2.E: Exercises
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Exercise \(\PageIndex{1}\)
For the following pairs of matrices, determine if the sum \(A + B\) is defined. If so, find the sum.
- \(A = \left [ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right ], B = \left [ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right ]\)
- \(A = \left [ \begin{array}{rrr} 2 & 1 & 2 \\ 1 & 1 & 0 \end{array} \right ], B = \left [ \begin{array}{rrr} -1 & 0 & 3\\ 0 & 1 & 4 \end{array} \right ]\)
- \(A = \left [ \begin{array}{rr} 1 & 0 \\ -2 & 3 \\ 4 & 2 \end{array} \right ], B = \left [ \begin{array}{rrr} 2 & 7 & -1 \\ 0 & 3 & 4 \end{array} \right ]\)
Exercise \(\PageIndex{2}\)
For each matrix \(A\), find the matrix \(-A\) such that \(A + (-A) = 0\).
- \(A = \left [ \begin{array}{rr} 1 & 2 \\ 2 & 1 \end{array} \right ]\)
- \(A = \left [ \begin{array}{rr} -2 & 3 \\ 0 & 2 \end{array} \right ]\)
- \(A = \left [ \begin{array}{rrr} 0 & 1 & 2 \\ 1 & -1 & 3 \\ 4 & 2 & 0 \end{array} \right ]\)
Exercise \(\PageIndex{3}\)
In the context of Proposition 2.1.1 , describe \(-A\) and \(0.\)
- Answer
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To get \(-A,\) just replace every entry of \(A\) with its additive inverse. The 0 matrix is the one which has all zeros in it.
2.1.2: Scalar Multiplication of Matrices
Exercise \(\PageIndex{4}\)
For each matrix \(A\), find the product \((-2)A, 0A,\) and \(3A\).
- \(A = \left [ \begin{array}{rr} 1 & 2 \\ 2 & 1 \end{array} \right ]\)
- \(A = \left [ \begin{array}{rr} -2 & 3 \\ 0 & 2 \end{array} \right ]\)
- \(A = \left [ \begin{array}{rrr} 0 & 1 & 2 \\ 1 & -1 & 3 \\ 4 & 2 & 0 \end{array} \right ]\)
Exercise \(\PageIndex{5}\)
Using only the properties given in Proposition 2.1.1 and Proposition 2.1.2 , show \(-A\) is unique.
- Answer
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Suppose \(B\) also works. Then \[-A=-A+\left( A+B\right) =\left( -A+A\right) +B=0+B=B\nonumber \]
Exercise \(\PageIndex{6}\)
Using only the properties given in Proposition 2.1.1 and Proposition 2.1.2 , show \(0\) is unique.
- Answer
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Suppose \(0^{\prime }\) also works. Then \(0^{\prime }=0^{\prime }+0=0.\)
Exercise \(\PageIndex{7}\)
Using only the properties given in Proposition 2.1.1 and Proposition 2.1.2 show \(0A=0.\) Here the \(0\) on the left is the scalar \(0\) and the \(0\) on the right is the zero matrix of appropriate size.
- Answer
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\(0A=\left( 0+0\right) A=0A+0A.\) Now add \(-\left( 0A\right)\) to both sides. Then \(0=0A\).
Exercise \(\PageIndex{8}\)
Using only the properties given in Proposition 2.1.1 and Proposition 2.1.2 , as well as previous problems show \(\left( -1\right) A=-A.\)
- Answer
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\(A+\left( -1\right) A=\left( 1+\left( -1\right) \right) A=0A=0.\) Therefore, from the uniqueness of the additive inverse proved in the above Problem \(\PageIndex{7}\), it follows that \(-A=\left( -1\right) A\).
2.2
Exercise \(\PageIndex{9}\)
Consider the matrices \(A =\left [ \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 1 & 7 \end{array} \right ], B=\left [ \begin{array}{rrr} 3 & -1 & 2 \\ -3 & 2 & 1 \end{array} \right ], C =\left [ \begin{array}{rr} 1 & 2 \\ 3 & 1 \end{array} \right ], \\ D=\left [ \begin{array}{rr} -1 & 2 \\ 2 & -3 \end{array} \right ], E=\left [ \begin{array}{r} 2 \\ 3 \end{array} \right ]\).
Find the following if possible. If it is not possible explain why.
- \(-3A\)
- \(3B-A\)
- \(AC\)
- \(CB\)
- \(AE\)
- \(EA\)
- Answer
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- \(\left [ \begin{array}{rrr} -3 & -6 & -9 \\ -6 & -3 & -21 \end{array} \right ]\)
- \(\left [ \begin{array}{rrr} 8 & -5 & 3 \\ -11 & 5 & -4 \end{array} \right ]\)
- Not possible
- \(\left [ \begin{array}{rrr} -3 & 3 & 4 \\ 6 & -1 & 7 \end{array} \right ]\)
- Not possible
- Not possible
Exercise \(\PageIndex{10}\)
Consider the matrices \(A =\left [ \begin{array}{rr} 1 & 2 \\ 3 & 2 \\ 1 & -1 \end{array} \right ], B=\left [ \begin{array}{rrr} 2 & -5 & 2 \\ -3 & 2 & 1 \end{array} \right ] , C =\left [ \begin{array}{rr} 1 & 2 \\ 5 & 0 \end{array} \right ], \\ D=\left [ \begin{array}{rr} -1 & 1 \\ 4 & -3 \end{array} \right ], E=\left [ \begin{array}{r} 1 \\ 3 \end{array} \right ]\)
Find the following if possible. If it is not possible explain why.
- \(-3A\)
- \(3B-A\)
- \(AC\)
- \(CA\)
- \(AE\)
- \(EA\)
- \(BE\)
- \(DE\)
- Answer
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- \(\left [ \begin{array}{rr} -3 & -6 \\ -9 & -6 \\ -3 & 3 \end{array} \right ]\)
- Not possible.
