Linear Algebra - Non-Homogeneous Cases

$\begin{align*}&\text{Find }\begin{bmatrix}x\y\end{bmatrix}\&\text{for the system: }\begin{bmatrix}1&16\-10&-11\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}136\-2\end{bmatrix}\end{align*}$

$\begin{bmatrix}-9.83\9.11\end{bmatrix}$

$\begin{bmatrix}-8.83\10.11\end{bmatrix}$

$\begin{bmatrix}-14.83\4.11\end{bmatrix}$

$\begin{bmatrix}-16.83\2.11\end{bmatrix}$

Explanation

$\begin{align*}&\text{When faced with a system of equations like the}\&\text{one represented by }\&\begin{bmatrix}1&16\-10&-11\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}136\-2\end{bmatrix}\&\text{a common approach is to manipulate coefficients to eliminate}\&\text{variables. This works, though there is another way to consider.}\&\text{If a matrix has an inverse,}\&\text{then the product of that matrix and its inverse is the identity matrix:}\&A^{-1}A=AA^{-1}=I\&\text{As such, if given an equation like: }Ax=B\&\text{we can isolate x as follows:}\&A^{-1}Ax=A^{-1}B\&x=A^{-1}B\&\text{With this property we can solve our system:}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-0.074&-0.107\0.067&0.007\end{bmatrix}\begin{bmatrix}136\-2\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-9.83\9.11\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Find }\begin{bmatrix}x\y\end{bmatrix}\&\text{for the system: }\begin{bmatrix}12&18\-20&6\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}7\-10\end{bmatrix}\end{align*}$

$\begin{bmatrix}0.51\0.05\end{bmatrix}$

$\begin{bmatrix}3.51\3.05\end{bmatrix}$

$\begin{bmatrix}4.51\4.05\end{bmatrix}$

$\begin{bmatrix}-8.49\-8.95\end{bmatrix}$

Explanation

$\begin{align*}&\text{When faced with a system of equations like the}\&\text{one represented by }\&\begin{bmatrix}12&18\-20&6\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}7\-10\end{bmatrix}\&\text{an approach often taught is to manipulate coefficients to eliminate}\&\text{variables. This approach works, though there is another to consider.}\&\text{If a matrix has an inverse, i.e. it has a determinant,}\&\text{then the product of that matrix and its inverse is the identity matrix:}\&A^{-1}A=AA^{-1}=I\&\text{It then stands to reason that if given an equation like: }Ax=B\&\text{We can isolate x as follows:}\&A^{-1}Ax=A^{-1}B\&x=A^{-1}B\&\text{With this property we can solve our system:}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}0.014&-0.042\0.046&0.028\end{bmatrix}\begin{bmatrix}7\-10\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}0.51\0.05\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Find }\begin{bmatrix}x\y\end{bmatrix}\&\text{for the system: }\begin{bmatrix}4&18\-16&16\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}20\-29\end{bmatrix}\end{align*}$

$\begin{bmatrix}2.39\0.58\end{bmatrix}$

$\begin{bmatrix}3.39\1.58\end{bmatrix}$

$\begin{bmatrix}-2.61\-4.42\end{bmatrix}$

$\begin{bmatrix}-5.61\-7.42\end{bmatrix}$

Explanation

$\begin{align*}&\text{When faced with a system of equations like the}\&\text{one represented by }\&\begin{bmatrix}4&18\-16&16\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}20\-29\end{bmatrix}\&\text{a common approach is to manipulate coefficients to eliminate}\&\text{variables. This works, though there is another way to consider.}\&\text{If a matrix has an inverse,}\&\text{then the product of that matrix and its inverse is the identity matrix:}\&A^{-1}A=AA^{-1}=I\&\text{As such, if given an equation like: }Ax=B\&\text{we can isolate x as follows:}\&A^{-1}Ax=A^{-1}B\&x=A^{-1}B\&\text{With this property we can solve our system:}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}0.045&-0.051\0.045&0.011\end{bmatrix}\begin{bmatrix}20\-29\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}2.39\0.58\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Solve the system of equations: }\begin{bmatrix}-5&-16\19&-3\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-167\87\end{bmatrix}\end{align*}$

