6. Find the values of x, y and z from the following equations:

(i) $\left[\begin{array}{ll}4\hfill & 3\hfill \\ x\hfill & 5\hfill \end{array}\right]=\left[\begin{array}{ll}y\hfill & z\hfill \\ 1\hfill & 5\hfill \end{array}\right]$

corresponding

By equating the elements of the matrices, we get,

x= 1

y= 4

z= 3.

(ii) $\left[\begin{array}{cc}\hfill x+y\hfill & \hfill 2\hfill \\ \hfill 5+z\hfill & \hfill xy\hfill \end{array}\right]=\left[\begin{array}{ll}6\hfill & 2\hfill \\ 5\hfill & 8\hfill \end{array}\right]$

By equating the corresponding elements of the matrices we get,

x+ y = 6 (I)

5 + Z = 5 $\Rightarrow z=5-5\Rightarrow z=0$

xy = 8

$\Rightarrow$ x $=\frac{8}{y}$ $\to (2)$

putting eq${}^n(2)$ in (1) we get

$\frac{8}{y}$ + y = 6.

$\Rightarrow$ 8 + y² = 6y

$\Rightarrow$ y² 6y + 8 = 0.

$\Rightarrow$ y² - 4y - 2y + 8 = 0

$\Rightarrow$ y (y-4) -2 (y-4) = 0

$\Rightarrow$ (y-4) (y-2) = 0

$\Rightarrow$ y= 4 0r y = 2.

When y = 4,x= 6-y = 6-4 = and z = 0.

Wheny = 2,x = 6-y = 6-2 = 4 and z = 0.

By equating the corresponding elements of the matrices we get,

x+ y + z = 9 -------(i)

x + z = 7 --------(ii)

y + z = 7 -------(iii)

Subtracting eqⁿ (3) from (1) and (2) from (1) we get,

x + y + z -y - z = 9 - 7 and x

...more

(i) $\left[\begin{array}{ll}4\hfill & 3\hfill \\ x\hfill & 5\hfill \end{array}\right]=\left[\begin{array}{ll}y\hfill & z\hfill \\ 1\hfill & 5\hfill \end{array}\right]$

corresponding

By equating the elements of the matrices, we get,

x= 1

y= 4

z= 3.

(ii) $\left[\begin{array}{cc}\hfill x+y\hfill & \hfill 2\hfill \\ \hfill 5+z\hfill & \hfill xy\hfill \end{array}\right]=\left[\begin{array}{ll}6\hfill & 2\hfill \\ 5\hfill & 8\hfill \end{array}\right]$

By equating the corresponding elements of the matrices we get,

x+ y = 6 (I)

5 + Z = 5 $\Rightarrow z=5-5\Rightarrow z=0$

xy = 8

$\Rightarrow$ x $=\frac{8}{y}$ $\to (2)$

putting eq${}^n(2)$ in (1) we get

$\frac{8}{y}$ + y = 6.

$\Rightarrow$ 8 + y² = 6y

$\Rightarrow$ y² 6y + 8 = 0.

$\Rightarrow$ y² - 4y - 2y + 8 = 0

$\Rightarrow$ y (y-4) -2 (y-4) = 0

$\Rightarrow$ (y-4) (y-2) = 0

$\Rightarrow$ y= 4 0r y = 2.

When y = 4,x= 6-y = 6-4 = and z = 0.

Wheny = 2,x = 6-y = 6-2 = 4 and z = 0.

By equating the corresponding elements of the matrices we get,

x+ y + z = 9 -------(i)

x + z = 7 --------(ii)

y + z = 7 -------(iii)

Subtracting eqⁿ (3) from (1) and (2) from (1) we get,

x + y + z -y - z = 9 - 7 and x + y + z - x - z = 9 - 5

$\Rightarrow$ x = 2 and y = 4 .

Putting x = 2 in eqⁿ (2)

2 + z = 5

$\Rightarrow$ z = 5 - 2 = 3.

So, x = 2,y = 4, z = 3.

