging rows if necessary, obtain a nonzero element in the first position in row 1. Replace row 1 by a multiple to make this element a 1.

  2. Use this element to produce zeros in the first position in the lower rows.

  3. If this produces rows that are zero to the left of the vertical bar, or all zeros, move these rows to the bottom. If there are no other rows, stop.

  4. If there are nonzero elements in rows below the first, move the row with the leftmost nonzero element to row 2. Replace row 2 by a multiple to make this element a 1.

  5. Use this element to produce zeros in the corresponding position in any rows below row 2 that are nonzero to the left of the vertical bar.

  6. Proceed as in steps 3 to 5 for any remaining rows.

\[ x+2y-2z=3 \]

31.4. Solve the system $ 2x + 5y = -7 $ by Gaussian elimination.

\[ 3x+7y-2z=-4 \]

The augmented matrix of the system is the matrix of Problem 31.2. Transforming to row-echelon form yields the matrix

\[ \begin{bmatrix}{{{1}}}&{{{2}}}&{{{-2}}}&{{{3}}} \\{{{0}}}&{{{1}}}&{{{4}}} \\{{{0}}}&{{{0}}}&{{{0}}}\end{bmatrix}which corresponds to the system\begin{aligned}x+2y-2z&=3\\ y+4z&=-13\\ 0&=0\end{aligned} \]

This system has an infinite number of solutions. Let z = r, r any real number. Then from equation (2), y = -13 - 4r. Substituting back into equation (1) yields:

\[ \begin{aligned}x+2(-13-4r)-2r&=3\\x&=10r+29\end{aligned} \]

Thus all solutions of the system can be written as $ (10r + 29, -13 - 4r, r) $ , r any real number.

31.5. Transform the matrix $ \[\begin{bmatrix}1&0&-1&1&2&2\\ 1&1&0&2&-3&-4\\ 2&0&-1&1&3&3\\ 0&-2&-1&-3&9&11\end{bmatrix}\]

$ to reduced row-echelon form.

The first element in row 1 is a 1. Use this to produce zeros in the first position in the lower rows:

\[ \begin{bmatrix}{{{1}}}&{{{0}}}&{{{-1}}}&{{{1}}}&{{{2}}} \\{{{1}}}&{{{1}}}&{{{0}}}&{{{2}}}&{{{-3}}} \\{{{2}}}&{{{0}}}&{{{-1}}}&{{{1}}}&{{{3}}} \\{{{0}}}&{{{-2}}}&{{{-1}}}&{{{-3}}}&{{{9}}}\end{bmatrix}\begin{array}{r}{{{2}}} \\{{{-4}}} \\{{{3}}} \\{{{11}}}\end{array}R_{2}+(-1)R_{1}\rightarrow R_{2}\begin{bmatrix}{{{1}}}&{{{0}}}&{{{-1}}}&{{{1}}}&{{{2}}} \\{{{0}}}&{{{1}}}&{{{1}}}&{{{1}}}&{{{-5}}} \\{{{0}}}&{{{0}}}&{{{1}}}&{{{-1}}}&{{{-1}}} \\{{{0}}}&{{{-2}}}&{{{-1}}}&{{{-3}}}&{{{9}}}\end{bmatrix}\begin{array}{r}{{{2}}} \\{{{-6}}} \\{{{-1}}} \\{{{11}}}\end{array} \]

Now the first nonzero element in row 2 is a 1. Use this to produce zeros in the position below it in the lower rows (only row 4 lacks a zero).

