The Determinant - Linear Algebra

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Calculate the determinant of matrix A where,

$A=\begin{bmatrix} -5&7 &10 \ -20&15 &2 \ 0& 3& 8 \end{bmatrix}$

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Calculating the determinant of a 3x3 matrix is more difficult than a 2x2 matrix. To calculate the determinant of a 3x3 matrix, we use the following $\left | A\right $|=a_{11}$$(a_{22}$$a_{33}$$-a_{23}$$a_{32}$$)-a_{12}$$(a_{21}$$a_{33}$$-a_{23}$$a_{31}$$)+a_{13}$$(a_{21}$$a_{32}$$-a_{22}$$a_{31}$)$

Calculate the determinant of matrix A where

$A = \begin{bmatrix} 5&4 \ 1&2 \ 0&7 \end{bmatrix}$

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The matrix must be square to calculate its determinant, therefore, it is not possible to calculate the determinant for this matrix.

Calculate the determinant of matrix A where,

$A=\begin{bmatrix} 4&1 \ 5& 3 \end{bmatrix}$

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To calculate the determinant of a 2x2 matrix, we can use the equation $a_${^{11}$}a_${^{22}$}-a_${^{12}$}a_${^{21}$}$

Calculate the determinant of matrix A where,

$A=\begin{bmatrix} 8&-8 \ 51&12 \end{bmatrix}$

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To calculate the determinant of a 2x2 matrix, we can use the equation $a_${^{11}$}a_${^{22}$}-a_${^{12}$}a_${^{21}$}$

Calculate the determinant of matrix A where,

$A=\begin{bmatrix} 8&6 \ 0&2 \end{bmatrix}$

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To calculate the determinant of a 2x2 matrix, we can use the equation $a_${^{11}$}a_${^{22}$}-a_${^{12}$}a_${^{21}$}$

Calculate the determinant of matrix A where,

$A=\begin{bmatrix} 2&5 &6 \ 2&1 &0 \ 5&3 &4 \end{bmatrix}$

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Calculating the determinant of a 3x3 matrix is more difficult than a 2x2 matrix. To calculate the determinant of a 3x3 matrix, we use the following $\left | A\right $|=a_{11}$$(a_{22}$$a_{33}$$-a_{23}$$a_{32}$$)-a_{12}$$(a_{21}$$a_{33}$$-a_{23}$$a_{31}$$)+a_{13}$$(a_{21}$$a_{32}$$-a_{22}$$a_{31}$)$

Calculate the determinant of matrix A.

$A =\begin{bmatrix} 5&3 \ 6&4 \end{bmatrix}$

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In order to find the determinant of a 2x2 matrix, compute $det A= $a_{11}$$a_{22}$$-a_{12}$$a_{21}$$ :

Calculate the determinant of matrix A.

$A=\begin{bmatrix} 2&5 &4 &7 &1 \ 8& 6&11 & 3&9 \4 & 1&3 &0 &6 \ 7& 0&2 &9 &1 \end{bmatrix}$

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It is not possible to calculate the determinant of this matrix because only square matrices (nxn) have determinants.

Calculate the determinant of matrix A.

$A=\begin{bmatrix} 1&5 &3 \ 0&-2 &7 \ 0&0 &-9 \end{bmatrix}$

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To find the determinant of a lower or upper triangular matrix, simply find the product of the diagonal entries of matrix A:

$\det(A) = (1)(-2)(-9) = 18$

Calculate the determinant of matrix A.

$A= \begin{bmatrix} 3& 0& 0&0 \- 6&2 & 0&0 \4 & 1& -1 &0 \ 7&0 & 8 &5 \end{bmatrix}$

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To find the determinant of a lower or upper triangular matrix, simply find the product of the diagonal entries of matrix A:

$\det(A) = (3)(2)(-1)(5) = -30$

Calculate the determinant of $\begin{bmatrix} 5&-2 \ 1&3 \end{bmatrix}$ .

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By definition,

$\begin{vmatrix} a&b \ c&d \end{vmatrix}= a(d)-b(c)$ ,

therefore,

$\begin{vmatrix} 5&-2 \ 1&3 \end{vmatrix}= 5(3)-1(-2)= 17$ .

Calculate the determinant of $\begin{bmatrix} 1&3 &1 \ 0& 2&0 \ 4&8 &2 \end{bmatrix}$

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For simplicity, we will find the determinant by expanding along the second row. Consider the following:

$\begin{vmatrix} 1&3 &1 \ 0& 2&0 \ 4&8 &2 \end{vmatrix}= -0[3(2)-1(8)]+2[1(2)-1(4)]-0[1(8)-3(4)]= -4$

Calculate the determinant of $\begin{bmatrix} 1&2 \ 3&6 \end{bmatrix}$ .

