Exercise 03 solutions
🔹 must know, 🔸 ideally should know, ⛑ challenging, 🔫 very challenging
🔹 3.2.23
In Exercises 23, use either elementary row or column operations, or cofactor expansion, to find the determinant by hand. Then use a software program or a graphing utility to verify your answer.
Solution (based on official solution)
First, we see that for the given matrix, there are no specific rules/formulas that could help us find its determinant. Therefore we apply the general approach and since second column is full of zeros except from one value, we can do co-factor expansion in the second column, therefore:
\[\left|\begin{array}{rrrr} 5 & 1 & 0 & 1 \\ 1 & 0 & -1 & -1 \\ 2 & 0 & 1 & 2 \\ -1 & 0 & 3 & 1 \end{array}\right|=-(1)\left|\begin{array}{rrr} 1 & -1 & -1 \\ 2 & 1 & 2 \\ -1 & 3 & 1 \end{array}\right|\]Notice that we multiply the \(3 \times 3\) matrix by \(-1\) which is the sign we get from the pattern matrix. Now, we need to find the determinant of the newly obtained \(3\times 3\) matrix. For this, we can use the special rule that adds first and second column as fourth and fifth columns and then we use sum/subtraction of product of diagonals to obtain the determinant. Therefore:
\[-(1)\left|\begin{array}{rrr} 1 & -1 & -1 \\ 2 & 1 & 2 \\ -1 & 3 & 1 \end{array}\right|=-(-8)=8\]To verify, you can use for instance numpy.linalg.det function.
🔹 3.2.25
In exercise 25, use elementary row or column operations to find the determinant.
Solution (based on the official solution)
This problem is very similar to the previous, but we are asked to solve it using different strategy. Essentially, our goal will be to reduce the original matrix into the triangular form and then use the special rule that the determinant can be obtained through the product of the diagonal terms. We have to remember that every time we transform matrix through some elementary row operation, there can be associated change to its determinant. Let’s start by \(R_2 - R_1 \Rightarrow R_2\):
For this operation, it applies that \(det(B) = det(A)\). We continue with \(R_3 - 4R_1 \Rightarrow R_3\) :
\[=\begin{aligned} \left|\begin{array}{rrr} 1 & 7 & -3 \\ 0 & -4 & 4 \\ 4 & 8 & 1 \end{array}\right| &=\left|\begin{array}{rrr} 1 & 7 & -3 \\ 0 & -4 & 4 \\ 0 & -20 & 13 \end{array}\right| \\ \end{aligned}\]Finally, we obtain the triangular matrix via \(R_3 - 5R_2 \Rightarrow R_3\):
\[=\left|\begin{array}{rrr} 1 & 7 & -3 \\ 0 & -4 & 4 \\ 0 & 0 & -7 \end{array}\right|=1(-4)(-7)=28\]⛑ 3.2.43
Prove the following property:
\[\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|+\left|\begin{array}{lll} b_{11} & a_{12} & a_{13} \\ b_{21} & a_{22} & a_{23} \\ b_{31} & a_{32} & a_{33} \end{array}\right|=\left|\begin{array}{lll} \left(a_{11}+b_{11}\right) & a_{12} & a_{13} \\ \left(a_{21}+b_{21}\right) & a_{22} & a_{23} \\ \left(a_{31}+b_{31}\right) & a_{32} & a_{33} \end{array}\right|\]Solution (based on the official solution)
If we look at the two determinant, we notice that they differ only in their first columns. Therefore, let’s do column expansion for both of these determinants in the first column:
\[=a_{11}\left|\begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right|-a_{21}\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array}\right|+a_{31}\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{22} & a_{23} \end{array}\right|+b_{11}\left|\begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right|-b_{21}\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array}\right|+b_{31}\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{22} & a_{23} \end{array}\right|\]Next, we can factor out common terms as follows:
\[=\left(a_{11}+b_{11}\right)\left|\begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right|-\left(a_{21}+b_{21}\right)\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array}\right|+\left(a_{31}+b_{31}\right)\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{22} & a_{23} \end{array}\right|\]Finally, we can convert this back to just one determinant:
\[=\left|\begin{array}{lll} \left(a_{11}+b_{11}\right) & a_{12} & a_{13} \\ \left(a_{21}+b_{21}\right) & a_{22} & a_{23} \\ \left(a_{31}+b_{31}\right) & a_{32} & a_{33} \end{array}\right|\]which proves the property. If it is hard for you to still understand what is going on, try to submit some concrete numbers instead of the algebraic terms.
🔫 3.2.47
Prove that \(det(B) = det(A)\) using the suggested steps.
Solution
See the official solution which actually has a nice explanation.