\[ \begin{bmatrix}{{{1}}}&{{{0}}}&{{{-1}}}&{{{1}}}&{{{2}}}&{{{2}}} \\{{{0}}}&{{{1}}}&{{{1}}}&{{{1}}}&{{{-5}}}&{{{-6}}} \\{{{0}}}&{{{0}}}&{{{1}}}&{{{-1}}}&{{{-1}}}&{{{-1}}} \\{{{0}}}&{{{-2}}}&{{{-1}}}&{{{-3}}}&{{{9}}}&{{{11}}}\end{bmatrix}\boldsymbol{R}_{4}+2\boldsymbol{R}_{2}\rightarrow\boldsymbol{R}_{4}\begin{bmatrix}{{{1}}}&{{{0}}}&{{{-1}}}&{{{1}}}&{{{2}}}&{{{2}}} \\{{{0}}}&{{{1}}}&{{{1}}}&{{{1}}}&{{{-5}}}&{{{-6}}} \\{{{0}}}&{{{0}}}&{{{1}}}&{{{-1}}}&{{{-1}}}&{{{-1}}} \\{{{0}}}&{{{0}}}&{{{1}}}&{{{-1}}}&{{{-1}}}&{{{-1}}}\end{bmatrix} \]

Now the first nonzero element in row 3 is a 1. Use this to produce a zero in the position below it in row 4.

\[ \begin{bmatrix}{{{1}}}&{{{0}}}&{{{-1}}}&{{{1}}}&{{{2}}}&{{{2}}} \\{{{0}}}&{{{1}}}&{{{1}}}&{{{1}}}&{{{-5}}}&{{{-6}}} \\{{{0}}}&{{{0}}}&{{{1}}}&{{{-1}}}&{{{-1}}}&{{{-1}}} \\{{{0}}}&{{{0}}}&{{{1}}}&{{{-1}}}&{{{-1}}}&{{{-1}}}\end{bmatrix}\boldsymbol{R}_{4}+(-1)\boldsymbol{R}_{3}\rightarrow\boldsymbol{R}_{4}\begin{bmatrix}{{{1}}}&{{{0}}}&{{{-1}}}&{{{1}}}&{{{2}}}&{{{2}}} \\{{{0}}}&{{{1}}}&{{{1}}}&{{{1}}}&{{{-5}}}&{{{-6}}} \\{{{0}}}&{{{0}}}&{{{1}}}&{{{-1}}}&{{{-1}}}&{{{-1}}} \\{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}\end{bmatrix} \]

This matrix is in row-echelon form. To produce reduced row-echelon form, use the leading 1 in each row to produce zeros in the corresponding position in the rows above, starting from the bottom row.

\[ \begin{bmatrix}{{{1}}}&{{{0}}}&{{{-1}}}&{{{1}}}&{{{2}}}&{{{2}}} \\{{{0}}}&{{{1}}}&{{{1}}}&{{{1}}}&{{{-5}}}&{{{-6}}} \\{{{0}}}&{{{0}}}&{{{1}}}&{{{-1}}}&{{{-1}}}&{{{-1}}} \\{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}\end{bmatrix}\begin{aligned}R_{1}+R_{3}&\rightarrow R_{1}\\ R_{2}+(-1)R_{3}&\rightarrow R_{2}\end{aligned}\begin{bmatrix}{{{1}}}&{{{0}}}&{{{0}}}&{{{0}}}&{{{1}}}&{{{1}}} \\{{{0}}}&{{{1}}}&{{{0}}}&{{{2}}}&{{{-4}}}&{{{-5}}} \\{{{0}}}&{{{0}}}&{{{1}}}&{{{-1}}}&{{{-1}}}&{{{-1}}} \\{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}\end{bmatrix} \]

This matrix is in reduced row-echelon form.

31.6. Solve by Gauss-Jordan elimination:

\[ \begin{aligned}x_{_{1}}&-x_{_{3}}+x_{_{4}}+2x_{_{5}}=2\\x_{_{1}}+x_{_{2}}&+2x_{_{4}}-3x_{_{5}}=-4\\2x_{_{1}}&-x_{_{3}}+x_{_{4}}+3x_{_{5}}=3\\-2x_{_{2}}-x_{_{3}}-3x_{_{4}}+9x_{_{5}}&=11\end{aligned} \]