- \(\left [ \begin{array}{rr} 11 & 2 \\ 13 & 6 \\ -4 & 2 \end{array} \right ]\)
- Not possible.
- \(\left [ \begin{array}{r} 7 \\ 9 \\ -2 \end{array} \right ]\)
- Not possible.
- Not possible.
- \(\left [ \begin{array}{r} 2 \\ -5 \end{array} \right ]\)
Exercise \(\PageIndex{11}\)
Let \(A=\left [ \begin{array}{rr} 1 & 1 \\ -2 & -1 \\ 1 & 2 \end{array} \right ]\), \(B=\left [ \begin{array}{rrr} 1 & -1 & -2 \\ 2 & 1 & -2 \end{array} \right ] ,\) and \(C=\left [ \begin{array}{rrr} 1 & 1 & -3 \\ -1 & 2 & 0 \\ -3 & -1 & 0 \end{array} \right ] .\) Find the following if possible.
- \(AB\)
- \(BA\)
- \(AC\)
- \(CA\)
- \(CB\)
- \(BC\)
- Answer
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- \(\left [ \begin{array}{rrr} 3 & 0 & -4 \\ -4 & 1 & 6 \\ 5 & 1 & -6 \end{array} \right ]\)
- \(\left [ \begin{array}{rr} 1 & -2 \\ -2 & -3 \end{array} \right ]\)
- Not possible
- \(\left [ \begin{array}{rr} -4 & -6 \\ -5 & -3 \\ -1 & -2 \end{array} \right ]\)
- \(\left [ \begin{array}{rrr} 8 & 1 & -3 \\ 7 & 6 & -6 \end{array} \right ]\)
Exercise \(\PageIndex{12}\)
Let \(A=\left [ \begin{array}{rr} -1 & -1 \\ 3 & 3 \end{array} \right ]\). Find all \(2\times 2\) matrices, \(B\) such that \(AB=0.\)
- Answer
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\[\begin{aligned} \left [ \begin{array}{rr} -1 & -1 \\ 3 & 3 \end{array} \right ] \left [ \begin{array}{cc} x & y \\ z & w \end{array} \right ] &=\left [ \begin{array}{cc} -x-z & -w-y \\ 3x+3z & 3w+3y \end{array} \right ] \\ &=\left [ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right ]\end{aligned}\] Solution is: \(w=-y,x=-z\) so the matrices are of the form \(\left [ \begin{array}{rr} x & y \\ -x & -y \end{array} \right ].\)
Exercise \(\PageIndex{13}\)
Let \(X=\left [ \begin{array}{rrr} -1 & -1 & 1 \end{array} \right ]\) and \(Y=\left [ \begin{array}{rrr} 0 & 1 & 2 \end{array} \right ] .\) Find \(X^{T}Y\) and \(XY^{T}\) if possible.
- Answer
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\(X^{T}Y = \left [ \begin{array}{rrr} 0 & -1 & -2 \\ 0 & -1 & -2 \\ 0 & 1 & 2 \end{array} \right ] , XY^{T} = 1\)
Exercise \(\PageIndex{14}\)
Let \(A=\left [ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right ] ,B=\left [ \begin{array}{rr} 1 & 2 \\ 3 & k \end{array} \right ] .\) Is it possible to choose \(k\) such that \(AB=BA?\) If so, what should \(k\) equal?
- Answer
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\[\begin{aligned} \left [ \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} \right ] \left [ \begin{array}{cc} 1 & 2 \\ 3 & k \end{array} \right ] &= \left [ \begin{array}{cc} 7 & 2k+2 \\ 15 & 4k+6 \end{array} \right ] \\ \left [ \begin{array}{cc} 1 & 2 \\ 3 & k \end{array} \right ] \left [ \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} \right ] &= \left [ \begin{array}{cc} 7 & 10 \\ 3k+3 & 4k+6 \end{array} \right ]\end{aligned}\] Thus you must have \(\begin{array}{c} 3k+3=15 \\ 2k+2=10 \end{array}\), Solution is: \(\left[ k=4\right]\)
Exercise \(\PageIndex{15}\)
Let \(A=\left [ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right ] ,B=\left [ \begin{array}{rr} 1 & 2 \\ 1 & k \end{array} \right ] .\) Is it possible to choose \(k\) such that \(AB=BA?\) If so, what should \(k\) equal?
- Answer
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\[\begin{aligned} \left [ \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} \right ] \left [ \begin{array}{cc} 1 & 2 \\ 1 & k \end{array} \right ] &= \left [ \begin{array}{cc} 3 & 2k+2 \\ 7 & 4k+6 \end{array} \right ] \\ \left [ \begin{array}{cc} 1 & 2 \\ 1 & k \end{array} \right ] \left [ \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} \right ] &= \left [ \begin{array}{cc} 7 & 10 \\ 3k+1 & 4k+2 \end{array} \right ]\end{aligned}\] However, \(7\neq 3\) and so there is no possible choice of \(k\) which will make these matrices commute.
Exercise \(\PageIndex{16}\)
Find \(2\times 2\) matrices, \(A\), \(B,\) and \(C\) such that \(A\neq 0,C\neq B,\) but \(AC=AB.\)
- Answer
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Let \(A=\left[\begin{array}{cc}1&-1 \\ -1&1\end{array}\right],\: B=\left[\begin{array}{cc}1&1\\1&1\end{array}\right],\: C=\left[\begin{array}{cc}2&2\\2&2\end{array}\right]\). \[\begin{aligned}\left[\begin{array}{cc}1&-1\\-1&1\end{array}\right]\left[\begin{array}{cc}1&1\\1&1\end{array}\right]&=\left[\begin{array}{cc}0&0\\0&0\end{array}\right] \\ \left[\begin{array}{cc}1&-1\\-1&1\end{array}\right]\left[\begin{array}{cc}2&2\\2&2\end{array}\right]&=\left[\begin{array}{cc}0&0\\0&0\end{array}\right] \end{aligned}\]
Exercise \(\PageIndex{17}\)
Give an example of matrices (of any size), \(A,B,C\) such that \(B\neq C\), \(A\neq 0,\) and yet \(AB=AC.\)
Exercise \(\PageIndex{18}\)
Find \(2 \times 2\) matrices \(A\) and \(B\) such that \(A \neq 0\) and \(B \neq 0\) but \(AB = 0\).