$\begin{bmatrix}5.93\8.58\end{bmatrix}$

$\begin{bmatrix}7.93\10.58\end{bmatrix}$

$\begin{bmatrix}-0.07\2.58\end{bmatrix}$

$\begin{bmatrix}-1.07\1.58\end{bmatrix}$

Explanation

$\begin{align*}&\text{When faced with a system of equations like the}\&\text{one represented by }\&\begin{bmatrix}-5&-16\19&-3\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-167\87\end{bmatrix}\&\text{a common approach is to manipulate coefficients to eliminate}\&\text{variables. This works, though there is another way to consider.}\&\text{If a matrix has an inverse,}\&\text{then the product of that matrix and its inverse is the identity matrix:}\&A^{-1}A=AA^{-1}=I\&\text{As such, if given an equation like: }Ax=B\&\text{we can isolate x as follows:}\&A^{-1}Ax=A^{-1}B\&x=A^{-1}B\&\text{With this, we can solve our system of equations:}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-0.009&0.050\-0.060&-0.016\end{bmatrix}\begin{bmatrix}-167\87\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}5.93\8.58\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Find }\begin{bmatrix}x\y\end{bmatrix}\&\text{for the system: }\begin{bmatrix}16&-11\-10&6\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-78\131\end{bmatrix}\end{align*}$

$\begin{bmatrix}-69.50\-94.00\end{bmatrix}$

$\begin{bmatrix}-72.50\-97.00\end{bmatrix}$

$\begin{bmatrix}-74.50\-99.00\end{bmatrix}$

$\begin{bmatrix}-76.50\-101.00\end{bmatrix}$

Explanation

$\begin{align*}&\text{When faced with a system of equations like the}\&\text{one represented by }\&\begin{bmatrix}16&-11\-10&6\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-78\131\end{bmatrix}\&\text{a common approach is to manipulate coefficients to eliminate}\&\text{variables. This works, though there is another way to consider.}\&\text{If a matrix has an inverse,}\&\text{then the product of that matrix and its inverse is the identity matrix:}\&A^{-1}A=AA^{-1}=I\&\text{As such, if given an equation like: }Ax=B\&\text{we can isolate x as follows:}\&A^{-1}Ax=A^{-1}B\&x=A^{-1}B\&\text{With this property we can solve our system:}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-0.429&-0.786\-0.714&-1.143\end{bmatrix}\begin{bmatrix}-78\131\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-69.50\-94.00\end{bmatrix}\end{align*}$

Give the partial fractions decomposition of $\frac{4}{x^{3}+ 4x^{2}- 3x -12 }$ .

$\frac{-4x+ 16 }{13 (x^{2}-3)}+ \frac{4}{13(x+ 4)}$

$\frac{4x- 16 }{13 (x^{2}-3)}+ \frac{4}{13(x+ 4)}$

$\frac{ 4x+ 16 }{13 (x^{2}-3)}+ \frac{4}{13(x+ 4)}$

$\frac{-4x+ 4 }{13 (x^{2}-3)}+ \frac{16}{13(x+ 4)}$

$\frac{4x-4 }{13 (x^{2}-3)}+ \frac{16}{13(x+ 4)}$

Explanation

The denominator of the fraction, $x^{3}+ 4x^{2}- 3x -12$ , can be factored as

$(x^{2} - 3 )(x+4)$ .

The partial fractions decomposition of a rational expression is the sum of fractions with these factors, with the numerator above each denominator being degree one less. Thus, the partial fractions decomposition of the given expression takes the form

$\frac{4}{x^{3}+ 4x^{2}- 3x -12 }$

$= \frac{4}{ (x^{2}-3)(x+ 4)}$

$= \frac{Ax+B}{x^{2}-3}+ \frac{C}{x+ 4}$

for some $A,B, C \in \mathbb{R}$ .

Express the fractions with a common denominator:

$\frac{4}{ (x^{2}-3) (x+ 4)} = \frac{(Ax+B)(x+ 4)}{(x^{2}-3) (x+ 4)}+ \frac{C(x^{2}-3)}{(x^{2}-3) (x+ 4)}$

Eliminate the denominators and perform some algebra:

$4 = (Ax+B)(x+ 4) + C(x^{2}-3)$

$4 = Ax^{2} +4Ax+ Bx + 4B + Cx^{2} -3C$

$4 = (A+C)x^{2} +(4A + B)x +( 4B -3C)$

Set the coefficients equal to form the linear system

This can be solved using Gauss-Jordan elimination on the augmented matrix

$\begin{bmatrix} 1 & 0 & 1 & 0 \ 4 & 1 & 0 & 0 \ 0 & 4 & -3 &4 \end{bmatrix}$

Perform the following row operations on the matrix to get it into reduced row-echelon form:

$-4R1+R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0 & 1 & 0 \ 0 & 1 & -4 & 0 \ 0 & 4 & -3 &4 \end{bmatrix}$