The augmented matrix of the system is the matrix of Problem 31.5. Transforming to reduced row-echelon form yields the matrix

\[ \begin{bmatrix}{{{1}}}&{{{0}}}&{{{0}}}&{{{0}}}&{{{1}}}&{{{1}}} \\{{{0}}}&{{{1}}}&{{{0}}}&{{{2}}}&{{{-4}}}&{{{-5}}} \\{{{0}}}&{{{0}}}&{{{1}}}&{{{-1}}}&{{{-1}}}&{{{-1}}} \\{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}\end{bmatrix}which corresponds to the system\begin{aligned}x_{1}&+x_{5}=&1\\ x_{2}&+2x_{4}-4x_{5}=&-5\\ x_{3}&-x_{4}&-x_{5}=&-1\end{aligned} \]

This system has an infinite number of solutions. Let $ x_{5}=r $ , $ x_{4}=s $ , r and s any real numbers. Then from equation (3), $ x_{3}=r+s-1 $ ; from equation (2), $ x_{2}=4r-2s-5 $ ; and from equation (1), $ x_{1}=1-r $ . Thus all solutions can be written as $ (1-r, 4r-2s-5, r+s-1, s, r) $ , r and s any real numbers.

31.7. Pumps A, B, and C, working together, can fill a tank in 2 hours. If only A and C are used, it would take 4 hours. If only B and C are used, it would take 3 hours. How long would it take for each to fill the tank, working separately?

Let $ t_{1} $ , $ t_{2} $ , and $ t_{3} $ be the times for pumps A, B, and C, respectively. Then the rate at which each pump works can be written as $ r_{1} = 1/t_{1} $ , $ r_{2} = 1/t_{2} $ , and $ r_{3} = 1/t_{3} $ . Using quantity of work = (rate)(time), the following tabular arrangement can be made:

RATE TIME QUANTITY OF WORK
Pump A $ r_{1} \(</td><td>2</td><td>\) 2r_{1} \(</td></tr><tr><td>Pump B</td><td>\) r_{2} \(</td><td>2</td><td>\) 2r_{2} \(</td></tr><tr><td>Pump C</td><td>\) r_{3} \(</td><td>2</td><td>\) 2r_{3} $

Thus, if all three machines working together can fill the tank in 2 hours,

\[ 2r_{_{1}}+2r_{_{2}}+2r_{_{3}}=1 \]

Similarly,

\[ \begin{aligned}&4r_{_{1}}+4r_{_{3}}=1\\&3r_{_{2}}+3r_{_{3}}=1\\ \end{aligned} \]

The system (1), (2), (3) has the augmented matrix

\[ \begin{bmatrix}{{{2}}}&{{{2}}}&{{{2}}}&{{{1}}} \\{{{4}}}&{{{0}}}&{{{4}}}&{{{1}}} \\{{{0}}}&{{{3}}}&{{{3}}}&{{{1}}}\end{bmatrix}which transforms to the reduced row echelon form\begin{bmatrix}{{{1}}}&{{{0}}}&{{{0}}}&{{{\left|\begin{array}{ccc}1/6}}} \\{{{0}}}&{{{1}}}&{{{0}}} \\{{{0}}}&{{{0}}}&{{{1}}}\end{array}\right|1/4\\1/12\end{bmatrix} \]

Thus, $ r_{1} = 1/6 $ job/hr, $ r_{2} = 1/4 $ job/hr, and $ r_{3} = 1/12 $ job/hr. Therefore, $ t_{1} = 6 $ hr for pump A to fill the tank, $ t_{2} = 4 $ hr for pump B to fill the tank, and $ t_{3} = 12 $ hr for pump C to fill the tank, working alone.

31.8. An investor has $800,000 that she wishes to divide among Certificates of Deposit (CDs) paying 6% interest, mutual funds paying 10% interest, growth stocks paying 12% interest, and venture capital paying 14% interest. Fox tax reasons, she wants to plan for an annual return of $78,000, and she wants to have the total of all other investments three times the amount invested in CDs. How should she divide her investment?

