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Compute the determinant in Exercise 3 using a cofactor expansion across the first row. Also, compute the determinant by a cofactor expansion down the second column.

3. \(\left| {\begin{aligned}{*{20}{c}}{\bf{2}}&{ - {\bf{2}}}&{\bf{3}}\\{\bf{3}}&{\bf{1}}&{\bf{2}}\\{\bf{1}}&{\bf{3}}&{ - {\bf{1}}}\end{aligned}} \right|\)

Short Answer

Expert verified

Thus, \(\left| {\begin{aligned}{*{20}{c}}2&{ - 2}&3\\3&1&2\\1&3&{ - 1}\end{aligned}} \right| = 0\).

Step by step solution

01

Write the determinant formula

The determinant computed by acofactor expansion across the ith row is

\(\det A = {a_{i1}}{C_{i1}} + {a_{i2}}{C_{i2}} + \cdots + {a_{in}}{C_{in}}\).

The determinant computed by a cofactor expansion down the jth column is

\(\det A = {a_{1j}}{C_{1j}} + {a_{2j}}{C_{2j}} + \cdots + {a_{nj}}{C_{nj}}\).

Here, A is an \(n \times n\) matrix, and \({C_{ij}} = {\left( { - 1} \right)^{i + j}}{A_{ij}}\).

02

Use the cofactor expansion across the first row

\(\begin{aligned}{c}\left| {\begin{aligned}{*{20}{c}}2&{ - 2}&3\\3&1&2\\1&3&{ - 1}\end{aligned}} \right| = {a_{11}}{C_{11}} + {a_{12}}{C_{12}} + {a_{13}}{C_{13}}\\ = {a_{11}}{\left( { - 1} \right)^{1 + 1}}\det {A_{11}} + {a_{12}}{\left( { - 1} \right)^{1 + 2}}\det {A_{12}} + {a_{13}}{\left( { - 1} \right)^{1 + 3}}\det {A_{13}}\\ = 2\left| {\begin{aligned}{*{20}{c}}1&2\\3&{ - 1}\end{aligned}} \right| - \left( { - 2} \right)\left| {\begin{aligned}{*{20}{c}}3&2\\1&{ - 1}\end{aligned}} \right| + 3\left| {\begin{aligned}{*{20}{c}}3&1\\1&3\end{aligned}} \right|\\ = 2\left( { - 7} \right) + 2\left( { - 5} \right) + 3\left( 8 \right)\\ = - 14 - 10 + 24\\ = 0\end{aligned}\)

03

Use the cofactor expansion down the second column

\(\begin{aligned}{c}\left| {\begin{aligned}{*{20}{c}}2&{ - 2}&3\\3&1&2\\1&3&{ - 1}\end{aligned}} \right| = {a_{12}}{C_{12}} + {a_{22}}{C_{22}} + {a_{32}}{C_{32}}\\ = {a_{12}}{\left( { - 1} \right)^{1 + 2}}\det {A_{12}} + {a_{22}}{\left( { - 1} \right)^{2 + 2}}\det {A_{22}} + {a_{32}}{\left( { - 1} \right)^{3 + 2}}\det {A_{32}}\\ = - \left( { - 2} \right)\left| {\begin{aligned}{*{20}{c}}3&2\\1&{ - 1}\end{aligned}} \right| + 1\left| {\begin{aligned}{*{20}{c}}2&3\\1&{ - 1}\end{aligned}} \right| - 3\left| {\begin{aligned}{*{20}{c}}2&3\\3&2\end{aligned}} \right|\\ = 2\left( { - 5} \right) + 1\left( { - 5} \right) - 3\left( { - 5} \right)\\ = - 10 - 5 + 15\\ = 0\end{aligned}\)

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Most popular questions from this chapter

(M) Read the documentation for your matrix program, and write the commands that will produce the following matrices (without keying in each entry of the matrix).

  1. A \({\bf{5}} \times {\bf{6}}\) matrix of zeros
  2. A \({\bf{3}} \times {\bf{5}}\) matrix of ones
  3. The \({\bf{6}} \times {\bf{6}}\) identity matrix
  4. A \({\bf{5}} \times {\bf{5}}\) diagonal matrix, with diagonal entries 3, 5, 7, 2, 4

Let \(A = \left[ {\begin{array}{*{20}{c}}B&{\bf{0}}\\{\bf{0}}&C\end{array}} \right]\), where \(B\) and \(C\) are square. Show that \(A\)is invertible if an only if both \(B\) and \(C\) are invertible.

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{\bf{5}}\\{ - {\bf{3}}}&{\bf{1}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{5}}}\\{\bf{3}}&k\end{aligned}} \right)\). What value(s) of \(k\), if any will make \(AB = BA\)?

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

7. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&Z\\{\bf{0}}&{\bf{0}}\\B&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

Suppose the first two columns, \({{\bf{b}}_1}\) and \({{\bf{b}}_2}\), of Bare equal. What can you say about the columns of AB(if ABis defined)? Why?

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