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# Using the properties of determinants, evaluate $\begin{vmatrix} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{vmatrix}$

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Toolbox:
• If each element of a row(or column)of a determinant is multiplied by a constant 'k' then |A|=k|A|.
• If each element of a row(or a column)of A is expressed as the sum of two or more terms,then the determinants of A can be expressed as the sum of determinants of two or more matrices.
Let $\Delta=\begin{vmatrix}x+4 & x &x\\x& x+4 & x\\x & x&x+4\end{vmatrix}$

Apply $C_1\rightarrow C_1+C_2+C_3$,we get

$\Delta=\begin{vmatrix}3x+4 & x &x\\3x+4& x+4 & x\\3x +4& x&x+4\end{vmatrix}$

Let us take (3x+4) as the common factor from $C_1$,then

$\Delta=(3x+4)\begin{vmatrix}1 & x &x\\1& x+4 & x\\1 &x &x+4\end{vmatrix}$

Apply $R_2\rightarrow R_2-R_1$,$R_3\rightarrow R_3-R_2$

$\Delta=(3x+4)\begin{vmatrix}1 & x &x\\0& 4 & 0\\0 &-4 &4\end{vmatrix}$

Now expanding along $R_1$ we get,

$\Delta=(3x+4)[1(16-0)-0+0]$

$\quad=16(3x+4)$

answered Mar 14, 2013

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