Let $ x_{1}= $ amount invested in CDs, $ x_{2}= $ amount invested in mutual funds, $ x_{3}= $ amount invested in growth stocks, and $ x_{4}= $ amount invested as venture capital. Form a table:

AMOUNT INVESTED RATE OF INTEREST INTEREST EARNED
CDs $ x_{1} $ 0.06 0.06 $ x_{1} \(</td></tr><tr><td>Mutual funds</td><td>\) x_{2} $ 0.1 0.1 $ x_{2} \(</td></tr><tr><td>Growth stocks</td><td>\) x_{3} $ 0.12 0.12 $ x_{3} \(</td></tr><tr><td>Venture capital</td><td>\) x_{4} $ 0.14 0.14 $ x_{4} $

Since the total investment is 800,000, $ x_{1} + x_{2} + x_{3} + x_{4} = 800,000 $ (1)

Since the total income is $78,000, 0.06x_{1} + 0.1x_{2} + 0.12x_{3} + 0.14x_{4} = 78,000

Since the total of other investments is to equal three times the amount invested in CDs,

\[ x_{2}+x_{3}+x_{4}=3x_{1},or in standard form,-3x_{1}+x_{2}+x_{3}+x_{4}=0 \]

The system (1), (2), (3) has the following augmented matrix:

\[ \begin{bmatrix}{{{1}}}&{{{1}}}&{{{1}}}&{{{1}}}&{{{800,000}}} \\{{{0.06}}}&{{{0.1}}}&{{{0.12}}}&{{{0.14}}}&{{{78,000}}} \\{{{-3}}}&{{{1}}}&{{{1}}}&{{{1}}}&{{{0}}}\end{bmatrix} \]

Transforming this to reduced row-echelon form yields:

\[ \begin{bmatrix}1&0&0&0&\left|\begin{array}{c}200,000\\ \end{array}\right.\\ 0&1&0&-1&\left|\begin{array}{c}300,000\\ \end{array}\right.\\ 0&0&1&2&\left|\begin{array}{c}300,000\\ \end{array}\right.\end{bmatrix} \]

This corresponds to the system of equations:

\[ \begin{aligned}x_{_{1}}&=200,000\\x_{_{2}}&-x_{_{4}}=300,000\\x_{_{3}}+2x_{_{4}}&=300,000\end{aligned} \]

This has an infinite number of solutions. Let $ x_{4}=r $ . Then all solutions can be written in the form

(200000, 300000 + r, 300000 - 2r, r). Thus the investor must put 200,000 into CDs, but has a wide range of further options meeting the given conditions. An amount r put into venture capital requires an amount 300,000 more in mutual funds, and an amount 300,000 - 2r in growth stocks. As long as these are all positive, the conditions of the problem are satisfied; for example, one solution would be to let r = 100,000, then $ x_{1} = 200,000 $ in CDs, $ x_{2} = 400,000 $ in mutual funds, $ x_{3} = 100,000 $ in growth stocks, and $ x_{4} = 100,000 $ in venture capital.

SUPPLEMENTARY PROBLEMS

31.9. Transform to row-echelon form:

\[ \begin{bmatrix}{{{2}}}&{{{5}}}&{{{3}}} \\{{{4}}}&{{{-2}}}&{{{-6}}}\end{bmatrix} \]

\[ \begin{aligned}(b)\begin{bmatrix}{{{2}}}&{{{5}}}&{{{3}}} \\{{{4}}}&{{{10}}}&{{{-6}}}\end{bmatrix}\end{aligned} \]

\[ \begin{aligned}(c)\begin{bmatrix}{{{2}}}&{{{5}}}&{{{3}}} \\{{{4}}}&{{{10}}}&{{{6}}}\end{bmatrix}\end{aligned} \]

\[ Ans.\quad(a)\begin{bmatrix}1&5/2\\ 0&1\end{bmatrix}\begin{aligned}3/2\\ 1\end{aligned};(b)\begin{bmatrix}1&5/2\\ 0&0\end{bmatrix}\begin{aligned}3/2\\ -12\end{aligned};(c)\begin{bmatrix}1&5/2\\ 0&0\end{bmatrix}\begin{aligned}3/2\\ 0\end{aligned} \]