- Answer
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Let \(A=\left[\begin{array}{cc}1&-1 \\ -1&1\end{array}\right],\: B=\left[\begin{array}{cc}1&1\\1&1\end{array}\right]\). \[\left[\begin{array}{cc}1&-1\\-1&1\end{array}\right]\left[\begin{array}{cc}1&1\\1&1\end{array}\right]=\left[\begin{array}{cc}0&0\\0&0\end{array}\right]\nonumber\]
Exercise \(\PageIndex{19}\)
Give an example of matrices (of any size), \(A,B\) such that \(A \neq 0\) and \(B \neq 0\) but \(AB=0.\)
Exercise \(\PageIndex{20}\)
Find \(2 \times 2\) matrices \(A\) and \(B\) such that \(A \neq 0\) and \(B \neq 0\) with \(AB \neq BA\).
- Answer
-
Let \(A=\left[\begin{array}{cc}0&1\\1&0\end{array}\right],\: B=\left[\begin{array}{cc}1&2\\3&4\end{array}\right]\). \[\begin{aligned}\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{cc}1&2\\3&4\end{array}\right]&=\left[\begin{array}{cc}3&4\\1&2\end{array}\right] \\ \left[\begin{array}{cc}1&2\\3&4\end{array}\right]\left[\begin{array}{cc}0&1\\1&0\end{array}\right]&=\left[\begin{array}{cc}2&1\\4&3\end{array}\right]\end{aligned}\]
Exercise \(\PageIndex{21}\)
Write the system \[\begin{array}{c} x_{1}-x_{2}+2x_{3} \\ 2x_{3}+x_{1} \\ 3x_{3} \\ 3x_{4}+3x_{2}+x_{1} \end{array}\nonumber \] in the form \(A\left [ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right ]\) where \(A\) is an appropriate matrix.
- Answer
-
\(A=\left [ \begin{array}{rrrr} 1 & -1 & 2 & 0 \\ 1 & 0 & 2 & 0 \\ 0 & 0 & 3 & 0 \\ 1 & 3 & 0 & 3 \end{array} \right ]\)
Exercise \(\PageIndex{22}\)
Write the system \[\begin{array}{c} x_{1}+3x_{2}+2x_{3} \\ 2x_{3}+x_{1} \\ 6x_{3} \\ x_{4}+3x_{2}+x_{1} \end{array}\nonumber \] in the form \(A\left [ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right ]\) where \(A\) is an appropriate matrix.
- Answer
-
\(A=\left [ \begin{array}{rrrr} 1 & 3 & 2 & 0 \\ 1 & 0 & 2 & 0 \\ 0 & 0 & 6 & 0 \\ 1 & 3 & 0 & 1 \end{array} \right ]\)
Exercise \(\PageIndex{23}\)
Write the system \[\begin{array}{c} x_{1}+x_{2}+x_{3} \\ 2x_{3}+x_{1}+x_{2} \\ x_{3}-x_{1} \\ 3x_{4}+x_{1} \end{array}\nonumber \] in the form \(A\left [ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right ]\) where \(A\) is an appropriate matrix.
- Answer
-
\(A=\left [ \begin{array}{rrrr} 1 & 1 & 1 & 0 \\ 1 & 1 & 2 & 0 \\ -1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 3 \end{array} \right ]\)
Exercise \(\PageIndex{24}\)
A matrix \(A\) is called idempotent if \(A^{2}=A.\) Let \[A= \left [ \begin{array}{rrr} 2 & 0 & 2 \\ 1 & 1 & 2 \\ -1 & 0 & -1 \end{array} \right ]\nonumber \] and show that \(A\) is idempotent.
2.3
Exercise \(\PageIndex{25}\)
For each pair of matrices, find the \((1,2)\)-entry and \((2,3)\)-entry of the product \(AB\).
- \(A = \left [ \begin{array}{rrr} 1 & 2 & -1 \\ 3 & 4 & 0 \\ 2 & 5 & 1 \end{array} \right ], B = \left [ \begin{array}{rrr} 4 & 6 & -2 \\ 7 & 2 & 1 \\ -1 & 0 & 0 \end{array} \right ]\)
- \(A = \left [ \begin{array}{rrr} 1 & 3 & 1 \\ 0 & 2 & 4 \\ 1 & 0 & 5 \end{array} \right ], B = \left [ \begin{array}{rrr} 2 & 3 & 0 \\ -4 & 16 & 1 \\ 0 & 2 & 2 \end{array} \right ]\)
2.4
Exercise \(\PageIndex{26}\)
Suppose \(A\) and \(B\) are square matrices of the same size. Which of the following are necessarily true?
- \(\left( A-B\right) ^{2}=A^{2}-2AB+B^{2}\)
- \(\left( AB\right) ^{2}=A^{2}B^{2}\)
- \(\left( A+B\right) ^{2}=A^{2}+2AB+B^{2}\)
- \(\left( A+B\right) ^{2}=A^{2}+AB+BA+B^{2}\)
- \(A^{2}B^{2}=A\left( AB\right) B\)
- \(\left( A+B\right) ^{3}=A^{3}+3A^{2}B+3AB^{2}+B^{3}\)
- \(\left( A+B\right) \left( A-B\right) =A^{2}-B^{2}\)
- Answer
-
- Not necessarily true.
- Not necessarily true.
- Not necessarily true.
- Necessarily true.
- Necessarily true.
- Not necessarily true.
- Not necessarily true.