🔹 3.4.1
In this exercises, find the adjoint of the matrix \(A\). Then use the adjoint to find the inverse of \(A\) (if possible).
\[A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\]Solution (based on the official solution)
First, we need to find the adjoint of matrix \(A\), recall that adjoint is transposed co-factor matrix. Since, we have \(2 \times 2\) matrix, then to get the co-factor for given position, we simply take the value across diagonal and then multiply it by corresponding sign. Therefore:
Then we do the transpose to get the adjoint of \(A\):
\[\operatorname{adj}(A)= C^T = \left[\begin{array}{rr} 4 & -3 \\ -2 & 1 \end{array}\right]^T=\left[\begin{array}{rr} 4 & -2 \\ -3 & 1 \end{array}\right]\]Now, we can use the adjoint to find the inverse of \(A\) using the following formula:
\[A^{-1}=\frac{1}{\operatorname{det}(A)} \operatorname{adj}(A)\]As you can see, we still need to find the inverse o \(A\). This should be easy since there is a close formula:
\[det(A) = 4 \times 1 - (-3) \times (-2) = -2\]Since \(det(A)\) in non-zero, we know that the inverse of \(A\) exists. So, now we plugin values to the formula:
\[A^{-1}=\frac{1}{|A|} \operatorname{adj}(A)=-\frac{1}{2}\left[\begin{array}{rr} 4 & -2 \\ -3 & 1 \end{array}\right]=\left[\begin{array}{rr} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{array}\right]\]By the way, you have already seen this formula last week session. (see the section Inverse of a matrix).
🔸 3.4.5
In this exercises, find the adjoint of the matrix \(A\). Then use the adjoint to find the inverse of \(A\) (if possible).
\[A=\left[\begin{array}{rrr} -3 & -5 & -7 \\ 2 & 4 & 3 \\ 0 & 1 & -1 \end{array}\right]\]Solution (based on the official solution)
This time, to find the co-factor matrix is quite cumbersome compare to the previous exercise, so:
\[C = \left[\begin{array}{rrr} -7 & 2 & 2 \\ -12 & 3 & 3 \\ 13 & -5 & -2 \end{array}\right]\]Therefore, adjoint is:
\[\operatorname{adj}(A)= C^T = \left[\begin{array}{rrr} -7 & -12 & 13 \\ 2 & 3 & -5 \\ 2 & 3 & -2 \end{array}\right]\]To find the determinant, we notice it is a \(3 \times 3\) matrix and therefore, we can append 1st and 2nd column as 4th and 5th columns and then get add/subtract together product of the diagonal elements which yields \(det(A) = -3\). Finally, we can write:
\[A^{-1}=\frac{1}{|A|} \operatorname{adj}(A)=-\frac{1}{3}\left[\begin{array}{rrr} -7 & -12 & 13 \\ 2 & 3 & -5 \\ 2 & 3 & -2 \end{array}\right]=\left[\begin{array}{rrr} \frac{7}{3} & 4 & -\frac{13}{3} \\ -\frac{2}{3} & -1 & \frac{5}{3} \\ -\frac{2}{3} & -1 & \frac{2}{3} \end{array}\right]\]🔹 3.4.9
In this exercise, use Cramer’s Rule to solve (if possible) the system of linear equations:
\[\begin{array}{r} x_1+2 x_2=5 \\ -x_1+x_2=1 \end{array}\]Solution (based on the official solution)
First, need to transform the given system into a coefficient matrix:
\[A=\left[\begin{array}{rr} 1 & 2 \\ -1 & 1 \end{array}\right]\]Then, we need to find determinant of this matrix using closed formula we get \(det(A) = 3\). Since the determinant is non-zero, we know that the system has precisely one unique solution. (see notes - Equivalent Conditions for a Nonsingular Matrix) Otherwise, we would not be able to use Cramer’s rule. Next, we also need to find determinants of matrices \(A_1, A_2\) where their corresponding columns get replaces by constants column vector, therefore:
\[\begin{aligned} &A_1=\left[\begin{array}{ll} 5 & 2 \\ 1 & 1 \end{array}\right], \quad\left|A_1\right|=3\\ &A_2=\left[\begin{array}{rr} 1 & 5 \\ -1 & 1 \end{array}\right], \quad\left|A_2\right|=6 \end{aligned}\]Now, we can plug these results into Cramer's formula and the solution for \(x_1, x_2\):
🔹 Review exercise 3.43
Let \(A\) and \(B\) be square matrices of order 4 such that \(det(A)=4\) and \(det(B)=2\).
Find:
(a) \(det(BA)\)
(b) \(det(B^2)\)
(c) \(det(2A)\)
(d) \(det((AB)^T)\)
(e) \(det(B^{-1})\)
Solution
We can solve this exercise using the algebraic rules:
(a) Product rule: \(det(AB) = det(A)det(B) = 8\)
(b) Product rule: \(det(B^2) = det(B)det(B) = 4\)
(c) Constant rule: \(det(2A) = 2^4 det(A) = 16 \times 4 = 64\)
(d) Transpose rule: \(det((AB)^T) = det(B^T A^T) = det(B^T) det(A^T) = det(B) det(A) = 8\)
(e) Inverse rule: \(det(B^{-1}) = \frac{1}{det(B)} = \frac{1}{2}\)
🔍 About the page
- Last updated: 14/09/2022
- Unless othewise stated, exercises come from the book:
Elementary Linear Algebra, International Metric Edition, Ron Larson - Purpose: This page was created as part of a preparation for my teaching assistant session in the course Linear algebra and optimization managed by
Rasmus Ejlers Møgelberg. Please note that unless othewise stated, the solutions on this page are my own, thus it should not be considered as official solutions provided as part of the course.