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Question: Let a matrix $A$ is given by $\begin{bmatrix} a & b & c \\ p & q & r \\ x & y & z \end{bmatrix}$ and...

Let a matrix AA is given by [abcpqrxyz]\begin{bmatrix} a & b & c \\ p & q & r \\ x & y & z \end{bmatrix} and suppose that det⁡(A)=6\det(A) = 6.

If B=[p+xq+yr+za+xb+yc+za+pb+qc+r]B = \begin{bmatrix} p+x & q+y & r+z \\ a+x & b+y & c+z \\ a+p & b+q & c+r \end{bmatrix} then

A

det⁡(B)=6\det(B) = 6

B

det⁡(B)=−6\det(B) = -6

C

det⁡(B)=12\det(B) = 12

D

det⁡(B)=−12\det(B) = -12

Answer

12

Explanation

Solution

The problem requires us to find the determinant of matrix BB given the determinant of matrix AA.

Let the given matrix AA be: A=[abcpqrxyz]A = \begin{bmatrix} a & b & c \\ p & q & r \\ x & y & z \end{bmatrix} We are given that det⁡(A)=6\det(A) = 6.

Let's denote the rows of matrix AA as row vectors: R1=[abc]R_1 = \begin{bmatrix} a & b & c \end{bmatrix} R2=[pqr]R_2 = \begin{bmatrix} p & q & r \end{bmatrix} R3=[xyz]R_3 = \begin{bmatrix} x & y & z \end{bmatrix}

So, A=[R1R2R3]A = \begin{bmatrix} R_1 \\ R_2 \\ R_3 \end{bmatrix}, and det⁡(A)=det⁡(R1,R2,R3)=6\det(A) = \det(R_1, R_2, R_3) = 6.

Now, let's write matrix BB in terms of these row vectors: B=[p+xq+yr+za+xb+yc+za+pb+qc+r]B = \begin{bmatrix} p+x & q+y & r+z \\ a+x & b+y & c+z \\ a+p & b+q & c+r \end{bmatrix} The rows of BB are: R1′=[p+xq+yr+z]=R2+R3R'_1 = \begin{bmatrix} p+x & q+y & r+z \end{bmatrix} = R_2 + R_3 R2′=[a+xb+yc+z]=R1+R3R'_2 = \begin{bmatrix} a+x & b+y & c+z \end{bmatrix} = R_1 + R_3 R3′=[a+pb+qc+r]=R1+R2R'_3 = \begin{bmatrix} a+p & b+q & c+r \end{bmatrix} = R_1 + R_2

So, B=[R2+R3R1+R3R1+R2]B = \begin{bmatrix} R_2 + R_3 \\ R_1 + R_3 \\ R_1 + R_2 \end{bmatrix}.

We need to calculate det⁡(B)\det(B). We will use properties of determinants involving row operations:

  1. If a row is multiplied by a scalar kk, the determinant is multiplied by kk.
  2. If a multiple of one row is added to another row, the determinant remains unchanged.
  3. If two rows are interchanged, the sign of the determinant changes.

Let's apply a sequence of row operations to BB to transform it into AA:

  1. Operation: R1′→R1′+R2′+R3′R'_1 \rightarrow R'_1 + R'_2 + R'_3 (Add all rows to the first row) The determinant remains unchanged. det⁡(B)=det⁡((R2+R3)+(R1+R3)+(R1+R2)R1+R3R1+R2)\det(B) = \det \begin{pmatrix} (R_2 + R_3) + (R_1 + R_3) + (R_1 + R_2) \\ R_1 + R_3 \\ R_1 + R_2 \end{pmatrix} det⁡(B)=det⁡(2(R1+R2+R3)R1+R3R1+R2)\det(B) = \det \begin{pmatrix} 2(R_1 + R_2 + R_3) \\ R_1 + R_3 \\ R_1 + R_2 \end{pmatrix}

  2. Operation: Factor out 2 from the first row. det⁡(B)=2det⁡(R1+R2+R3R1+R3R1+R2)\det(B) = 2 \det \begin{pmatrix} R_1 + R_2 + R_3 \\ R_1 + R_3 \\ R_1 + R_2 \end{pmatrix}

  3. Operation: R2′→R2′−R1′R'_2 \rightarrow R'_2 - R'_1 (Subtract the current first row from the second row) The determinant remains unchanged. det⁡(B)=2det⁡(R1+R2+R3(R1+R3)−(R1+R2+R3)R1+R2)\det(B) = 2 \det \begin{pmatrix} R_1 + R_2 + R_3 \\ (R_1 + R_3) - (R_1 + R_2 + R_3) \\ R_1 + R_2 \end{pmatrix} det⁡(B)=2det⁡(R1+R2+R3−R2R1+R2)\det(B) = 2 \det \begin{pmatrix} R_1 + R_2 + R_3 \\ -R_2 \\ R_1 + R_2 \end{pmatrix}

  4. Operation: R3′→R3′−R1′R'_3 \rightarrow R'_3 - R'_1 (Subtract the current first row from the third row) The determinant remains unchanged. det⁡(B)=2det⁡(R1+R2+R3−R2(R1+R2)−(R1+R2+R3))\det(B) = 2 \det \begin{pmatrix} R_1 + R_2 + R_3 \\ -R_2 \\ (R_1 + R_2) - (R_1 + R_2 + R_3) \end{pmatrix} det⁡(B)=2det⁡(R1+R2+R3−R2−R3)\det(B) = 2 \det \begin{pmatrix} R_1 + R_2 + R_3 \\ -R_2 \\ -R_3 \end{pmatrix}

  5. Operation: Factor out -1 from the second row and -1 from the third row. det⁡(B)=2×(−1)×(−1)det⁡(R1+R2+R3R2R3)\det(B) = 2 \times (-1) \times (-1) \det \begin{pmatrix} R_1 + R_2 + R_3 \\ R_2 \\ R_3 \end{pmatrix} det⁡(B)=2det⁡(R1+R2+R3R2R3)\det(B) = 2 \det \begin{pmatrix} R_1 + R_2 + R_3 \\ R_2 \\ R_3 \end{pmatrix}

  6. Operation: R1′→R1′−R2′−R3′R'_1 \rightarrow R'_1 - R'_2 - R'_3 (Subtract the current second and third rows from the first row) The determinant remains unchanged. det⁡(B)=2det⁡((R1+R2+R3)−R2−R3R2R3)\det(B) = 2 \det \begin{pmatrix} (R_1 + R_2 + R_3) - R_2 - R_3 \\ R_2 \\ R_3 \end{pmatrix} det⁡(B)=2det⁡(R1R2R3)\det(B) = 2 \det \begin{pmatrix} R_1 \\ R_2 \\ R_3 \end{pmatrix}

We recognize that det⁡(R1R2R3)\det \begin{pmatrix} R_1 \\ R_2 \\ R_3 \end{pmatrix} is simply det⁡(A)\det(A). Given det⁡(A)=6\det(A) = 6. Therefore, det⁡(B)=2×det⁡(A)=2×6=12\det(B) = 2 \times \det(A) = 2 \times 6 = 12.