2.5
Exercise \(\PageIndex{27}\)
Consider the matrices \(A =\left [ \begin{array}{rr} 1 & 2 \\ 3 & 2 \\ 1 & -1 \end{array} \right ], B=\left [ \begin{array}{rrr} 2 & -5 & 2 \\ -3 & 2 & 1 \end{array} \right ], C =\left [ \begin{array}{rr} 1 & 2 \\ 5 & 0 \end{array} \right ], \\ D=\left [ \begin{array}{rr} -1 & 1 \\ 4 & -3 \end{array} \right ], E=\left [ \begin{array}{r} 1 \\ 3 \end{array} \right ]\)
Find the following if possible. If it is not possible explain why.
- \(-3A{^T}\)
- \(3B - A^{T}\)
- \(E^{T}B\)
- \(EE^{T}\)
- \(B^{T}B\)
- \(CA^{T}\)
- \(D^{T}BE\)
- Answer
-
- \(\left [ \begin{array}{rrr} -3 & -9 & -3 \\ -6 & -6 & 3 \end{array} \right ]\)
- \(\left [ \begin{array}{rrr} 5 & -18 & 5 \\ -11 & 4 & 4 \end{array} \right ]\)
- \(\left [ \begin{array}{rrr} -7 & 1 & 5 \end{array} \right ]\)
- \(\left [ \begin{array}{rr} 1 & 3 \\ 3 & 9 \end{array} \right ]\)
- \(\left [ \begin{array}{rrr} 13 & -16 & 1\\ -16 & 29 & -8 \\ 1 & -8 & 5 \end{array} \right ]\)
- \(\left [ \begin{array}{rrr} 5 & 7 & -1 \\ 5 & 15 & 5 \end{array} \right ]\)
- Not possible.
Exercise \(\PageIndex{28}\)
Let \(A\) be an \(n\times n\) matrix. Show \(A\) equals the sum of a symmetric and a skew symmetric matrix.
- Hint
-
Show that \(\frac{1}{2}\left( A^{T}+A\right)\) is symmetric and then consider using this as one of the matrices.
Exercise \(\PageIndex{29}\)
Show that the main diagonal of every skew symmetric matrix consists of only zeros. Recall that the main diagonal consists of every entry of the matrix which is of the form \(a_{ii}\).
- Answer
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If \(A\) is symmetric then \(A=-A^{T}.\) It follows that \(a_{ii}=-a_{ii}\) and so each \(a_{ii}=0\).
Exercise \(\PageIndex{30}\)
Prove 3 from Lemma 2.5.1 . That is, show that for an \(m \times n\) matrix \(A\), an \(m \times n\) matrix \(B\), and scalars \(r, s\), the following holds: \[\left( rA + sB \right) ^T = rA^{T} + sB^{T}\nonumber \]
2.6
Exercise \(\PageIndex{31}\)
Prove that \(I_{m}A=A\) where \(A\) is an \(m\times n\) matrix.
- Answer
-
\(\left( I_{m}A\right) _{ij}\equiv \sum_{j}\delta _{ik}A_{kj}=A_{ij}\)
Exercise \(\PageIndex{32}\)
Suppose \(AB=AC\) and \(A\) is an invertible \(n\times n\) matrix. Does it follow that \(B=C?\) Explain why or why not.
- Answer
-
Yes \(B=C\). Multiply \(AB = AC\) on the left by \(A^{-1}\).
Exercise \(\PageIndex{33}\)
Suppose \(AB=AC\) and \(A\) is a non invertible \(n\times n\) matrix. Does it follow that \(B=C\)? Explain why or why not.
Exercise \(\PageIndex{34}\)
Give an example of a matrix \(A\) such that \(A^{2}=I\) and yet \(A\neq I\) and \(A\neq -I.\)
- Answer
-
\(A = \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array} \right ]\)
2.7
Exercise \(\PageIndex{35}\)
Let \[A=\left [ \begin{array}{rr} 2 & 1 \\ -1 & 3 \end{array} \right ]\nonumber \] Find \(A^{-1}\) if possible. If \(A^{-1}\) does not exist, explain why.
- Answer
-
\(\left [ \begin{array}{rr} 2 & 1 \\ -1 & 3 \end{array} \right ]^{-1}= \left [ \begin{array}{rr} \frac{3}{7} & -\frac{1}{7} \\ \frac{1}{7} & \frac{2}{7} \end{array} \right ]\)
Exercise \(\PageIndex{36}\)
Let \[A=\left [ \begin{array}{rr} 0 & 1 \\ 5 & 3 \end{array} \right ]\nonumber \] Find \(A^{-1}\) if possible. If \(A^{-1}\) does not exist, explain why.
- Answer
-
\(\left [ \begin{array}{cc} 0 & 1 \\ 5 & 3 \end{array} \right ]^{-1}= \left [ \begin{array}{cc} -\frac{3}{5} & \frac{1}{5} \\ 1 & 0 \end{array} \right ]\)
Exercise \(\PageIndex{37}\)
Add exercises text here.Let \[A=\left [ \begin{array}{rr} 2 & 1 \\ 3 & 0 \end{array} \right ]\nonumber \] Find \(A^{-1}\) if possible. If \(A^{-1}\) does not exist, explain why.
- Answer
-
\(\left [ \begin{array}{cc} 2 & 1 \\ 3 & 0 \end{array} \right ]^{-1}= \left [ \begin{array}{cc} 0 & \frac{1}{3} \\ 1 & -\frac{2}{3} \end{array} \right ]\)
Exercise \(\PageIndex{38}\)
Let \[A=\left [ \begin{array}{rr} 2 & 1 \\ 4 & 2 \end{array} \right ]\nonumber \] Find \(A^{-1}\) if possible. If \(A^{-1}\) does not exist, explain why.
- Answer
-
\(\left [ \begin{array}{cc} 2 & 1 \\ 4 & 2 \end{array} \right ]^{-1}\) does not exist. The of this matrix is \(\left [ \begin{array}{cc} 1 & \frac{1}{2} \\ 0 & 0 \end{array} \right ]\)
Exercise \(\PageIndex{39}\)
Let \(A\) be a \(2\times 2\) invertible matrix, with \(A=\left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] .\) Find a formula for \(A^{-1}\) in terms of \(a,b,c,d\).