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By definition,

$\begin{vmatrix} a&b \ c&d \end{vmatrix}= ad-bc \Rightarrow \begin{vmatrix} 1&2 \ 3&6 \end{vmatrix} = 1(6)- 2(3) = 0$ .

$A = \begin{bmatrix} 3&1 &0 \ 0&a &12 \ 1& 0 & a \end{bmatrix}$

is a singular matrix for what values of ?

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A matrix is singular - without an inverse - if and only if its determinant is equal to 0.

One way to calculate the determinant of a three-by-three matrix is to add the products of the three diagonals going from upper-left to lower-right, then subtract the products of the three diagonals going from upper-right to lower left. From the diagram below:

we see that the products of the three upper-left to lower-right diagonals are:

$3 \cdot a \cdot a = $3a^{2}$$

$1 \cdot 12 \cdot 1 = 12$

$0 \cdot 0 \cdot 0 = 0$

From the diagram below:

we see that the products of the three upper-right to lower-left diagonals are:

$0 \cdot a \cdot 1 = 0$

$3 \cdot 12 \cdot 0 = 0$

$1 \cdot 0 \cdot a = 0$

Add the first three and subtract the last three:

This must be equal to 0, so set it as such, and solve for :

$a = \pm $\sqrt{-4}$ = \pm 2i$

is a five-by-five matrix with determinant 100.

Give the determinant of .

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The transpose of a square matrix has the same determinant as the original matrix, so

$\det S ^{T}= \det S = 100$ .

Given: a matrix such that $\det K = $\frac{3}{4}$$ .

Give $\det( $K^{T}$ ) ^{-1}$ .

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The determinant of the transpose of a matrix is equal to that of the original matrix; the determinant of the inverse of a matrix is equal to the reciprocal of that of the original matrix. Therefore,

$\det( $K^{T}$ ) ^{-1} = (\det $K^{T}$ ) ^{-1} = (\det K ) ^{-1} =\left ( $\frac{3}{4}$ \right $)^{-1}$= $\frac{4}{3}$$ .

Given: matrix such that $\det C = $\frac{3}{5}$$ .

Evaluate $\det $C^{T}$C$ .

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The determinant of is equal to that of the transpose of ; also, the determinant of the matrix product of two matrices is equal to the product of the determinants. Therefore,

$\det $C^{T}$C$

$= \det $C^{T}$ \cdot \det C$

$= \det C \cdot \det C$

$= $\frac{3}{5}$ \cdot $\frac{3}{5}$$

$= $\frac{9}{25}$$

Consider the matrix

$A = \begin{bmatrix} - 1& 2 & 0 \ 3& y& -2 \ 0& -5& 2 \end{bmatrix}$

where is a real number.

Evaluate so that the minor of this matrix is equal to 12.

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The minor of the matrix is the determinant of the matrix formed when Row 3 and Column 1 of are struck out. This is shown below:

The minor is therefore equal to

$\begin{vmatrix} 2 & 0\ y& -2 \end{vmatrix}$

This is equal to the product of the upper-left and lower-right elements minus the product of the upper-right to lower-right elements, which is

Therefore, regardless of the value of , the minor cannot be equal to 12.

Consider the matrix

$A = \begin{bmatrix} - 1& 2 & 0 \ 3& y& -2 \ 2& -5& 2 \end{bmatrix}$

where is a real number.

Evaluate so that the minor of this matrix is equal to 9.

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The minor of the matrix is the determinant of the matrix formed when Row 1 and Column 3 of are struck out. This is shown below:

The minor is therefore equal to

$\begin{vmatrix} 3& y\ 2& -5 \end{vmatrix}$

This is equal to the product of the upper-left and lower-right elements minus the product of the upper-right to lower-right elements, which is

Set this equal to 9 and solve for :

Consider the matrix

$A = \begin{bmatrix} x& 3&4 \ -x&4 &-1 \ 1&2 & 3 \end{bmatrix}$

where is a real number.

Evaluate so that the minor of this matrix is equal to 77.

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The minor of the matrix is the determinant of the matrix formed when Row 3 and Column 3 of are struck out. This is shown below:

The minor is therefore equal to

$\begin{vmatrix} x& 3\ -x& 4 \end{vmatrix}$

This is equal to the product of the upper-left and lower-right elements minus the product of the upper-right to lower-right elements, which is

Set this equal to 77 and solve for :