$-4R2+R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & 1 & 0 \ 0 & 1 & -4 & 0 \ 0 & 0 & 13 &4 \end{bmatrix}$

$\frac{1}{13}R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & 1 & 0 \ \ 0 & 1 & -4 & 0 \ \ 0 & 0 & 1 & \frac{4}{13} \end{bmatrix}$

$-R3 + R1 \rightarrow R1$

$4R3 + R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0 & 0 & - \frac{4}{13} \ \ 0 & 1 & 0 & \frac{16}{13} \ \ 0 & 0 & 1 & \frac{4}{13} \end{bmatrix}$

$A = - \frac{4}{13} ,B = \frac{16}{13} , C= \frac{4}{13}$ , so the partial fractions decomposition is

$\frac{4}{x^{3}+ 4x^{2}- 3x -12 } = \frac{- \frac{4}{13}x+ \frac{16}{13} }{x^{2}-3}+ \frac{\frac{4}{13}}{x+ 4}$ ,

or, simplified,

$\frac{4}{x^{3}+ 4x^{2}- 3x -12 } = \frac{-4x+ 16 }{13 (x^{2}-3)}+ \frac{4}{13(x+ 4)}$ .

It is recommended that you use a calculator with matrix arithmetic capability for this question.

The graph of a quartic (degree four) polynomial passes through the following five points:

, , , ,

What is the cubic term of ?

Round to four decimal digits.

$0.0104x^{3}$

$-0.1484x^{3}$

$2.4469x^{3}$

$-0.2917x^{3}$

None of the other choices gives the correct response.

Explanation

A quartic polynomial takes the form

$p(x) = a_{4}x^{4} + a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0}$

where the $a_{n}$ are the coefficients. The value we are trying to identify is $a_{4}$ , the quartic term.

An equation can be formed from each given ordered pair by substituting the abscissa for and the ordinate for . The problem can be simplified somewhat by noting that, since the -intercept of the graph of is given as , $a_{0} = -2$ . Therefore, we are looking for the coefficients of

$p(x) = a_{4}x^{4} + a_{3}x^{3} + a_{2}x^{2} + a_{1}x -2$ .

The equations formed from the other four points are

$256 a_{4} -64a_{3} + 16a_{2} - 4a_{1} -2 = 0$

$16 a_{4} -8a_{3} + 4a_{2} - 2a_{1} -2 = 6$

$16 a_{4} +8a_{3} + 4a_{2} +2a_{1} -2 = 5$

$256 a_{4} +64a_{3} + 16a_{2} + 4a_{1} -2 = -1$

Adding two to both sides of each equation:

$256 a_{4} -64a_{3} + 16a_{2} - 4a_{1} =2$

$16 a_{4} -8a_{3} + 4a_{2} - 2a_{1} =8$

$16 a_{4} +8a_{3} + 4a_{2} +2a_{1} =7$

$256 a_{4} +64a_{3} + 16a_{2} + 4a_{1} = 1$

This is a system of four linear equations in four variables, which can be rewritten as the matrix multiplication equation , where is the matrix of coefficients, , the column matrix of variables, and the matrix of constants; this equation is

$\begin{bmatrix} 256 & -64 & 16 & -4 \ 16 & -8 & 4 & -2 \ 16 & 8 & 4 & 2 \ 256 & 64 & 16 & 4 \end{bmatrix}\begin{bmatrix} a_{4} \a_{3} \ a _{2} \ a_{1} \end{bmatrix}=\begin{bmatrix}2 \8\7\1 \end{bmatrix}$

We are trying to calculate ; since , $x = A^{-1}b$ , or

$\begin{bmatrix} a_{4} \a_{3} \ a _{2} \ a_{1} \end{bmatrix}=\begin{bmatrix} 256 & -64 & 16 & -4 \ 16 & -8 & 4 & -2 \ 16 & 8 & 4 & 2 \ 256 & 64 & 16 & 4 \end{bmatrix} ^{-1}\begin{bmatrix}2 \8\7\1 \end{bmatrix}$

This calculates to

$\begin{bmatrix} a_{4} \a_{3} \ a _{2} \ a_{1} \end{bmatrix}= \begin{bmatrix}-0.1484 \ 0.0104\2.4469\-0.2917 \end{bmatrix}$ .

Since the cubic term is requested, select $a_{3} = 0.0104$ . The correct response is $0.0104 x^{3}$ .