- Answer
-
\(\left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ]^{-1}= \left [ \begin{array}{cc} \frac{d}{ad-bc} & -\frac{b}{ad-bc} \\ -\frac{c}{ad-bc} & \frac{a}{ad-bc} \end{array} \right ]\)
Exercise \(\PageIndex{40}\)
Let \[A=\left [ \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 0 & 2 \end{array} \right ]\nonumber \] Find \(A^{-1}\) if possible. If \(A^{-1}\) does not exist, explain why.
- Answer
-
\(\left [ \begin{array}{ccc} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 0 & 2 \end{array} \right ]^{-1}= \left [ \begin{array}{rrr} -2 & 4 & -5 \\ 0 & 1 & -2 \\ 1 & -2 & 3 \end{array} \right ]\)
Exercise \(\PageIndex{41}\)
Let \[A=\left [ \begin{array}{rrr} 1 & 0 & 3 \\ 2 & 3 & 4 \\ 1 & 0 & 2 \end{array} \right ]\nonumber \] Find \(A^{-1}\) if possible. If \(A^{-1}\) does not exist, explain why.
- Answer
-
\(\left [ \begin{array}{ccc} 1 & 0 & 3 \\ 2 & 3 & 4 \\ 1 & 0 & 2 \end{array} \right ]^{-1}= \left [ \begin{array}{rrr} -2 & 0 & 3 \\ 0 & \frac{1}{3} & -\frac{2}{3} \\ 1 & 0 & -1 \end{array} \right ]\)
Exercise \(\PageIndex{42}\)
Let \[A=\left [ \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 4 & 5 & 10 \end{array} \right ]\nonumber \] Find \(A^{-1}\) if possible. If \(A^{-1}\) does not exist, explain why.
- Answer
-
The reduced row-echelon form is \(\left [ \begin{array}{ccc} 1 & 0 & \frac{5}{3} \\ 0 & 1 & \frac{2}{3} \\ 0 & 0 & 0 \end{array} \right ]\). There is no inverse.
Exercise \(\PageIndex{43}\)
Let \[A=\left [ \begin{array}{rrrr} 1 & 2 & 0 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 1 & -3 & 2 \\ 1 & 2 & 1 & 2 \end{array} \right ]\nonumber \] Find \(A^{-1}\) if possible. If \(A^{-1}\) does not exist, explain why.
- Answer
-
\(\left [ \begin{array}{rrrr} 1 & 2 & 0 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 1 & -3 & 2 \\ 1 & 2 & 1 & 2 \end{array} \right ]^{-1}= \left [ \begin{array}{rrrr} -1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ 3 & \frac{1}{2} & - \frac{1}{2} & - \frac{5}{2} \\ -1 & 0 & 0 & 1 \\ -2 & - \frac{3}{4} & \frac{1}{4} & \frac{9}{4} \end{array} \right ]\)
Exercise \(\PageIndex{44}\)
Using the inverse of the matrix, find the solution to the systems:
- \[\left [ \begin{array}{rr} 2 & 4 \\ 1 & 1 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{r} 1 \\ 2 \end{array} \right ]\nonumber \]
- \[\left [ \begin{array}{rr} 2 & 4 \\ 1 & 1 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{r} 2 \\ 0 \end{array} \right ]\nonumber \]
Now give the solution in terms of \(a\) and \(b\) to \[\left [ \begin{array}{rr} 2 & 4 \\ 1 & 1 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array}\right ] = \left [ \begin{array}{c} a \\ b \end{array} \right ]\nonumber \]
Exercise \(\PageIndex{45}\)
Using the inverse of the matrix, find the solution to the systems:
- \[\left [ \begin{array}{rrr} 1 & 0 & 3 \\ 2 & 3 & 4 \\ 1 & 0 & 2 \end{array} \right ] \left [ \begin{array}{c} x \\ y \\ z \end{array} \right ] =\left [ \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right ]\nonumber \]
- \[\left [ \begin{array}{rrr} 1 & 0 & 3 \\ 2 & 3 & 4 \\ 1 & 0 & 2 \end{array} \right ] \left [ \begin{array}{c} x \\ y \\ z \end{array} \right ] =\left [ \begin{array}{r} 3 \\ -1 \\ -2 \end{array} \right ]\nonumber \]
Now give the solution in terms of \(a,b,\) and \(c\) to the following: \[\left [ \begin{array}{rrr} 1 & 0 & 3 \\ 2 & 3 & 4 \\ 1 & 0 & 2 \end{array} \right ] \left [ \begin{array}{c} x \\ y \\ z \end{array} \right ] =\left [ \begin{array}{c} a \\ b \\ c \end{array} \right ]\nonumber \]
- Answer
-
- \(\left [ \begin{array}{c} x \\ y \\ z \end{array} \right ] =\left [ \begin{array}{c} 1 \\ -\frac{2}{3} \\ 0 \end{array} \right ]\)
- \(\left [ \begin{array}{c} x \\ y \\ z \end{array} \right ] = \left [ \begin{array}{r} -12 \\ 1 \\ 5 \end{array} \right ]\)
- \(\left [ \begin{array}{c} x \\ y \\ z \end{array} \right ] = \left [ \begin{array}{c} 3c-2a \\ \frac{1}{3}b-\frac{2}{3}c \\ a-c \end{array} \right ]\)
Exercise \(\PageIndex{46}\)
Show that if \(A\) is an \(n\times n\) invertible matrix and \(X\) is a \(n\times 1\) matrix such that \(AX=B\) for \(B\) an \(n\times 1\) matrix, then \(X=A^{-1}B\).
- Answer
-
Multiply both sides of \(AX=B\) on the left by \(A^{-1}\).
Exercise \(\PageIndex{47}\)
Prove that if \(A^{-1}\) exists and \(AX=0\) then \(X=0\).