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<p><strong>(i)</strong>&nbsp;<span title="Click to copy mathml"><math><mrow><mrow><mo>[</mo><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mn>4</mn></mtd><mtd columnalign="left"><mn>3</mn></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mi>x</mi></mtd><mtd columnalign="left"><mn>5</mn></mtd></mtr></mtable></mrow><mo>]</mo></mrow><mo>=</mo><mrow><mo>[</mo><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mi>y</mi></mtd><mtd columnalign="left"><mi>z</mi></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mn>1</mn></mtd><mtd columnalign="left"><mn>5</mn></mtd></mtr></mtable></mrow><mo>]</mo></mrow></mrow></math></span></p><p>corresponding</p><p>By equating the elements of the matrices, we get,</p><p>x= 1</p><p>y= 4</p><p>z= 3.</p><p><strong>(ii)</strong>&nbsp;<span title="Click to copy mathml"><math><mrow><mrow><mo>[</mo><mrow><mtable><mtr columnalign="center"><mtd columnalign="center"><mi>x</mi><mo>+</mo><mi>y</mi></mtd><mtd columnalign="center"><mn>2</mn></mtd></mtr><mtr columnalign="center"><mtd columnalign="center"><mn>5</mn><mo>+</mo><mi>z</mi></mtd><mtd columnalign="center"><mi>x</mi><mi>y</mi></mtd></mtr></mtable></mrow><mo>]</mo></mrow><mo>=</mo><mrow><mo>[</mo><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mn>6</mn></mtd><mtd columnalign="left"><mn>2</mn></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mn>5</mn></mtd><mtd columnalign="left"><mn>8</mn></mtd></mtr></mtable></mrow><mo>]</mo></mrow></mrow></math></span></p><p>By equating the corresponding elements of the matrices we get,</p><p>x+ y = 6 (I)</p><p>5 + Z = 5&nbsp;<span title="Click to copy mathml"><math><mrow><mo>⇒</mo><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>z</mi><mo>=</mo><mn>5</mn><mo>−</mo><mn>5</mn><mo>⇒</mo><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>z</mi><mo>=</mo><mn>0</mn></mrow></math></span></p><p>xy = 8</p><p><span title="Click to copy mathml"><math><mrow><mo>⇒</mo></mrow></math></span>&nbsp;x&nbsp;<span title="Click to copy mathml"><math><mrow><mo>=</mo><mfrac><mrow><mn>8</mn></mrow><mrow><mi>y</mi></mrow></mfrac></mrow></math></span>&nbsp;<span title="Click to copy mathml"><math><mrow><mo>→(2)</mo></mrow></math></span></p><p>putting eq<span title="Click to copy mathml"><math><mrow><msup><mrow></mrow><mrow><mi>n</mi></mrow></msup><mo stretchy="false">(</mo><mn>2</mn><mo stretchy="false">)</mo></mrow></math></span>&nbsp;in (1) we get</p><p><span title="Click to copy mathml"><math><mrow><mfrac><mrow><mn>8</mn></mrow><mrow><mi>y</mi></mrow></mfrac></mrow></math></span>&nbsp;+ y = 6.</p><p><span title="Click to copy mathml"><math><mrow><mo>⇒</mo></mrow></math></span> 8 + <em>y</em>² = 6y</p><p><span title="Click to copy mathml"><math><mrow><mo>⇒</mo></mrow></math></span> <em>y</em>² 6y + 8 = 0.</p><p><span title="Click to copy mathml"><math><mrow><mo>⇒</mo></mrow></math></span> <em>y</em>² - 4y - 2y + 8 = 0</p><p><span title="Click to copy mathml"><math><mrow><mo>⇒</mo></mrow></math></span> y (y-4) -2 (y-4) = 0</p><p><span title="Click to copy mathml"><math><mrow><mo>⇒</mo></mrow></math></span> (y-4) (y-2) = 0</p><p><span title="Click to copy mathml"><math><mrow><mo>⇒</mo></mrow></math></span>&nbsp;y= 4 0r y = 2.</p><p>When y = 4,x= 6-y = 6-4 = and z = 0.</p><p>Wheny = 2,x = 6-y = 6-2 = 4 and z = 0.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1732685340phpVUFUzz_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1732685340phpVUFUzz.jpeg" alt="" width="181" height="96"></picture></div></div><p>By equating the corresponding elements of the matrices we get,</p><p>x+ y + z = 9 -------(i)</p><p>x + z = 7 --------(ii)</p><p>y + z = 7 -------(iii)</p><p>Subtracting eqⁿ (3) from (1) and (2) from (1) we get,</p><p>x + y + z -y - z = 9 - 7 and x + y + z - x - z = 9 - 5</p><p><span title="Click to copy mathml"><math><mrow><mo>⇒</mo></mrow></math></span>&nbsp;x = 2 and y = 4 .</p><p>Putting x = 2 in eqⁿ (2)</p><p>2 + z = 5</p><p><span title="Click to copy mathml"><math><mrow><mo>⇒</mo></mrow></math></span> z = 5 - 2 = 3.</p><p>So, x = 2,y = 4, z = 3.</p>

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