Expand: \[\left| \begin{matrix}
1 & -7 & 3 \\
5 & -6 & 0 \\
1 & 2 & -3 \\
\end{matrix} \right|\]
Answer
364.8k+ views
Hint: To find the value of a given matrix, use the formula for expansion of value of the matrix and substitute the values to get the value of expansion.
We have the matrix\[\left| \begin{matrix}
1 & -7 & 3 \\
5 & -6 & 0 \\
1 & 2 & -3 \\
\end{matrix} \right|\]. We observe that this is a \[3\times 3\]matrix.
Matrix is a rectangular array of numbers, symbols or expressions, arranged in rows and columns. Provided that two matrices have the same size (each matrix has the same number of rows and same number of columns as the other), two matrices can be added or subtracted element by element. The rule for matrix multiplication is that two matrices can be multiplied only when the number of columns in the first matrix equals the number of rows in the second one. Also, matrix multiplication is not commutative.
Any matrix is represented as \[A={{\left[ {{a}_{ij}} \right]}_{m\times n}}\] which has \[m\] rows and \[n\] columns.
We know that the formula for expansion of the matrix \[\left| \begin{matrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{matrix} \right|\]is\[a\left( ei-hf \right)-b\left( di-fg \right)+c\left( dh-ge \right)\].
Substituting the value \[a=1,b=-7,c=3,d=5,e=-6,f=0,g=1,h=2,i=-3\], we have \[\left| \begin{matrix}
1 & -7 & 3 \\
5 & -6 & 0 \\
1 & 2 & -3 \\
\end{matrix} \right|=1\left[ \left( -6 \right)\left( -3 \right)-\left( 0 \right)\left( 2 \right) \right]-\left( -7 \right)\left[ \left( 5 \right)\left( -3 \right)-\left( 0 \right)\left( 1 \right) \right]+3\left[ \left( 5 \right)\left( 2 \right)-\left( 1 \right)\left( -6 \right) \right]\].
Simplifying the above equation, we have\[\left| \begin{matrix}
1 & -7 & 3 \\
5 & -6 & 0 \\
1 & 2 & -3 \\
\end{matrix} \right|=1\left( 18-0 \right)+7\left( -15-0 \right)+3\left( 10+6 \right)=18-105+48=-39\].
Thus, we have the value of the given matrix as\[-39\].
This method of matrix expansion of matrix is called cofactor expansion. It is an expression for the weighted sum of the determinants of the sub matrices of the given matrix. For a matrix of the form \[A={{\left[ {{a}_{ij}} \right]}_{n\times n}}\], we define \[{{C}_{ij}}={{\left( -1 \right)}^{i+j}}{{M}_{ij}}\] as the cofactor of matrix \[A\] where \[{{M}_{ij}}\] is the \[i,j\] minor of \[A\] .
We define the determinant of \[A\] as \[\left| A \right|={{a}_{i1}}{{C}_{i1}}+{{a}_{i2}}{{C}_{i2}}+...+{{a}_{in}}{{C}_{in}}={{a}_{1i}}{{C}_{1i}}+{{a}_{2i}}{{C}_{2i}}+...{{a}_{ni}}{{C}_{ni}}\]. Thus, we observe that we can expand a matrix in multiple ways along its various rows and columns.
Note: We must keep in mind that we can expand any matrix along various rows and columns. But we will get the same value of the matrix by expanding along any row or column.
We have the matrix\[\left| \begin{matrix}
1 & -7 & 3 \\
5 & -6 & 0 \\
1 & 2 & -3 \\
\end{matrix} \right|\]. We observe that this is a \[3\times 3\]matrix.
Matrix is a rectangular array of numbers, symbols or expressions, arranged in rows and columns. Provided that two matrices have the same size (each matrix has the same number of rows and same number of columns as the other), two matrices can be added or subtracted element by element. The rule for matrix multiplication is that two matrices can be multiplied only when the number of columns in the first matrix equals the number of rows in the second one. Also, matrix multiplication is not commutative.
Any matrix is represented as \[A={{\left[ {{a}_{ij}} \right]}_{m\times n}}\] which has \[m\] rows and \[n\] columns.
We know that the formula for expansion of the matrix \[\left| \begin{matrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{matrix} \right|\]is\[a\left( ei-hf \right)-b\left( di-fg \right)+c\left( dh-ge \right)\].
Substituting the value \[a=1,b=-7,c=3,d=5,e=-6,f=0,g=1,h=2,i=-3\], we have \[\left| \begin{matrix}
1 & -7 & 3 \\
5 & -6 & 0 \\
1 & 2 & -3 \\
\end{matrix} \right|=1\left[ \left( -6 \right)\left( -3 \right)-\left( 0 \right)\left( 2 \right) \right]-\left( -7 \right)\left[ \left( 5 \right)\left( -3 \right)-\left( 0 \right)\left( 1 \right) \right]+3\left[ \left( 5 \right)\left( 2 \right)-\left( 1 \right)\left( -6 \right) \right]\].
Simplifying the above equation, we have\[\left| \begin{matrix}
1 & -7 & 3 \\
5 & -6 & 0 \\
1 & 2 & -3 \\
\end{matrix} \right|=1\left( 18-0 \right)+7\left( -15-0 \right)+3\left( 10+6 \right)=18-105+48=-39\].
Thus, we have the value of the given matrix as\[-39\].
This method of matrix expansion of matrix is called cofactor expansion. It is an expression for the weighted sum of the determinants of the sub matrices of the given matrix. For a matrix of the form \[A={{\left[ {{a}_{ij}} \right]}_{n\times n}}\], we define \[{{C}_{ij}}={{\left( -1 \right)}^{i+j}}{{M}_{ij}}\] as the cofactor of matrix \[A\] where \[{{M}_{ij}}\] is the \[i,j\] minor of \[A\] .
We define the determinant of \[A\] as \[\left| A \right|={{a}_{i1}}{{C}_{i1}}+{{a}_{i2}}{{C}_{i2}}+...+{{a}_{in}}{{C}_{in}}={{a}_{1i}}{{C}_{1i}}+{{a}_{2i}}{{C}_{2i}}+...{{a}_{ni}}{{C}_{ni}}\]. Thus, we observe that we can expand a matrix in multiple ways along its various rows and columns.
Note: We must keep in mind that we can expand any matrix along various rows and columns. But we will get the same value of the matrix by expanding along any row or column.
Last updated date: 28th Sep 2023
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Total views: 364.8k
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