- Answer
-
Multiply on both sides on the left by \(A^{-1}.\) Thus \[0=A^{-1}0=A^{-1}\left( AX\right) =\left( A^{-1}A\right) X=IX = X\nonumber \]
Exercise \(\PageIndex{49}\)
Show that if \(A\) is an invertible \(n\times n\) matrix, then so is \(A^{T}\) and \(\left( A^{T}\right) ^{-1}=\left( A^{-1}\right) ^{T}.\)
- Answer
-
You need to show that \(\left( A^{-1}\right) ^{T}\) acts like the inverse of \(A^{T}\) because from uniqueness in the above problem, this will imply it is the inverse. From properties of the transpose, \[\begin{aligned} A^{T}\left( A^{-1}\right) ^{T} &=\left( A^{-1}A\right) ^{T}=I^{T}=I \\ \left( A^{-1}\right) ^{T}A^{T} &=\left( AA^{-1}\right) ^{T}=I^{T}=I\end{aligned}\] Hence \(\left( A^{-1}\right) ^{T}=\left( A^{T}\right) ^{-1}\) and this last matrix exists.
Exercise \(\PageIndex{50}\)
Show \(\left( AB\right) ^{-1}=B^{-1}A^{-1}\) by verifying that \[AB\left( B^{-1}A^{-1}\right) =I\nonumber \] and \[B^{-1}A^{-1}\left( AB\right) =I\nonumber \] Hint: Use Problem \(\PageIndex{48}\) .
- Answer
-
\(\left( AB\right) B^{-1}A^{-1}=A\left( BB^{-1}\right) A^{-1}=AA^{-1}=I\) \(B^{-1}A^{-1}\left( AB\right) =B^{-1}\left( A^{-1}A\right) B=B^{-1}IB=B^{-1}B=I\)
Exercise \(\PageIndex{51}\)
Show that \(\left( ABC\right) ^{-1}=C^{-1}B^{-1}A^{-1}\) by verifying that \[\left( ABC\right) \left( C^{-1}B^{-1}A^{-1}\right) =I\nonumber \] and \[\left( C^{-1}B^{-1}A^{-1}\right)\left( ABC\right) =I\nonumber \] Hint: Use Problem \(\PageIndex{48}\) .
- Answer
-
The proof of this exercise follows from the previous one.
Exercise \(\PageIndex{52}\)
If \(A\) is invertible, show \(\left( A^{2}\right) ^{-1}=\left( A^{-1}\right) ^{2}.\) Hint: Use Problem \(\PageIndex{48}\) .
- Answer
-
\(A^{2}\left( A^{-1}\right) ^{2}=AAA^{-1}A^{-1}=AIA^{-1}=AA^{-1}=I\) \(\left( A^{-1}\right) ^{2}A^{2}=A^{-1}A^{-1}AA=A^{-1}IA=A^{-1}A=I\)
Exercise \(\PageIndex{53}\)
If \(A\) is invertible, show \(\left( A^{-1}\right) ^{-1}=A.\) Hint: Use Problem \(\PageIndex{48}\) .
- Answer
-
\(A^{-1}A=AA^{-1}=I\) and so by uniqueness, \(\left( A^{-1}\right) ^{-1}=A\).
2.8
Exercise \(\PageIndex{54}\)
Let \(A = \left [ \begin{array}{rr} 2 & 3 \\ 1 & 2 \end{array}\right ]\). Suppose a row operation is applied to \(A\) and the result is \(B = \left [ \begin{array}{rr} 1 & 2 \\ 2 & 3 \end{array}\right ]\). Find the elementary matrix \(E\) that represents this row operation.
Exercise \(\PageIndex{55}\)
Let \(A = \left [ \begin{array}{rr} 4 & 0 \\ 2 & 1 \end{array}\right ]\). Suppose a row operation is applied to \(A\) and the result is \(B = \left [ \begin{array}{rr} 8 & 0 \\ 2 & 1 \end{array}\right ]\). Find the elementary matrix \(E\) that represents this row operation.
Exercise \(\PageIndex{56}\)
Let \(A = \left [ \begin{array}{rr} 1 & -3 \\ 0 & 5 \end{array}\right ]\). Suppose a row operation is applied to \(A\) and the result is \(B = \left [ \begin{array}{rr} 1 & -3 \\ 2 & -1 \end{array}\right ]\). Find the elementary matrix \(E\) that represents this row operation.
Exercise \(\PageIndex{57}\)
Let \(A = \left [ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 5 & 1 \\ 2 & -1 & 4 \end{array}\right ]\). Suppose a row operation is applied to \(A\) and the result is \(B = \left [ \begin{array}{rrr} 1 & 2 & 1\\ 2 & -1 & 4 \\ 0 & 5 & 1 \end{array}\right ]\).
- Find the elementary matrix \(E\) such that \(EA = B\).
- Find the inverse of \(E\), \(E^{-1}\), such that \(E^{-1}B = A\).
Exercise \(\PageIndex{58}\)
Let \(A = \left [ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 5 & 1 \\ 2 & -1 & 4 \end{array}\right ]\). Suppose a row operation is applied to \(A\) and the result is \(B = \left [ \begin{array}{rrr} 1 & 2 & 1\\ 0 & 10 & 2 \\ 2 & -1 & 4 \end{array}\right ]\).
- Find the elementary matrix \(E\) such that \(EA = B\).
- Find the inverse of \(E\), \(E^{-1}\), such that \(E^{-1}B = A\).
Exercise \(\PageIndex{59}\)
Let \(A = \left [ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 5 & 1 \\ 2 & -1 & 4 \end{array}\right ]\). Suppose a row operation is applied to \(A\) and the result is \(B = \left [ \begin{array}{rrr} 1 & 2 & 1\\ 0 & 5 & 1 \\ 1 & -\frac{1}{2} & 2 \end{array}\right ]\).
- Find the elementary matrix \(E\) such that \(EA = B\).
- Find the inverse of \(E\), \(E^{-1}\), such that \(E^{-1}B = A\).