$\begin{align*}&\text{Find the values of }\begin{bmatrix}x\y\end{bmatrix}\&\text{for the function: }\begin{bmatrix}7&15\0&-18\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-8\6\end{bmatrix}\end{align*}$

$\begin{bmatrix}-0.43\-0.33\end{bmatrix}$

$\begin{bmatrix}2.57\2.67\end{bmatrix}$

$\begin{bmatrix}-5.43\-5.33\end{bmatrix}$

$\begin{bmatrix}-8.43\-8.33\end{bmatrix}$

Explanation

$\begin{align*}&\text{When faced with a system of equations like the}\&\text{one represented by }\&\begin{bmatrix}7&15\0&-18\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-8\6\end{bmatrix}\&\text{an approach often taught is to manipulate coefficients to eliminate}\&\text{variables. This approach works, though there is another to consider.}\&\text{If a matrix has an inverse, i.e. it has a determinant,}\&\text{then the product of that matrix and its inverse is the identity matrix:}\&A^{-1}A=AA^{-1}=I\&\text{It then stands to reason that if given an equation like: }Ax=B\&\text{We can isolate x as follows:}\&A^{-1}Ax=A^{-1}B\&x=A^{-1}B\&\text{Using this, we can solve for our values:}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}0.143&0.119\0.000&-0.056\end{bmatrix}\begin{bmatrix}-8\6\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-0.43\-0.33\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Find }\begin{bmatrix}x\y\z\end{bmatrix}\&\text{for the system: }\begin{bmatrix}-3&-15&6\15&-9&14\-18&-5&-10\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-14\5\-8\end{bmatrix}\end{align*}$

$\begin{bmatrix}5.086\-2.8548\-6.9274\end{bmatrix}$

$\begin{bmatrix}8.086\0.14516\-3.9274\end{bmatrix}$

$\begin{bmatrix}11.086\3.1452\-0.92742\end{bmatrix}$

$\begin{bmatrix}-2.914\-10.8548\-14.9274\end{bmatrix}$

Explanation

$- ¬\begin{align*}&\text{When faced with a system of equations like the}\&\text{one represented by }\&\begin{bmatrix}-3&-15&6\15&-9&14\-18&-5&-10\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-14\5\-8\end{bmatrix}\&\text{an approach often taught is to manipulate coefficients to eliminate}\&\text{variables. There is another to consider.}\&\text{Note that if a matrix has an inverse, }\&\text{then the product of that matrix and its inverse is the identity matrix:}\&A^{-1}A=AA^{-1}=I\&\text{Knowing this, it stands to reason that if given an equation like: }Ax=B\&\text{We can isolate x as follows:}\&A^{-1}Ax=A^{-1}B\&x=A^{-1}B\&\text{With this property we can solve our system:}\&\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-0.43011&0.48387&0.41935\0.27419&-0.37097&-0.35484\0.6371&-0.68548&-0.67742\end{bmatrix}\begin{bmatrix}-14\5\-8\end{bmatrix}\&\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}5.086\-2.8548\-6.9274\end{bmatrix}\end{align*}$

$\begin{align*}&\text{Find }\begin{bmatrix}x\y\end{bmatrix}\&\text{for the system: }\begin{bmatrix}12&-6\-2&-5\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}7\3\end{bmatrix}\end{align*}$

$\begin{bmatrix}0.24\-0.69\end{bmatrix}$

$\begin{bmatrix}2.24\1.31\end{bmatrix}$

$\begin{bmatrix}5.24\4.31\end{bmatrix}$

$\begin{bmatrix}7.24\6.31\end{bmatrix}$

Explanation

$\begin{align*}&\text{When faced with a system of equations like the}\&\text{one represented by }\&\begin{bmatrix}12&-6\-2&-5\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}7\3\end{bmatrix}\&\text{an approach often taught is to manipulate coefficients to eliminate}\&\text{variables. This approach works, though there is another to consider.}\&\text{If a matrix has an inverse, i.e. it has a determinant,}\&\text{then the product of that matrix and its inverse is the identity matrix:}\&A^{-1}A=AA^{-1}=I\&\text{It then stands to reason that if given an equation like: }Ax=B\&\text{We can isolate x as follows:}\&A^{-1}Ax=A^{-1}B\&x=A^{-1}B\&\text{With this property we can solve our system:}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}0.069&-0.083\-0.028&-0.167\end{bmatrix}\begin{bmatrix}7\3\end{bmatrix}\&\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}0.24\-0.69\end{bmatrix}\end{align*}$

Non-Homogeneous Cases

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