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Chapter 2: Problem 19

Let \(A\) and \(B\) be \(k \times k\) matrices and let \\[ M=\left(\begin{array}{ll} O & B \\ A & O \end{array}\right) \\] Show that \(\operatorname{det}(M)=(-1)^{k} \operatorname{det}(A) \operatorname{det}(B)\).

Short Answer

Expert verified
To find the determinant of the block matrix \(M=\left(\begin{array}{ll} O & B \\\ A & O \end{array}\right)\), we can perform row operations to transform M into an upper triangular block matrix \(M'=\left(\begin{array}{ll} A & O \\\ O & B \end{array}\right)\). Since we only swapped k rows, the determinant of M is related to the determinant of M' by a factor of \((-1)^k\). Finally, we can compute the determinant of M' as the product of the determinants of its diagonal blocks, which results in the expression \(\operatorname{det}(M) = (-1)^k * \operatorname{det}(A) * \operatorname{det}(B)\).

Step by step solution

01

Write down matrix M

We are given matrix M in the form: \[ M=\left(\begin{array}{ll} O & B \\\ A & O \end{array}\right) \] Where A and B are k脳k matrices and O is a k脳k zero matrix.
02

Perform row operations on M to obtain M' (a transformed matrix)

Let's perform the following row operations to transform matrix M: 1. Swap the first k rows of M with its last k rows. 2. Perform any other necessary row operations to obtain an upper triangular block matrix. The resulting matrix M' after these operations will look like: \[ M'=\left(\begin{array}{ll} A & O \\\ O & B \end{array}\right) \]
03

Calculate the determinant of M'

The determinant of matrix M' is the product of the determinants of its diagonal blocks: \[ \operatorname{det}(M') = \operatorname{det}(A) * \operatorname{det}(B) \]
04

Relate the determinant of M' to the determinant of M

Since we only swapped rows in the process of transforming M to M', the determinant of M is related to the determinant of M' by a factor of (-1)^(number of row swaps). In this case, we swapped k rows, so the factor is (-1)^k: \[ \operatorname{det}(M) = (-1)^k * \operatorname{det}(M') \]
05

Combine results and conclude

Substituting the expression for det(M') from Step 3 into the equation above, we get: \[ \operatorname{det}(M) = (-1)^k * \operatorname{det}(A) * \operatorname{det}(B) \] This concludes the proof. The determinant of the given block matrix M is indeed equal to (-1)^k * det(A) * det(B).

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Most popular questions from this chapter

Let \(A\) and \(B\) be \(2 \times 2\) matrices. (a) \(\operatorname{Does} \operatorname{det}(A+B)=\operatorname{det}(A)+\operatorname{det}(B) ?\) (b) \(\operatorname{Does} \operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B) ?\) (c) \(\operatorname{Does} \operatorname{det}(A B)=\operatorname{det}(B A) ?\) Justify your answers.

Use Cramer's rule to solve each of the following systems: (a) \(\quad x_{1}+2 x_{2}=3\) (b) \(2 x_{1}+3 x_{2}=2\) \(3 x_{1}-x_{2}=1\) \(3 x_{1}+2 x_{2}=5\) (c) \(2 x_{1}+x_{2}-3 x_{3}=0\) \(4 x_{1}+5 x_{2}+x_{3}=8\) \(-2 x_{1}-x_{2}+4 x_{3}=2\) (d) \(\quad x_{1}+3 x_{2}+x_{3}=1\) \(2 x_{1}+x_{2}+x_{3}=5\) \(-2 x_{1}+2 x_{2}-x_{3}=-8\) (e) \(x_{1}+x_{2}\) \(=0\) \(x_{2}+x_{3}-2 x_{4}=1\) \(x_{1}+2 x_{3}+x_{4}=0\) \(x_{1}+x_{2}+x_{4}=0\)

Suppose that a \(3 \times 3\) matrix \(A\) factors into a product \\[ \left[\begin{array}{ccc} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{array}\right]\left[\begin{array}{ccc} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{array}\right] \\] Determine the value of det(A).

Let \(A\) be a nonsingular \(n \times n\) matrix with a nonzero cofactor \(A_{n n},\) and set \\[ c=\frac{\operatorname{det}(A)}{A_{n n}} \\] Show that if we subtract \(c\) from \(a_{n n}\), then the resulting matrix will be singular.

Use determinants to determine whether the following \(2 \times 2\) matrices are nonsingular: (a) \(\left(\begin{array}{ll}3 & 5 \\ 2 & 4\end{array}\right)\) (b) \(\left(\begin{array}{ll}3 & 6 \\ 2 & 4\end{array}\right)\) (c) \(\left(\begin{array}{rr}3 & -6 \\ 2 & 4\end{array}\right)\)
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