Exercise \(\PageIndex{60}\)
Let \(A = \left [ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 5 & 1 \\ 2 & -1 & 4 \end{array}\right ]\). Suppose a row operation is applied to \(A\) and the result is \(B = \left [ \begin{array}{rrr} 1 & 2 & 1\\ 2 & 4 & 5 \\ 2 & -1 & 4 \end{array}\right ]\).
- Find the elementary matrix \(E\) such that \(EA = B\).
- Find the inverse of \(E\), \(E^{-1}\), such that \(E^{-1}B = A\).
2.10
Exercise \(\PageIndex{61}\)
Find an \(LU\) factorization of \(\left [ \begin{array}{rrr} 1 & 2 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 3 \end{array} \right ] .\)
- Answer
-
\[\left [ \begin{array}{ccc} 1 & 2 & 0 \\ 2 & 1 & 3 \\ 1 & 2 & 3 \end{array} \right ] = \left [ \begin{array}{ccc} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 0 & 1 \end{array} \right ] \left [ \begin{array}{rrr} 1 & 2 & 0 \\ 0 & -3 & 3 \\ 0 & 0 & 3 \end{array} \right ]\nonumber \]
Exercise \(\PageIndex{62}\)
Find an \(LU\) factorization of \(\left [ \begin{array}{rrrr} 1 & 2 & 3 & 2 \\ 1 & 3 & 2 & 1 \\ 5 & 0 & 1 & 3 \end{array} \right ] .\)
- Answer
-
\[\left [ \begin{array}{cccc} 1 & 2 & 3 & 2 \\ 1 & 3 & 2 & 1 \\ 5 & 0 & 1 & 3 \end{array} \right ] = \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 5 & -10 & 1 \end{array} \right ] \left [ \begin{array}{rrrr} 1 & 2 & 3 & 2 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & -24 & -17 \end{array} \right ]\nonumber \]
Exercise \(\PageIndex{63}\)
Find an \(LU\) factorization of the matrix \(\left [ \begin{array}{rrrr} 1 & -2 & -5 & 0 \\ -2 & 5 & 11 & 3 \\ 3 & -6 & -15 & 1 \end{array} \right ] .\)
- Answer
-
\[\left [ \begin{array}{rrrr} 1 & -2 & -5 & 0 \\ -2 & 5 & 11 & 3 \\ 3 & -6 & -15 & 1 \end{array} \right ] = \left [ \begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 3 & 0 & 1 \end{array} \right ] \left [ \begin{array}{rrrr} 1 & -2 & -5 & 0 \\ 0 & 1 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{array} \right ]\nonumber \]
Exercise \(\PageIndex{64}\)
Find an \(LU\) factorization of the matrix \(\left [ \begin{array}{rrrr} 1 & -1 & -3 & -1 \\ -1 & 2 & 4 & 3 \\ 2 & -3 & -7 & -3 \end{array} \right ] .\)
- Answer
-
\[\left [ \begin{array}{rrrr} 1 & -1 & -3 & -1 \\ -1 & 2 & 4 & 3 \\ 2 & -3 & -7 & -3 \end{array} \right ] = \left [ \begin{array}{rrr} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 2 & -1 & 1 \end{array} \right ] \left [ \begin{array}{rrrr} 1 & -1 & -3 & -1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{array} \right ]\nonumber \]
Exercise \(\PageIndex{65}\)
Find an \(LU\) factorization of the matrix \( \ \ \left [ \begin{array}{rrrr} 1 & -3 & -4 & -3 \\ -3 & 10 & 10 & 10 \\ 1 & -6 & 2 & -5 \end{array} \right ] .\)
- Answer
-
\[\left [ \begin{array}{rrrr} 1 & -3 & -4 & -3 \\ -3 & 10 & 10 & 10 \\ 1 & -6 & 2 & -5 \end{array} \right ] = \left [ \begin{array}{rrr} 1 & 0 & 0 \\ -3 & 1 & 0 \\ 1 & -3 & 1 \end{array} \right ] \left [ \begin{array}{rrrr} 1 & -3 & -4 & -3 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 1 \end{array} \right ]\nonumber \]
Exercise \(\PageIndex{66}\)
Find an \(LU\) factorization of the matrix \(\left [ \begin{array}{rrrr} 1 & 3 & 1 & -1 \\ 3 & 10 & 8 & -1 \\ 2 & 5 & -3 & -3 \end{array} \right ] .\)
- Answer
-
\[\left [ \begin{array}{rrrr} 1 & 3 & 1 & -1 \\ 3 & 10 & 8 & -1 \\ 2 & 5 & -3 & -3 \end{array} \right ] = \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 2 & -1 & 1 \end{array} \right ] \left [ \begin{array}{rrrr} 1 & 3 & 1 & -1 \\ 0 & 1 & 5 & 2 \\ 0 & 0 & 0 & 1 \end{array} \right ]\nonumber \]
Exercise \(\PageIndex{67}\)
Find an \(LU\) factorization of the matrix \(\left [ \begin{array}{rrr} 3 & -2 & 1 \\ 9 & -8 & 6 \\ -6 & 2 & 2 \\ 3 & 2 & -7 \end{array} \right ] .\)
- Answer
-
\[\left [ \begin{array}{rrr} 3 & -2 & 1 \\ 9 & -8 & 6 \\ -6 & 2 & 2 \\ 3 & 2 & -7 \end{array} \right ] = \left [ \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 3 & 1 & 0 & 0 \\ -2 & 1 & 1 & 0 \\ 1 & -2 & -2 & 1 \end{array} \right ] \left [ \begin{array}{rrr} 3 & -2 & 1 \\ 0 & -2 & 3 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right ]\nonumber \]
Exercise \(\PageIndex{68}\)
Find an \(LU\) factorization of the matrix \(\left [ \begin{array}{rrr} -3 & -1 & 3 \\ 9 & 9 & -12 \\ 3 & 19 & -16 \\ 12 & 40 & -26 \end{array} \right ] .\)
Exercise \(\PageIndex{69}\)
Find an \(LU\) factorization of the matrix \(\left [ \begin{array}{rrr} -1 & -3 & -1 \\ 1 & 3 & 0 \\ 3 & 9 & 0 \\ 4 & 12 & 16 \end{array} \right ] .\)
- Answer
-
\[\left [ \begin{array}{rrr} -1 & -3 & -1 \\ 1 & 3 & 0 \\ 3 & 9 & 0 \\ 4 & 12 & 16 \end{array} \right ] = \left [ \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ -3 & 0 & 1 & 0 \\ -4 & 0 & -4 & 1 \end{array} \right ] \left [ \begin{array}{rrr} -1 & -3 & -1 \\ 0 & 0 & -1 \\ 0 & 0 & -3 \\ 0 & 0 & 0 \end{array} \right ]\nonumber \]