31.10. Solve, using the information from the previous problem:

\[ \begin{aligned}&2x+5y=3\\&4x-2y=-6\\ \end{aligned} \]

  1. $ 2x + 5y = 3 $

$ 4x + 10y = -6 $

\[ \begin{aligned}2x+5y&=3\\ 4x+10y&=6\end{aligned} \]

Ans. (a) $ (-1, 1) $ ; (b) no solution: (c) $ (, r) $ , r any real number.

31.11. Transform to reduced row-echelon form:

\[ \begin{bmatrix}{{{2}}}&{{{3}}}&{{{8}}} \\{{{3}}}&{{{-1}}}&{{{12}}} \\{{{5}}}&{{{2}}}&{{{20}}}\end{bmatrix} \]

\[ \begin{bmatrix}{{{2}}}&{{{3}}}&{{{-4}}}&{{{8}}} \\{{{3}}}&{{{4}}}&{{{-5}}}&{{{6}}} \\{{{1}}}&{{{1}}}&{{{-1}}}&{{{2}}}\end{bmatrix} \]

\[ \begin{bmatrix}{{{1}}}&{{{3}}}&{{{4}}}&{{{5}}}&{{{1}}} \\{{{3}}}&{{{5}}}&{{{2}}}&{{{6}}}&{{{7}}} \\{{{4}}}&{{{8}}}&{{{6}}}&{{{11}}}&{{{8}}}\end{bmatrix} \]

\[ \begin{aligned}&(a)\begin{bmatrix}{{{1}}}&{{{0}}}&{{{4}}} \\{{{0}}}&{{{1}}}&{{{0}}} \\{{{0}}}&{{{0}}}&{{{0}}}\end{bmatrix};(b)\begin{bmatrix}{{{1}}}&{{{0}}}&{{{1}}}&{{{2}}} \\{{{0}}}&{{{1}}}&{{{-2}}}&{{{0}}} \\{{{0}}}&{{{0}}}&{{{0}}}&{{{4}}}\end{bmatrix};(c)\begin{bmatrix}{{{1}}}&{{{0}}}&{{{-7/2}}}&{{{-7/4}}}&{{{4}}} \\{{{0}}}&{{{1}}}&{{{5/2}}}&{{{9/4}}}&{{{-1}}} \\{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}&{{{0}}}\end{bmatrix}\end{aligned} \]

32.12. Solve, using the information from the previous problem:

\[ \begin{aligned}2x+3y&=8\\ (a)\quad3x-y&=12\\ 5x+2y&=20\end{aligned} \]

\[ \begin{aligned}2x+3y-4z&=8\\ (b)3x+4y-5z&=6\\ x+y-z&=2\end{aligned} \]

\[ \begin{aligned}x_{1}+3x_{2}+4x_{3}+5x_{4}&=1\\3x_{1}+5x_{2}+2x_{3}+6x_{4}&=7\\4x_{1}+8x_{2}+6x_{3}+11x_{4}&=8\end{aligned} \]

Ans. (a) (4, 0); (b) no solution; (c) $ (4 + + , -1 - - , s, r) $ , r and s any real numbers

31.13. A mixture of 140 pounds of nuts is to be made from almonds costing $4 per pound, pecans costing $6 per pound, and brazil nuts costing $7.50 per pound. If the mixture will sell for $5.50 per pound, what possible combinations of nuts can be made?

Ans. If t = number of pounds of brazil nuts, then any combination of 105 - 1.75t pounds of pecans and $ 35 + 0.75t $ pounds of almonds for which all three are positive; thus, 0 < t < 60, 105 > 105 - 1.75t > 0, and $ 35 < 35 + 0.75t < 80 $ .