Exercise \(\PageIndex{70}\)
Find the \(LU\) factorization of the coefficient matrix using Dolittle’s method and use it to solve the system of equations. \[\begin{array}{c} x+2y=5 \\ 2x+3y=6 \end{array}\nonumber \]
- Answer
-
An \(LU\) factorization of the coefficient matrix is \[\left [ \begin{array}{cc} 1 & 2 \\ 2 & 3 \end{array} \right ] = \left [ \begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array} \right ] \left [ \begin{array}{cc} 1 & 2 \\ 0 & -1 \end{array} \right ]\nonumber \] First solve \[\left [ \begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array} \right ] \left [ \begin{array}{c} u \\ v \end{array} \right ] =\left [ \begin{array}{c} 5 \\ 6 \end{array} \right ]\nonumber \] which gives \(\left [ \begin{array}{c} u \\ v \end{array} \right ] =\) \(\left [ \begin{array}{r} 5 \\ -4 \end{array} \right ] .\) Then solve \[\left [ \begin{array}{rr} 1 & 2 \\ 0 & -1 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{r} 5 \\ -4 \end{array} \right ]\nonumber \] which says that \(y=4\) and \(x=-3.\)
Exercise \(\PageIndex{71}\)
Find the \(LU\) factorization of the coefficient matrix using Dolittle’s method and use it to solve the system of equations. \[\begin{array}{c} x+2y+z=1 \\ y+3z=2 \\ 2x+3y=6 \end{array}\nonumber \]
- Answer
-
An \(LU\) factorization of the coefficient matrix is \[\left [ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 2 & 3 & 0 \end{array} \right ] = \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & -1 & 1 \end{array} \right ] \left [ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array} \right ]\nonumber \] First solve \[\left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & -1 & 1 \end{array} \right ] \left [ \begin{array}{c} u \\ v \\ w \end{array} \right ] =\left [ \begin{array}{c} 1 \\ 2 \\ 6 \end{array} \right ]\nonumber \] which yields \(u=1,v=2,w=6\). Next solve \[\left [ \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array} \right ] \left [ \begin{array}{c} x \\ y \\ z \end{array} \right ] =\left [ \begin{array}{c} 1 \\ 2 \\ 6 \end{array} \right ]\nonumber \] This yields \(z=6,y=-16,x=27.\)
Exercise \(\PageIndex{72}\)
Find the \(LU\) factorization of the coefficient matrix using Dolittle’s method and use it to solve the system of equations. \[\begin{array}{c} x+2y+3z=5 \\ 2x+3y+z=6 \\ x-y+z=2 \end{array}\nonumber \]
Exercise \(\PageIndex{73}\)
Find the \(LU\) factorization of the coefficient matrix using Dolittle’s method and use it to solve the system of equations. \[\begin{array}{c} x+2y+3z=5 \\ 2x+3y+z=6 \\ 3x+5y+4z=11 \end{array}\nonumber \]
- Answer
-
An \(LU\) factorization of the coefficient matrix is \[\left [ \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 5 & 4 \end{array} \right ] = \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 1 & 1 \end{array} \right ] \left [ \begin{array}{rrr} 1 & 2 & 3 \\ 0 & -1 & -5 \\ 0 & 0 & 0 \end{array} \right ]\nonumber \] First solve \[\left [ \begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 1 & 1 \end{array} \right ] \left [ \begin{array}{c} u \\ v \\ w \end{array} \right ] =\left [ \begin{array}{c} 5 \\ 6 \\ 11 \end{array} \right ]\nonumber \] Solution is: \(\left [ \begin{array}{c} u \\ v \\ w \end{array} \right ] =\) \(\left [ \begin{array}{c} 5 \\ -4 \\ 0 \end{array} \right ] .\) Next solve \[\left [ \begin{array}{rrr} 1 & 2 & 3 \\ 0 & -1 & -5 \\ 0 & 0 & 0 \end{array} \right ] \left [ \begin{array}{c} x \\ y \\ z \end{array} \right ] =\left [ \begin{array}{c} 5 \\ -4 \\ 0 \end{array} \right ]\nonumber \] Solution is: \(\left [ \begin{array}{c} x \\ y \\ z \end{array} \right ] =\left [ \begin{array}{c} 7t-3 \\ 4-5t \\ t \end{array} \right ] ,t\in \mathbb{R}\).
Exercise \(\PageIndex{74}\)
Is there only one \(LU\) factorization for a given matrix? Hint: Consider the equation \[\left [ \begin{array}{rr} 0 & 1 \\ 0 & 1 \end{array} \right ] =\left [ \begin{array}{rr} 1 & 0 \\ 1 & 1 \end{array} \right ] \left [ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right ] .\nonumber \] Look for all possible \(LU\) factorizations.
- Answer
-
Sometimes there is more than one \(LU\) factorization as is the case in this example. The given equation clearly gives an \(LU\) factorization. However, it appears that the following equation gives another \(LU\) factorization. \[\left [ \begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array} \right ] =\left [ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right ] \left [ \begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array} \right ]\nonumber \]